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Dynamics

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**Things to Remember from Last Year**

Newton's Three Laws of Motion Inertial Reference Frames Mass vs. Weight Forces we studied: weight / gravity normal force tension friction (kinetic and static) Drawing Free Body Diagrams Problem Solving

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**Newton's Laws of Motion ΣF = ma**

1. An object maintains its velocity (both speed and direction) unless acted upon by a nonzero net force. 2. Newton’s second law is the relation between acceleration and force. 3. Whenever one object exerts a force on a second object, the second object exerts an equal force in the opposite direction on the first object. ΣF = ma

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**Which of Newton's laws best explains why motorists should buckle-up?**

1 Which of Newton's laws best explains why
motorists should buckle-up? A First Law B Second Law C Third Law D Law of Gravitation

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**speed remained the same, but it is turning to the right.**

2 You are standing in a moving bus, facing forward,
and you suddenly fall forward. You can infer from
this that the bus's A velocity decreased. B velocity increased. C speed remained the same, but it is turning
to the right. D speed remained the same, but it is turning
to the left.

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**normal force due to your contact with the floor of the bus**

3 You are standing in a moving bus, facing forward,
and you suddenly fall forward as the bus comes to
an immediate stop. What force caused you to fall
forward? A gravity B normal force due to your contact with the floor
of the bus C force due to friction between you and the floor
of the bus D There is not a force leading to your fall.

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**slowly slow down, then stop.**

4 When the rocket engines on the spacecraft are
suddenly turned off, while traveling in empty
space, the starship will A stop immediately. B slowly slow down, then stop. C go faster. D move with constant speed

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5 A net force F accelerates a mass m with an
acceleration a. If the same net force is applied to
mass 2m, then the acceleration will be A 4a B 2a C a/2 D a/4

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6 A net force F acts on a mass m and produces an
acceleration a. What acceleration results if a net
force 2F acts on mass 4m? A a/2 B 8a C 4a D 2a

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**The table pushing up on the object with force mg**

7 An object of mass m sits on a flat table. The Earth
pulls on this object with force mg, which we will
call the action force. What is the reaction force? A The table pushing up on the object with force
mg B The object pushing down on the table with
force mg C The table pushing down on the floor with force
mg D The object pulling upward on the Earth with
force mg

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**the force on the truck is greater then the force on the car**

8 A 20-ton truck collides with a 1500-lb car and
causes a lot of damage to the car. Since a lot of
damage is done on the car A the force on the truck is greater then the
force on the car B the force on the truck is equal to the force
on the car C the force on the truck is smaller than the
force on the car D the truck did not slow down during the
collision

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**Inertial Reference Frames**

Newton's laws are only valid in inertial reference
frames: An inertial reference frame is one in which Newton’s
first law is valid. This excludes rotating and
accelerating frames.

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Mass and Weight MASS is the measure of the inertia of an object, the
resistance of an object to accelerate. WEIGHT is the force exerted on that object by
gravity. Close to the surface of the Earth, where the
gravitational force is nearly constant, the weight is: Mass is measured in kilograms, weight Newtons. FG = mg

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**Normal Force and Weight**

FN mg The Normal Force,
FN, is ALWAYS perpendicular to
the surface. Weight, mg, is
ALWAYS directed
downward.

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**Mass is less, weight is same**

9 The acceleration due to gravity is lower on the
Moon than on Earth. Which of the following is
true about the mass and weight of an astronaut on
the Moon's surface, compared to Earth? A Mass is less, weight is same B Mass is same, weight is less C Both mass and weight are less D Both mass and weight are the same

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10 A 14 N brick is sitting on a table. What is the
normal force supplied by the table? A 14 N upwards B 28 N upwards C 14 N downwards D 28 N downwards

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Kinetic Friction Friction forces are
ALWAYS parallel to
the surface exerting
them. Kinetic friction is
always directed
opposite to
direction the object
is sliding and has
magnitude: fK = μkFN fK v

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11 A 4.0kg brick is sliding on a surface. The
coefficient of kinetic friction between the surfaces
is What it the size of the force of friction? A 8.0 B 8.8 C 9.0 D 9.8

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**A brick is sliding to the right on a horizontal surface. **

12 A brick is sliding to the right on a horizontal
surface. What are the directions of the two surface forces:
the friction force and the normal force? A right, down B right, up C left, down D left, up

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Static Friction Static friction is
always equal and
opposite the Net
Applied Force
acting on the object
(not including
friction). Its magnitude is: fS ≤ μSFN fS FAPP

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13 A 4.0 kg brick is sitting on a table. The coefficient
of static friction between the surfaces is
What is the largest force that can be applied
horizontally to the brick before it begins to slide? A 16.33 B 17.64 C 17.98 D 18.12

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14 A 4.0kg brick is sitting on a table. The coefficient
of static friction between the surfaces is If a
10 N horizontal force is applied to the brick, what
will be the force of friction and will the brick
move? A 16.12, no B 17.64, no C 16.12, yes D 17.64, yes

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Tension Force FT mg a When a cord or rope pulls on an
object, it is said to be under
tension, and the force it exerts is
called a tension force, FT.

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15 A crane is lifting a 60 kg load at a constant
velocity. Determine the tension force in the cable. A 568 N B 578 N C 504 N D 600 N

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16 A system of two blocks is accelerated by an
applied force of magnitude F on the frictionless
horizontal surface. The tension in the string
between the blocks is: A 6F 6 kg 4 kg F B 4F C 3/5 F D 1/6 F E 1/4 F

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Two Dimensions

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**Adding Orthogonal Forces**

Since forces are vectors, they add as vectors. The simplest case is if the forces are perpendicular
(orthogonal) with another. Let's add a 50 N force at 0o with a 40 N force at 90o.

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**Adding Orthogonal Forces**

1. Draw the first force
vector, 50 N at 0o,
beginning at the origin. 180o 90o 270o 0o

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**Adding Orthogonal Forces**

1. Draw the first force
vector, 50 N at 0o, beginning at the origin. 2. Draw the second force
vector, 40 N at 90o, with its tail at the tip of the first
vector. 180o 90o 270o 0o

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**Adding Orthogonal Forces**

1. Draw the first force
vector, 50 N at 0o, beginning at the origin. 2. Draw the second force
vector, 40 N at 90o, with its tail at the tip of the first
vector. 3. Draw Fnet from the tail of the first force to the tip of
the last force. 180o 90o 270o 0o

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**Adding Orthogonal Forces**

We get the magnitude of the resultant from the
Pythagorean Theorem. c2 = a2 + b2 Fnet2 = (50N)2 + (40N)2 = 2500N N2 = 4100N2 Fnet = (4100N2)1/2 = 64 N 180o 90o 270o 0o

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**Adding Orthogonal Forces**

We get the direction of the net Force from the inverse tangent. tan(θ) = opp / adj tan(θ) = (40N) / (50N) tan(θ) = 4/5 tan(θ) = 0.8 θ = tan-1(0.80) = 39o Fnet = 64 N at 39o 180o 90o 270o 0o

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**Decomposing Forces into Orthogonal Components**

Let's decompose a 120 N
force at 34o into its x and y
components. F θ 0o

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**Decomposing Forces into Orthogonal Components**

We have Fnet, which is the hypotenuse, so let's first
find the adjacent side by
using cosθ = adj / hyp cosθ = Fx / Fnet Fx = Fnet cosθ = 120N cos34o = 100 N F θ 0o Fx

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**Decomposing Forces into Orthogonal Components**

Now let's find the opposite
side by using sinθ = adj / hyp sinθ = Fy / Fnet Fy = Fnet sinθ = 120N sin34o = 67 N Fy F θ 0o Fx

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**Adding non-Orthogonal Forces**

Now that we know how to add orthogonal forces. And we know now to break forces into orthogonal
components, so we can make any force into two
orthogonal forces. We combine those two steps to add any number of
forces together at any different angles.

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**Adding non-Orthogonal Forces**

Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o First, we will do it graphically, to show the
principle for how we will do it analytically. Then, we will do it analytically, which is easy once
you see why it works that way.

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**Adding non-Orthogonal Forces**

First, here are the
vectors all drawn
from the origin. F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

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**Adding non-Orthogonal Forces**

Now we arrange
them tail to tip. F3 F2 F1 F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

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**Adding non-Orthogonal Forces**

Then we draw in the
Resultant, the sum of the vectors. F3 F2 F F F1 F2 F3 F1 F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

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**Adding non-Orthogonal Forces**

F1x F2x F3x F1y F2y F3y F Now, we graphically
break each vector
into its orthogonal
components. F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o

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**Adding non-Orthogonal Forces**

Then, we remove
the original vectors
and note that the
sum of the
components is the
same as the sum of
the vectors. F3y F3x F F2y F2x F1y F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o F1x

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**Adding non-Orthogonal Forces**

This is the key result: The sum of the x-
components of the
vectors equals the x-
component of the
Resultant. As it is for the y-
components. F3y F F2y Fy F1y F1x F2x F3x F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Fx

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**Adding non-Orthogonal Forces**

This explains how we will add vectors analytically. 1. Write the magnitude and direction (from 0 to 360 degrees)
of each vector. 2. Compute the values of the x and y components. 3. Add the x components to find the x component of the
Resultant. 4. Repeat for the y-components. 5. Use Pythagorean Theorem to find the magnitude of the
Resultant. 6. Use Inverse Tangent to find the Resultant's direction.

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**Adding non-Orthogonal Forces**

First make a table and enter what you know. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp.
(Fcosθ) y-comp.
(Fsinθ) F1 F2 F3 Fnet

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**Adding non-Orthogonal Forces**

Then calculate the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp.
(Fcosθ) y-comp.
(Fsinθ) F1 8.0 N 50o F2 6.5 N 75o F3 8.4 N 30o Fnet

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**Adding non-Orthogonal Forces**

Then add up the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp.
(Fcosθ) y-comp.
(Fsinθ) F1 8.0 N 50o 5.14 N 6.13 N F2 6.5 N 75o 1.68 N 6.28 N F3 8.4 N 30o 7.27 N 4.2 N Fnet

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**Adding non-Orthogonal Forces**

Then use Pythagorean Theorem and tan-1 to determine Fnet. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp.
(Fcosθ) y-comp.
(Fsinθ) F1 8.0 N 50o 5.14 N 6.13 N F2 6.5 N 75o 1.68 N 6.28 N F3 8.4 N 30o 7.27 N 4.20 N Fnet 14.09 N 16.61 N

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**Adding non-Orthogonal Forces**

Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp.
(Fcosθ) y-comp.
(Fsinθ) F1 8.0 N 50o 5.14 N 6.13 N F2 6.5 N 75o 1.68 N 6.28 N F3 8.4 N 30o 7.27 N 4.20 N Fnet 22.9 N 14.09 N 16.61 N

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**Adding non-Orthogonal Forces**

Now use the same procedure to add these forces. DO NOT IGNORE NEGATIVE SIGNS F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o

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**Adding non-Orthogonal Forces**

F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 21 N 60o F2 65 N 150o F3 45 N 330o Fnet

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**Answers Force F1 F2 F3 Fnet Magnitude Direction 21 N 60o 65 N 150o**

F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 21 N 60o 10.50 N 8.66 N F2 65 N 150o N 32.50 N F3 45 N 330o 38.97 N N Fnet 131 N 290o -6.82 N 18.66 N

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**I. The vector sum of the three forces must equal zero.**

17 Three forces act on an object. Which of the
following is true in order to keep the object in
translational equilibrium? I. The vector sum of the three forces must equal zero. II. The magnitude of the three forces must be equal. III. All three forces must be parallel. A I only B II only C I and III only D II and III only E I, II, and III

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**Force and friction acting on an object**

Previously, we solved
problems with multiple
forces, but they were
either parallel or
perpendicular. For instance, draw the
free body diagram of
the case where a box
is being pulled along a
surface, with friction,
at constant speed.

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**Force and friction acting on an object**

FN mg FAPP f Now find the
acceleration given
that the applied
force is 20 N, the
box has a mass of
3.0kg, and the
coefficient of
kinetic friction is
0.20.

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**Force and friction acting on an object**

FN mg FAPP f x - axis y - axis ΣF = ma FAPP - fk = ma FAPP - μkFN = ma FAPP - μkmg = ma a = (FAPP - μkmg)/m a = (20N - (0.20)(30N))/3.0kg a = (20N - 6.0N)/3.0kg a = (14N)/3.0kg a = 4.7 m/s2 ΣF = ma FN - mg = 0 FN = mg FN = (3.0kg)(10m/s2) FN = 30N FAPP = 20N m = 3.0kg μk = 0.20

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**Forces at angles acting on an object**

Now we will solve
problems where the
forces act at an angle so
that it is not parallel or
perpendicular with one
another. First we do a free body
diagram, just as we did
previously.

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**Forces at angles acting on an object**

The next, critical, step is
to choose axes. Previously, we always
used vertical and
horizontal axes, since
one axis lined up with
the forces...and the
acceleration. FN mg FAPP f Now, we must choose axes so that all the acceleration is
along one axis, and there is no acceleration along the other. You always have to ask, "In which direction could this object
accelerate?" Then make one axis along that direction, and the
other perpendicular to that. What's the answer in this case?

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**Forces at angles acting on an object**

This time vertical and
horizontal axes still
work...since we assume
the box will slide along
the surface. However, if this
assumption is wrong,
we will get answers that
do not make sense, and
we will have to
reconsider our choice. FN mg FAPP f y X

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**Forces at angles acting on an object**

y X Now we have to break
any forces that do not
line up with our axes
into components that
do. In this case, FAPP, must be broken into Fx and Fy FN FAPP f mg

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**Forces at angles acting on an object**

FN mg FAPP f y X Fy Fx Once we do that,
we can now
proceed as we did
previously, using
each component
appropriately.

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**Force and friction acting on an object**

FN mg f y X FAPP Fy Fx Let's use our work
to find the
acceleration if the
applied force is 20 N
at 37o above horizontal, the box
has a mass of 3.0kg,
and the coefficient
of kinetic friction is
0.20.

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**Force and friction acting on an object**

FAPP = 20N at 37o m = 3.0kg μk = 0.20 x - axis y - axis ΣF = ma Fx - fk = ma Fx - μkFN = ma Fcosθ - μkmg = ma a = (Fcosθ - μkFN)/m a = (20N cos37o - (0.20)(18N))/3.0kg a = (16N - 3.6N)/3.0kg a = (12.4N)/3.0kg a = 4.1 m/s2 ΣF = ma FN + Fy - mg = 0 FN = mg - Fy FN = mg - Fsinθ FN = (3.0kg)(10m/s2) - (20N)(sin37o) FN = 30N - 12N FN = 18N Note that FN is lower due to the force helping
to support the object FN mg f y X FAPP FAPPy FAPPx

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**Normal Force and Friction**

Friction was reduced
because the Normal
Force was reduced; the
box's weight, mg, was
supported by the y-
component of the force
plus the Normal
Force...so the Normal
Force was
lowered...lowering
friction. mg FAPP FAPPy FN Just looking at the y-axis ΣF = ma FN + Fy - mg = 0 FN = mg - Fy FN = mg - Fsinθ FN Fy mg

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**Normal Force and Friction**

y X FN What would happen
with both the Normal
Force and Friction in
the case that the object
is being pushed along
the floor by a downward
angled force? f FAPP mg

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**Normal Force and Friction**

y X FN In this case the pushing
force is also pushing
the box into the surface,
increasing the Normal
Force as well as friction. f Fx Fy FAPP mg Just looking at the y-axis ΣF = ma FN - Fy - mg = 0 FN = mg + Fy FN = mg + Fsinθ FN mg Fy

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**18 A mg B mg sin(θ) C mg cos(θ) D mg + F sin(θ) E mg – F sin(θ)**

The normal force on the box is: Fapp θ A mg B mg sin(θ) C mg cos(θ) D mg + F sin(θ) E mg – F sin(θ)

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**19 A B μ(mg - Fsin(θ)) C μ(mg + Fcos(θ)) D E μmg**

The frictional force on the box is: A μ(mg + Fsin(θ)) Fapp θ B μ(mg - Fsin(θ)) C μ(mg + Fcos(θ)) D μ(mg - Fcos(θ)) E μmg

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**20 v m A mg - Fapp cosθ B C mg D mg + Fapp sinθ E mg + Fapp cosθ**

A block of mass m is pulled along a horizontal
surface at constant speed v by a force Fapp , which acts at an angle of θ with the horizontal. The
coefficient of kinetic friction between the block
and the surface is μ. The normal force exerted on the block by the
surface is: Fapp θ v m A mg - Fapp cosθ B mg - Fapp sinθ C mg D mg + Fapp sinθ E mg + Fapp cosθ

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**The friction force on the block is:**

21 A block of mass m is pulled along a horizontal
surface at constant speed v by a force Fapp , which acts at an angle of θ with the horizontal. The
coefficient of kinetic friction between the block and
the surface is μ. The friction force on the block is: Fapp θ v m A μ(mg - Fapp cosθ) B μ(mg - Fapp sinθ) C μmg D μ(mg + Fapp sinθ) E μ(mg + Fapp cosθ)

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**Normal Force and Weight**

FN mg Previously we dealt
mostly with
horizontal (or,
rarely, vertical
surfaces). In that
case FN, and mg were always along
the same axis. Now we will look at
the more general
case.

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Inclined Plane On the picture, draw
the free body
diagram for the
block. Show the weight
and the normal
force.

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**Inclined Plane FN is ALWAYS perpendicular to the surface.**

mg is ALWAYS
directed downward. But now, they are
neither parallel or
perpendicular to one
another. FN mg

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Choosing Axes Previously, we used
vertical and horizontal
axes. That worked
because problems
always resulted in an
acceleration that was
along one of those
axes. We must choose axes
along which all the
acceleration is along
one axis...and none
along the other. FN mg

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Choose of Axes In this case, the block
can only accelerate
along the surface of the
plane. Even if there is no
acceleration in a
problem, this is the
ONLY POSSIBLE
direction of acceleration. So we rotate our x-y
axes to line up with the
surface of the plane. y X FN mg a

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**Inclined Plane Problems**

By the way, here's one way to see that θ is the both the angle of incline and the angle between mg and the new y-axis. Let's name the angle of the
inclined plane θ and the angle between mg and the x-axis θ' and show that θ = θ' Since the angles in a triangle add
to 180o, and the bottom left angle is 90o, that means: α + θ = 90o y-axis FN mg α θ' θ

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**Inclined Plane Problems**

Now look at the angles in the upper left corner of the triangle. The surface of the inclined plane
forms an angle of 180o, so 90o + θ' + α = 180o θ' + α = 90o But we already showed that α + θ = 90o θ' + α = 90o = α + θ θ' = θ y-axis FN mg α θ' θ

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**FN Inclined Plane Problems a mg**

We need to resolve all forces which don't align with an
axis into components that will. In this case, we resolve mg into its x and y components. y X FN mg mg sin θ mg cos θ θ a y X FN mg a θ θ

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**Now we just use Newton's Second Law, which is true for each axis. **

Inclined Plane Problems Now we just use Newton's Second Law, which is true for each
axis. x - axis y - axis FN mg mg sin θ mg cos θ θ a y X ΣFx = max mgsinθ = max a = gsinθ down the plane ΣFy = may FN - mgcosθ = 0 FN = mgcosθ

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**θ ΣFx = max mg sin θ = ma g sin θ = a a = g sin θ**

Answer y x FN mg mg sin θ mg cos θ θ ΣFx = max mg sin θ = ma g sin θ = a a = g sin θ a = 10 m/s2 sin (30o) a = 10 m/s2 (0.5) a = 5 m/s2 θ A 5 kg block slides down a frictionless
incline at an angle of 30 degrees. a) Draw a free body diagram. b) Find its acceleration. (Use g = 10 m/s2)

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**Inclined Plane Problems with Friction**

We can add kinetic friction to our inclined plane example. The
kinetic friction points opposite the direction of motion. We now have a second vector along the x axis. fk points in the negative direction (recall fk = μk FN) y x FN mg mg sinθ mg cosθ θ fk a FN mg θ fk a

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**Inclined Plane Problems with Friction**

x - axis y - axis ΣFy = may FN = mg cosθ ΣFx = max mgsinθ - fk = ma mgsinθ - μkFN = ma mgsinθ - μkmgcosθ = ma gsinθ - μkgcosθ = a a = gsinθ - μkgcosθ a = g(sinθ - μk cosθ) y x FN mg mg sinθ mg cosθ θ fk a

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**Inclined Plane Problems with Friction**

The general solution for objects sliding down an incline is: a = g(sinθ - μkcosθ) Note, that if there is no friction: μk = 0 and we get our previous result for a frictionless plane: a = gsinθ

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**If the object is sliding with constant velocity: a = 0**

Inclined Plane Problems with Friction a = g(sinθ - μkcosθ) If the object is sliding with constant velocity: a = 0 y x FN a = g(sinθ - μkcosθ) 0 = g(sinθ - μkcosθ) 0 = sinθ - μkcosθ μk cosθ = sinθ μk = sinθ / cosθ μk = tanθ fk a = 0 mg cosθ mg θ mg sinθ

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**Inclined Plane with Static Friction**

We just showed that for an object
sliding with constant velocity down
an inclined plane that: μk = tanθ Similarly, substituting μs for μk (at the maximum static force; the
largest angle of incline before the
object begins to slide) than: μs = tanθmax But this requires that the MAXIMUM
ANGLE of incline, θmax, be used to determine μs. y x FN fs a = 0 mg cosθ mg θ mg sinθ

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Answer y x FN mg mg sin θ mg cos θ θ fk θ ΣF = ma mg sin θ - fK = 0 mg sin θ = fK mg sin θ = μ FN mg sin θ = μ mg cos θ μ = mg sin θ / mg cos θ μ = tan θ μ = tan 30 = 0.58 A 5 kg block slides down a incline at an
angle of 30o with a constant speed. a) Draw a free body diagram. b) Find the coefficient of friction between the block and the incline. (Use g = 10 m/s2)

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Answer FN mg mg sin θ mg cos θ θ Fapp fK a = 0 y-direction ΣF = ma FN - mg cos θ = 0 FN = mg cos θ θ x-direction ΣF = ma Fapp - mg sin θ - fk = ma Fapp - mg sin θ -μk FN = 0 Fapp = mg sin θ + μk mg cos θ Fapp = 38 N A 5 kg block is pulled up an incline at
an angle of 30o at a constant velocity
The coefficient of friction between the
block and the incline is 0.3. a) Draw a free body diagram. b) Find the applied force. (Use g = 10 m/s2)

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**FN Fapp fK mg x-axis ΣF = ma Fapp - mg sin θ - fk = ma**

mg cos θ θ Fapp fK a y-axis ΣF = ma FN - mg cos θ = 0 FN = mg cos θ x-axis ΣF = ma Fapp - mg sin θ - fk = ma Fapp - mg sin θ - μk FN = ma Fapp - mg sin θ - μk mg cos θ = ma a = (Fapp/m) - g sin θ - μk g cos θ a = 0.4 m/s2 A 5 kg block is pulled UP an incline at an angle of 30
degrees with a force of 40 N.
The coefficient of friction
between the block and the
incline is 0.3. a) Draw a free body diagram. b) Find the block's acceleration. (Use g = 10 m/s2)

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**Inclined Plane Problems**

If a mass, m, slides down a frictonless inclined plane, we have this general setup: y x FN mg θ a It is helpful to rotate
our reference frame so
that the +x axis is
parallel to the inclined
plane and the +y axis
points in the direction
of FN.

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**FN fs mg a = 0 x-direction ΣF = ma fs = mg sin θ**

mg cos θ θ fs a = 0 y-direction ΣF = ma FN - mg cos θ = 0 FN = mg cos θ x-direction ΣF = ma fs = mg sin θ μs mg cos θ = mg sin θ μs cos θ = sin θ μs = sin θ / cos θ μs = tan θ θ = tan-1 μs θ = 21.8o A 5 kg block remains
stationary on an incline. The
coefficients of static and
kinetic friction are 0.4 and
0.3, respectively. a) Draw a free body diagram. b) Determine the angle that
the block will start to move. (Use g = 10 m/s2)

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22 A block of mass m slides down a rough
incline as shown. Which free body
diagram correctly shows the forces acting
on the block? A B C D E f W N f W N f W N f W N f W N

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23 A block with a mass of 15 kg slides down
a 43° incline as shown above with an
acceleration of 3 m/s2. What is the normal force N applied by the inclined plane on the block? A 95 N B 100 N C 105 N D 110 N E 115 N

93
24 A block with a mass of 15 kg slides down a
43° incline as shown above with an
acceleration of 3 m/s2. The magnitude of the frictional force along the plane is nearly: A 56 N B 57 N C 58 N D 59 N E 60 N

94
Static Equilibrium There is a whole field of problems called
"Statics" that has to do with cases where no
acceleration occurs, objects remain at rest. Anytime we construct something (bridges,
buildings, houses, etc.) we want them to
remain stationary, not accelerate. So this is a
very important field. The two types of static equilibrium are with
respect to linear and rotational acceleration, a
balancing of force and of torque (forces that
cause objects to rotate). We'll look at them in
that order.

95
Tension Force Previously, we did problems where a rope
supporting an object exerted a vertical force
straight upward, along the same axis as the
force mg was pulling it down. That led to
the simple case that if a = 0, then FT = mg FT mg

96
25 A uniform rope of weight 20 N hangs from a hook
as shown above. A box of mass 60 kg is
suspended from the rope. What is the tension in
the rope? A 20 N throughout the rope B 120 N throughout the rope C 200 N throughout the rope D 600 N throughout the rope E It varies from 600 N at the
bottom of the rope to 620 N
at the top.

97
Tension Force But it is possible for two (or more) ropes to
support an object (a = 0) by exerting forces
at angles. In that case: The vertical components of the force
exerted by each rope must add up to mg. And the horizontal components must add to
zero. T2 T1 mg

98
Tension Force y X So we need to break the forces into
components that align with our axes. T2 T1 mg

99
**Tension Force y One result can be seen just from this diagram.**

mg T1x T2x y X T2y T1y One result can be seen just from this
diagram. In order for the forces along the x-axis to
cancel out, the more vertical Tension must
be much larger than the Tension which is at
a greater angle to the vertical. In problems with two supporting ropes or
wires, the more vertical has a greater
tension.

100
Tension Force 50o 20o mg T1x T2x y X T2y T1y Let's calculate the tension in two
ropes if T1 is at an angle of 50o from
the vertical and T2 is at an angle of
20o from the vertical and they are
supporting a 8.0 kg mass. Note, that in this case the horizontal
component is given by Tsinθ, where before it was given by Tcosθ. To know which function to use you
MUST draw the diagram and find the
sides opposite and adjacent to the
given angle.

101
**Tension Force y y - axis x - axis T2y T1y T1x T2x X mg ΣFx = max = 0**

T1sinθ1 = T2sinθ2 T1 = T2sinθ2/sinθ1 T1 = T2 (sin20o/sin50o) T1 = T2 (0.34/0.77) T1 = 0.44 T2 T1 = 0.44 (64N) T1 = 28N ΣFy = may = 0 T1y + T2y - mg = 0 T1cosθ1 + T2cosθ2 = mg (0.44T2)cos(50o) + T2cos(20o) = mg (0.44T2)(0.64) + T2(0.94) =(8kg)(9.8m/s2) 1.22 T2 = 78N T2 = 78N / 1.22 T2 = 64N

102
Tension Force θ θ y X If the ropes form equal angles to the
vertical, the tension in each must
also be equal, otherwise the x-
components of Tension would not
add to zero. Ty Ty θ θ Tx Tx mg

103
**Tension Force x - axis y - axis y Ty Ty Tx Tx X mg ΣFy = may = 0**

θ θ y X ΣFy = may = 0 T1y + T2y - mg = 0 Tcosθ + Tcosθ = mg 2Tcosθ = mg T = mg / (2cosθ) Note that the tension
rises as cosθ becomes smaller...which occurs as
θ approaches 90o. It goes to infinity at 90o, which shows that the
ropes can never be
perfectly horizontal. ΣFx = max = 0 T1x - T2x = 0 Tsinθ = Tsinθ which just confirms
that if the angles are
equal, the tensions
are equal Ty Ty θ θ Tx Tx mg

104
26 A lamp of mass m is suspended from two ropes of
unequal length as shown above. Which of the
following is true about the tensions T1 and T2 in the cables? T1 T2 A T1 > T2 B T1 = T2 C T1 < T2 D T1 - T2 = mg E T1 + T2 = mg

105
27 A large mass m is suspended from two massless
strings of an equal length as shown below. The
tension force in each string is: θ m A ½ mg cos(θ) B 2 mg cos(θ) C mg cos(θ) D mg/cos(θ) E mg/2cos(θ)

106
**Torque and Rotational Equilibrium**

Forces causes objects to linearly accelerate. Torque causes objects to rotationally accelerate. Rotational dynamics is a major topic of AP Physics C:
Mechanics. It isn't particularly difficult, but for AP B, we
only need to understand the static case, where all the
torques cancel and add to zero. First we need to know what torque is.

107
**Torque and Rotational Equilibrium**

Until now we've treated objects as points, we haven't been
concerned with their shape or extension in space. We have assumed that any applied force acts through the
center of the object and it is free to accelerate. That does
not result in rotation, just linear acceleration. But if the force acts on an object so that it causes the
object to rotate around its center of mass...or around a
pivot point, that force has exerted a torque on the object.

108
**Torque and Rotational Equilibrium**

A good example is opening a door,
making a door rotate. The door
does not accelerate in a straight
line, it rotates around its hinges. Think of the best direction and
location to push on a heavy door to
get it to rotate and you'll have a
good sense of how torque works. Which force (blue arrow) placed at
which location would create the
most rotational acceleration of the
green door about the black hinge?

109
**Torque and Rotational Equilibrium**

The maximum torque is obtained from: The largest force At the greatest distance from the pivot At an angle to the line to the pivot that is closest to perpendicular Mathematically, this becomes: τ = Frsinθ τ (tau) is the symbol for torque; F is the applied force r is the distance from the pivot θ is the angle of the force to a line to the pivot

110
**Torque and Rotational Equilibrium**

τ = Frsinθ When r decreases, so does
the torque for a given
force. When r = 0, τ = 0. When θ differs from 90o, the torque decreases.
When θ = 0o or 180o, τ = 0. F θ r

111
**Two Physical Interpretations of Torque These are all equivalent.**

τ = F(rsinθ) The torque due to F acting at
angle θ can be replaced by F
acting at a smaller distance
rsinθ. τ = r(Fsinθ) The torque due to F acting
at angle θ can be replaced
by a smaller perpendicular
force Fsinθ acting at r. Fsinθ rsinθ F These are all
equivalent. F θ θ r F Fsinθ F r θ r rsinθ

112
**Rotational Equilibrium**

When the sum of the torques on an object is zero, the
object is in rotational equilibrium. Define counter clockwise (CCW) as the positive direction
for rotation and clockwise (CW) as the negative. For instance, what perpendicular force, F, must be
applied at a distance of 7.0 m for the pivot to exactly
offset a 20N force acting at a distance of 4.0m from the
pivot of a door at an angle of 30o?

113
**Rotational Equilibrium**

Στ = 0 F1r1sinθ1 + F2r2sinθ2 =0 F1r1sinθ1 = - F2r2sinθ2 F1 = - F2r2sinθ2 / (r1sinθ1) F1 = - (-20N)(4.0m)sin(30o) / ((7m)sin90o) F1 = (80N-m)(0.50) / (7m) F1 = 5.7 N 30o 4.0m F1

114
**Rotational Equilibrium**

4kg 2kg What mass must be added at distance 4.0m to put the
above apparatus into equilibrium?

115
**Rotational Equilibrium**

4kg 2kg Στ = 0 F1r1sinθ1 + F2r2sinθ2 - F3r3sinθ2 = 0 m1gr1 + m2gr2 - m3gr3 = 0 m1r1 + m2r2 - m3r3 = 0 m3 = (m1r1 + m2r2) / r3 m3 = ((4kg)(3m) +(2kg)(1m)) / 4m F1 = (14kg-m)) / (4m) F1 = 3.5 kg

116
12 kg 15 kg A 12 kg load hangs from one end of a rope that passes over a small frictionless
pulley. A 15 kg counterweight is suspended from the other end of the rope. The
system is released from rest. a. Draw a free-body diagram for each object showing all applied forces in
relative scale. Next to each diagram show the direction of the acceleration
of that object. b. Find the acceleration each mass. c. What is the tension force in the rope? d. What distance does the 12 kg load move in the first 3 s? e. What is the velocity of 15 kg mass at the end of 5 s?

117
**m1 = 12 kg m2 = 15 kg c. FT = m1g + m1a or FT = m2g - m2a**

= 12kg(10 m/s2) + 12kg(1.11 m/s2) = 120 N N = N FT = m2g - m2a = 15kg(10 m/s2) - 15kg(1.11 m/s2) = 150 N N = N b. ΣF = ma ΣF = FT - m1g = m1a FT = m1g + m1a ΣF = FT - m2g = -m2a FT = m2g - m2a FT = FT m1g + m1a = m2g - m2a m1a + m2a = m2g - m1g a(m1 + m2) = g(m2 - m1) a = (g(m2 - m1)) / (m1 + m2) = (10 m/s2 (15kg - 12kg)) / (12kg + 15kg) = 30 m/s2 * kg / 27kg a = 1.11 m/s2

118
**m1 = 12 kg m2 = 15 kg d. x = xo + vot + 1/2 at2 xo = 0 vo = 0**

x = 1/2 at2 = 1/2 (1.11 m/s2)(3s)2 = 5 m e. v = vo + at v = at = (1.11 m/s2)(5s) = 5.55 m/s

119
500 g 300 g A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction
between the block and the surface is The block is connected by a massless
string to the second block with a mass of 300 g. The string passes over a light
frictionless pulley as shown above. The system is released from rest. a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g
masses. Include all forces and draw them to relative scale. Draw the
expected direction of acceleration next to each free-body diagram. b. Use Newton’s Second Law to write an equation for the 500 g mass. c. Use Newton’s Second Law to write an equation for the 300 g mass. d. Find the acceleration of the system by simultaneously solving the system
of two equations. e. What is the tension force in the string?

120
**= ((0.3kg)(10 m/s2) - (0.25)(0.5kg)(10m/s2)) / (0.5kg +0.3kg) **

m1 = 500 g = 0.5 kg m2 = 300 g = 0.3 kg μ = 0.25 d. FT - f = m1a FT - m2g = - m2a FT = f + m1a FT = m2g - m2a f + m1a = m2g - m2a m1a + m2a = m2g - f a(m1 + m2) = m2g - f a = (m2g - f) / (m1 + m2) = (m2g - μm1g) / (m1 + m2) = ((0.3kg)(10 m/s2) - (0.25)(0.5kg)(10m/s2)) / (0.5kg +0.3kg) = ( ) kg*m/s2 / (0.8 kg) = m/s2 e. FT = f + m1a or FT = m2g - m2a FT = μm1g + m1a FT = (0.25)(0.5kg)(10m/s2) + 0.5kg( m/s2) = 2.34 N FT = 0.3kg(10m/s2) - 0.3kg( m/s2) = 2.34 N a. FT FN m2g m1g f a b. ΣF = ma x-direction ΣF = FT - f = m1a f = μFn f = μm1g y-direction ΣF = ma = 0 ΣF = FN - m1g = 0 FN = m1g c. ΣF = ma ΣF = FT - m2g = - m2a

121
F A 10-kilogram block is pushed along a rough horizontal surface by a
constant horizontal force F as shown above. At time t = 0, the velocity v of
the block is 6.0 meters per second in the same direction as the force. The
coefficient of sliding friction is 0.2. Assume g = 10 meters per second
squared. a. Calculate the force F necessary to keep the velocity constant.

122
**a. Calculate the force F necessary to keep the velocity constant.**

m = 10 kg v = 6 m/s μ = 0.2 g = 10 m/s2 a. Calculate the force F necessary to keep the velocity constant. x y ΣF = ma ΣF = FN - mg = 0 Fapp - fk = 0 FN = mg Fapp = fk Fapp = μFN Fapp = μmg Fapp = (0.2)(10kg)(10m/s2) Fapp = 20 N FN fk Fapp mg

123
**A helicopter holding a 70-kilogram package suspended from a rope 5**

A helicopter holding a 70-kilogram package suspended from a rope 5.0 meters
long accelerates upward at a rate of 5 m/s2. Neglect air resistance on the
package. a. On the diagram, draw and label all of the forces acting on the package. b. Determine the tension in the rope. c. When the upward velocity of the helicopter is 30 meters per second, the
rope is cut and the helicopter continues to accelerate upward at 5 m/s2.
Determine the distance between the helicopter and the package 2.0
seconds after the rope is cut.

124
**m = 70 kg a = 5 m/s2 x = 5 m ΣF = ma FT - mg = ma FT = mg + ma**

a. On the diagram, draw and label all of the forces acting on the package. b. Determine the tension in the rope. mg FT a ΣF = ma FT - mg = ma FT = mg + ma FT = m (g+a) FT = (70kg) (10 m/s2 + 5 m/s2) FT = (70kg) (15 m/s2) FT = 1050 N

125
ahelicopter apackage m = 70 kg a = 5 m/s2 x = 5 m v = 30 m/s c. When the upward velocity of the helicopter is 30
meters per second, the rope is cut and the helicopter
continues to accelerate upward at 5.2 m/s2. Determine the distance between the helicopter and the package 2
seconds after the rope is cut. Method 1 Use the reference frame of the package after it is released. The relative acceleration of the helicopter is given by: arelative = ahelicopter - apackage arelative = 5 m/s2 - (-10 m/s2) arelative = 15 m/s2 Their relative initial velocity is zero, since they are connected together The helicopter is 5.0 m above the package when released, the length of the rope. y = y0 + v0 + 1/2at2 y = 5m /2(15 m/s2)(2s)2 y= 35m

126
ahelicopter apackage m = 70 kg a = 5 m/s2 x = 5 m v = 30 m/s c. When the upward velocity of the helicopter is 30
meters per second, the rope is cut and the helicopter
continues to accelerate upward at 5.2 m/s2. Determine the distance between the helicopter and the package 2
seconds after the rope is cut. Method 2 Use the reference frame of the earth, but use the initial height of the package as zero. Package Helicopter y = y0 + v0t+ 1/2at2 y = y0 + v0t+ 1/2at2 y = 0 + (30 m/s)(2s) + 1/2(-10 m/s2)(2s)2 y = 5m + (30 m/s)(2s) + 1/2(+5 m/s2)(2s)2 y= 60 m - 20 m y= 5m + 60 m + 10 m y= 40 m y= 75 m Their separation is just the difference in their positions Δy = yhelicopter - ypackage Δy = 75m - 40m Δy = 35m

127
**a. The acceleration of the 4-kilogram block **

1 kg 2 kg 4 kg Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings,
one of which passes over a frictionless pulley of negligible mass, as shown above.
Calculate each of the following. a. The acceleration of the 4-kilogram block b. The tension in the string supporting the 4-kilogram block c. The tension in the string connected to the l-kilogram block

128
**a. The acceleration of the 4-kilogram block ΣF = ma m1: m2: m3: **

m1 = 1 kg m2 = 2 kg m3 = 4 kg g = 9.8 m/s2 a. The acceleration of the 4-kilogram block ΣF = ma m1: m2: m3: FT1 = m1a FT2 - FT1 = m2a FT2 - m3g = -m3a FT2 - m1a = m2a FT2 - m3g = -m3a FT2 = m1a + m2a FT2 = m3g - m3a m1a + m2a = m3g - m3a m1a + m2a + m3a = m3g a(m1 + m2 + m3) = m3g a = m3g / (m1 + m2 + m3) a = (4kg)(9.8 m/s2) / (1kg + 2kg + 4kg) a = 5.6 m/s2 4 kg 1 kg 2 kg FN m1g m2g m3g FT1 FT2 a

129
m1 = 1 kg m2 = 2 kg m3 = 4 kg g = 9.8 m/s2 b. The tension in the string supporting the 4-kilogram block FT2 = m1a + m2a FT2 = 1kg(5.6 m/s2) + 2kg(5.6 m/s2) FT2 = 16.8 N or FT2 = m3g - m3a FT2 = 4kg(9.8 m/s2) - 4kg(5.6 m/s2) c. The tension in the string connected to the l-kilogram block FT1 = m1a FT1 = 1kg(5.6 m/s2) FT1 = 5.6 N

130
**a. Calculate the acceleration of the 10-kilogram block in case I. **

10 kg 5 kg Case II A 10-kilogram block rests initially on a table as shown in cases I and II above.
The coefficient of sliding friction between the block and the table is The
block is connected to a cord of negligible mass, which hangs over a massless
frictionless pulley. In case I a force of 50 newtons is applied to the cord. In case
II an object of mass 5 kilograms is hung on the bottom of the cord. Use g = 10
meters per second squared. a. Calculate the acceleration of the 10-kilogram block in case I. b. On the diagrams below, draw and label all the forces acting on each block in
case II c. Calculate the acceleration of the 10-kilogram block in case II. d. Calculate the tension force in case II.

131
**a. Calculate the acceleration of the 10-kilogram block in case I. **

m1 = 10 kg m2 = 5 kg μ = 0.2 FApp = 50 N g = 10 m/s2 a. Calculate the acceleration of the 10-kilogram block in case I. ΣF = ma FApp - fK = ma FApp - μmg = ma a = (FApp - μmg)/m a = (50N - 0.2(10kg)(10m/s2))/10kg a = 3 m/s2 y ΣF = ma ΣF = FN - mg = 0 FN = mg fk = μFN fk = μmg b. On the diagrams below, draw and label all the forces acting on each block in
case II FN m1g f FT m2g 5 kg 10 kg

132
m1 = 10 kg m2 = 5 kg μ = 0.2 FApp = 50 N g = 10 m/s2 c. Calculate the acceleration of the 10-kilogram block in case II. ΣF = ma FT - f = m1a FT - m2g = -m2a FT - μm1g = m1a FT = m2g - m2a FT = μm1g + m1a μm1g + m1a = m2g - m2a m1a + m2a = m2g - μm1g a(m1 + m2) = g(m2 - μm1) a = g(m2 - μm1)/(m1 + m2) = 10 m/s2(5kg - 0.2*10kg)/(5kg +10kg) a = 2 m/s2 d. Calculate the tension force in case II. FT - m2g = -m2a FT = m2g - m2a FT = m2(g - a) = 5kg (10 m/s2 - 2 m/s2) = 40 N

133
**c. The magnitude of the acceleration of M1 **

In the system shown above, the block of mass M1 is on a rough horizontal table. The
string that attaches it to the block of mass M2 passes over a frictionless pulley of
negligible mass. The coefficient of kinetic friction mk between M1 and the table is less
than the coefficient of static friction ms a. On the diagram below, draw and identify all the forces acting on the block of mass
M1. b. In terms of M1 and M2 determine the minimum value of ms that will prevent the
blocks from moving. The blocks are set in motion by giving M2 a momentary downward push. In terms of
M1, M2, mk, and g, determine each of the following: c. The magnitude of the acceleration of M1 d. The tension in the string.

134
M1, M2, mk, ms, g a. On the diagram below, draw and identify all the forces acting on the block of
mass M1. b. In terms of M1 and M2 determine the minimum value of ms that will prevent the blocks from moving. The pulley system is not moving, so acceleration is zero. ΣF = ma x-direction for M1: FT - f = 0 FT = f y-direction for M1: FN - M1g = 0 FN = M1g y-direction for M2: FT - M2g = 0 FT = M2g f = μsFN or msFN in this case f = msM1g = M2g ms = M2 / M1 M1g FN FT f

135
**The blocks are set in motion by giving M2 a momentary downward push**

The blocks are set in motion by giving M2 a momentary downward push. In terms of M1, M2, mk, and g, determine each of the following: c. The magnitude of the acceleration of M1 ΣF = ma x-direction for M1: FT - f = M1a FT = f + M1a y-direction for M1: FN - M1g = 0 FN = M1g f = μkFN or mkFN in this case f = mkM1g y-direction for M2: FT - M2g = -M2a FT = M2g - M2a FT = f + M1a, FT = M2g - M2a mkM1g + M1a = M2g - M2a M1a + M2a = M2g - mkM1g a(M1 + M2) = M2g - mkM1g a = M2g - mkM1g / (M1 + M2) d. The tension in the string. FT = f + M1a or FT = M2g - M2a FT = mkM1g + M1(M2g - mkM1g / (M1 + M2) ) or FT = M2g - M2(M2g - mkM1g / (M1 + M2) )

136
**a. What is the tension FT in the string. **

m1 m2 2 m 60o Two 10-kilogram boxes are connected by a massless string that passes over a
massless frictionless pulley as shown above. The boxes remain at rest, with the
one on the right hanging vertically and the one on the left 2.0 meters from the
bottom of an inclined plane that makes an angle of 60° with the horizontal. The
coefficients of kinetic friction and static friction between the Ieft-hand box and the
plane are 0.15 and 0.30, respectively. You may use g = 10 m/s2, sin 60° = 0.87, and cos 60° = 0.50. a. What is the tension FT in the string. b. On the diagram above, draw and label all the forces acting on the box that is
on the plane. c. Determine the magnitude of the frictional force acting on the box on the
plane. d. The string is then cut and the left-hand box slides down the inclined plane.
What is the magnitude of its acceleration?

137
m1 = 10 kg m2 = 10 kg x = 2 m θ = 60o , sin 60° = 0.87, cos 60° = 0.50 μk = 0.15, μs = 0.30 g = 10 m/s2 a. What is the tension FT in the string. The acceleration is 0 because the system is not moving. ΣF = ma x-direction for m1: FT - f - m1gsinθ = 0 FT = f + m1gsinθ y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μsFN f = μsm1gcosθ FT = f + m1gsinθ FT = μsm1gcosθ + m1gsinθ FT = (0.30)(10kg)(10 m/s2)(0.50) + (10kg)(10 m/s2)(0.87) FT = 102 N

138
**y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μsFN **

m1 = 10 kg m2 = 10 kg x = 2 m θ = 60o , sin 60° = 0.87, cos 60° = 0.50 μk = 0.15, μs = 0.30 g = 10 m/s2 b. On the diagram above, draw and label all the forces acting on the box that is on the
plane. c. Determine the magnitude of the frictional force acting on the box on the plane. y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μsFN f = μsm1gcosθ f = 0.30(10kg)(10 m/s2)(0.50) f = 15 N m1 FN f m1g FT 60o

139
**x-direction of m1: f - m1gsinθ = -m1a **

m1 = 10 kg m2 = 10 kg x = 2 m θ = 60o , sin 60° = 0.87, cos 60° = 0.50 μk = 0.15, μs = 0.30 g = 10 m/s2 d. The string is then cut and the left-hand box slides down the inclined plane. What is
the magnitude of its acceleration? ΣF = ma x-direction of m1: f - m1gsinθ = -m1a y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μkFN = μkm1gcosθ m1a = m1gsinθ - f a = (m1gsinθ - μkm1gcosθ) / m1 a = ((10kg)(10 m/s2)(0.87) - (0.15)(10kg)(10 m/s2)(0.50)) / 10kg a = 7.95 m/s2 m1 m2 60o a f FN m1g

140
m1 m2 M v θ Masses ml and m2, are connected by a light string. They are further connected
to a block of mass M by another light string that passes over a pulley. Blocks l
and 2 move with a constant velocity v down the inclined plane, which makes an
angle θ with the horizontal. The kinetic frictional force on block 1 is f and that on
block 2 is 2f. a. On the figure above, draw and label all the forces on block m1, m2, and M. Express your answers to each of the following in terms of ml, m2, g, θ, and f. b. Determine the coefficient of kinetic friction between the inclined plane and
block 1. c. Determine the value of the suspended mass M that allows blocks 1 and 2 to
move with constant velocity down the plane. d. The string between blocks 1 and 2 is now cut. Determine the acceleration
of block 1 while it is on the inclined plane.

141
m1 m2 M v θ a. On the figure above, draw and label all the forces on block m1, m2, and M. m1 v m1g FT1 f FN a = 0 m2 m2g FT2 M Mg θ

142
**Express your answers to each of the following in terms of ml, m2, g, θ, and f.**

b. Determine the coefficient of kinetic friction between the inclined plane and block 1. f = μFN ΣF = ma FN - m1gcosθ = 0 FN = m1gcosθ f = μm1gcosθ μ = f /(m1gcosθ) c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane. m1: ΣF = ma M: FT2 - Mg = 0 FT1 - m1gsinθ - f = 0 FT2 = Mg FT1 = m1gsinθ m2: ΣF = ma FT2 - FT1 - m2gsinθ - 2f = 0 Mg - m1gsinθ - f - m2gsinθ - 2f = 0 Mg = m1gsinθ + m2gsinθ + 3f M = (m1gsinθ + m2gsinθ + 3f) / g

143
**d. The string between blocks 1 and 2 is now cut**

d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1
while it is on the inclined plane. x-direction: ΣF = ma f - m1gsinθ = -m1a m1gsinθ - f = m1a a = (m1gsinθ - f) / m1 m1 m1g f FN a θ

144
Calculus Based

145
Fluid Resistance When first learning dynamics discussed cases in which an
object moves with any friction. Then we began to take the
frictional force into account between an object and its
surface. Now we are going to begin taking into account air
resistance. v Fresistance

146
**Fluid Resistance No resistance v = vo + gt vo = 0 a is constant**

y No resistance v = vo + gt vo = 0 a is constant v increasing mg = ma a = g

147
Fluid Resistance If we do not take into account air resistance then the
object will fall with a constant acceleration. If air resistance however is taken into account then
the objects acceleration will eventually be cancelled
out by the drag force. Since the acceleration will essentially become zero, at
some point in time your velocity will be constant.

148
**Fluid Resistance With resistance**

FR a=g vo=0 In this example the equation for
the resistive force is given as F = kv, but in other problems the
resisitive force equation can be
different. When the falling object stops
accelerating it reaches its terminal
velocity.

149
**If the equation for the force due to air resistance **

28 If the equation for the force due to air resistance is kv2. What is the terminal velocity reached by the object? A B C D E

150
**Velocity with respect to Time**

151
**Graphical Representation of the Fluid Resistance Equation**

x t v t a t

152
29 Which of these is the graph that best represents the
general velocity versus time for air resistance? v t A B v t t v C D v t

153
We have learned before that distance is the integral of
velocity so by integrating the velocity equation we were
able to come up with the distance equation.

154
30 Which of these is the graph that best
represents the general position versus time
for air resistance? t x B x t A x t C D x t

155
**The final equation shows how acceleration approaches zero.**

156
31 Which of these is the graph that best represents
the general acceleration versus time for air
resistance? a t t a A B a t C D a t

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