# Dynamics.

## Presentation on theme: "Dynamics."— Presentation transcript:

Dynamics

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

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

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

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.

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.

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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.

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

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

Two Dimensions

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.

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

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

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

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

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

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

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

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

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.

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.

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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.

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.

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

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.

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?

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

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

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.

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.

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

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

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

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

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)

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

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

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)

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.

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

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

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

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

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

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

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

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

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)

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

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

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

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

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

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)

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)

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)

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.

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)

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

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

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

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.

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

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.

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

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

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.

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.

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

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

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

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

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)

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.

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.

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?

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

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

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

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?

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

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

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

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?

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

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

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?

= ((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

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.

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

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.

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

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

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

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

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

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

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.

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

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

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.

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

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) )

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?

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

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

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

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.

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

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

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

Calculus Based

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

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

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.

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.

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

Velocity with respect to Time

Graphical Representation of the Fluid Resistance Equation
x t v t a t

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

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.

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

The final equation shows how acceleration approaches zero.

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