1. Dynamics Problems

2. Warm Up
The system below is in equilibrium. If the scale is calibrated in N, what does it read?
Since the tension is distributed over the entire (mass-less) rope, the scale will read (9.8N/kg)(5kg) = 49N
5kg
5kg

3. Warm Up Determine the Normal force in each of the following: F W F F m

4. Warm Up
Write F=ma for each scenario: m F a F W F m

5. Example Diagram Free body Diagram
A pumpkin of unknown mass is suspended by a cord attached to the ceiling and pushed away from vertical. When a 24.0 N force is applied to the pumpkin at an angle of to horizontal, the pumpkin will remain in equilibrium when the cord makes an angle of with the vertical.
(A) What is the tension in the cord when the pumpkin is in equilibrium ?   
(B) What is the mass of the pumpkin ?
Diagram
Free body Diagram

6. (A) What is the tension in the cord when the pumpkin is in equilibrium ? 
Since the pumpkin is in static equilibrium, we need only look at the horizontal components

7. (B) What is the mass of the pumpkin ?
Since the pumpkin is in static equilibrium, we need only look at the vertical components

8. Example Diagram Free body Diagram
The tension in the horizontal rope is 30N
A) Determine the weight of the object
Diagram
Free body Diagram
30N
400
500
500

9. A) Determine the weight of the object
500
The weight is in static equilibrium, so the appropriate net component forces must be zero.
Horizontal Component
Vertical Component
The weight of the mass is 36N

10. Weight on a Wire Diagram Free body Diagram
A rope extends between two poles. A 80N weight hangs from it as per the diagram.
A) Determine the tension in both parts of the rope.
Diagram
Free body Diagram
100
150
80N T1 T2

11. A) Determine the tension in both parts of the rope.
The weight is in static equilibrium, so the appropriate net component forces must be zero.
Horizontal Component
Vertical Component

12. Example Free body Diagram Block Ring
The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?.
Free body Diagram
Block
Ring
35N
7.0N

13. The system in the diagram is just on the verge of slipping. If a 7
The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?.
The weight is just in static equilibrium, so the appropriate net component forces must be zero.
Ring
Block
From Block:
From ring:
Combining:

14. Atwood’s Machine Example:
Masses m1 = 10 kg and m2 = 20kg are attached to an ideal massless string and hung as shown around an ideal massless pulley.
What are the tensions in the string T1 and T2 ?
Find the accelerations, a1 and a2, of the masses.
Fixed Pulley T1 T2 m1 a1 m2 a2

15. Draw free body diagrams for each object
Applying Newton’s Second Law:
T1 - m1g = m1a (a)
T2 - m2g = -m2a2
=> -T2 + m2g = +m2a2 (b)
But T1 = T2 = T
since pulley is ideal
and a1 = -a2 =a
since the masses are
connected by the string
m2g
m1g
Free Body Diagrams T1 T2 a1 a2

16. Solve for Acceleration
-m1g + T = m1 a (a)
-T + m2g = m2 a (b)
Two equations and two unknowns
we can solve for both unknowns (T and a).
Add (b) + (a):
g(m2 – m1 ) = a(m1+ m2 )
m2g
m1g
Free Body Diagrams T1 T2 a1 a2

17. Solve for T -m1g + T = m1 a (a) -T + m2g = m2 a (b)
Plug a into (b) and Solve for T
m2g
m1g
Free Body Diagrams T1 T2 a1 a2

18. Atwood Machine Review
So we find: m1 m2 a T

19. Example
A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box?

20. Pulling a Box (Part 1)
A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box?
Free body Diagram +y +x a

21. Pulling a Box (Part 1)
A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? +y +x a
Vertical Forces
Horizontal Forces
Solving

22. Example
Two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes?

23. Pulling a Box (Part 2)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes?
Free body Diagram +y +x a

24. Pulling a Box (Part 2) + Forces
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? a +y +x
Forces
4.00 kg Box
6.00 kg Box
Adding to eliminate T and find a +

25. Pulling a Box (Part 2)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? a +y +x
Solve for Acceleration
Now for Tension
We could have used the other tension formula from Box 2 and obtained the same answer

26. Example
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is connected by a 1kg rope and they are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes?
1.00kg

27. Pulling a Box (Part 3)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes?
Free body Diagram
1.00 kg
Because the rope has mass, the two ends will experience different tensions +y +x a

28. Pulling a Box (Part 3) Forces
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? a +y +x
Forces
4.00 kg Box
6.00 kg Box
Using F=ma for the system to find a

29. Pulling a Box (Part 3)
A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? a +y +x
Solve for Acceleration
Now for T1
Now for T2

30. Example
A worker drags a 38.0 kg box along the floor by pulling on a rope attached to the box. The coefficient of friction between the floor and the box is us=0.450 and uk=0.410.
a) What are the force of friction and acceleration when the worker applies a horizontal force of 150N?
b) What are the force of friction and acceleration when the worker applies a horizontal force of 190N?
38kg

31. Solution (Free Body Diagram)
a) What are the force of friction and acceleration of the worker applies a horizontal force of 150N?
The Normal force up
Friction to the left
The applied force of tension to the right
38kg
The force of gravity down +y +x

32. Solution (Vector Components)
What are the force of friction and acceleration of the worker applies a horizontal force of 150N?
38kg +y +x
To determine if the box will move, we must find the maximum static friction and compare it to the applied force.
Since the applied force by the worker is only 150N, the box will not move

33. Solution (Vector Components)
b) What are the force of friction and acceleration of the worker applies a horizontal force of 190N?
38kg +y +x
Since the applied force is greater than 168N from part a), we will have an acceleration in the x direction. So we will apply Newton’s 2nd Law in the horizontal direction.
FK=(0.410)(372.4N)=153N
The acceleration of the box is m/s2 [E]

34. Example
A train of three masses is pulled along a frictionless surface. Calculate the tensions in the ropes. T1 T2
8 kg
5 kg
13 kg
30 N
We can find the acceleration of the train by treating the three masses as one unit. T2 T1
Tension in rope T2 or T1 F
Tension in rope T1

35. Example
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40?
300

36. Solution (Free Body Diagram)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40?
The Normal force up
Friction to the left
The applied force of tension at 300
300
Tension broken down into components
The force of gravity down

37. Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40?
300 +y +x
Vertical Components
Horizontal Components
Since we have a constant velocity, acceleration is 0
We will need FN, so solve for FN

38. Solution (Force Components)
A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40?
300 +y +x
Solve for FT

39. Example
A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s.

40. Solution (Free Body Diagram)
A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y
Choose axis orientation to match the direction of motion and the normal to the surface
Normal is perpendicular to the surface
Object
Decompose gravity into axis components
Force of friction opposes direction of motion
Force of gravity is straight down

41. Remember to solve for FN because we will need it later
Solution (Force Vectors)
A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y
y direction
x direction
Remember to solve for FN because we will need it later

42. Example 9: Solution (Force Vectors)
A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +x +y
Acceleration
Speed

43. Example
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.

44. Solution (Free Body Diagram)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. +y +x a
Box B
Normal Force
Friction from A
Friction from Table
Object B
Force of Gravity from A and B
Tension
Applied Force

45. Solution (Free Body Diagram)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. +y +x a
Normal Force
Box A
Friction
Object
Tension
Force of Gravity from A only +y +x a

46. Solution (Force Vectors)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. A B a +y +x
x-direction for Box A
x-direction for Box B
This was determined using Box A

47. Solution (Force Vectors)
Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. A B a +y +x

48. Example
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?
1.5 kg
2.0
1.0

49. Example (free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?
3 Blocks taken as a Single Unit
Normal Force
1.5 kg
2.0
1.0
1.5 kg
2.0
1.0
Friction
Object
Applied +y +x
Force of Gravity a

50. Example (force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?
1.5 kg
2.0
1.0 +y +x a

51. Example (free body diagram)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?
2.0 kg Block and 1.0 kg taken as a Single Unit
Since we are considering 2.0kg and 1.0kg block as a unit, then the Force is the push of 1.5 kg block on the combined block
Normal Force
Friction
Object
2.0 kg
Applied
1.5 kg
1.0 kg
Force of Gravity +y +x a

52. Example (force vectors)
How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?
2.0 kg
1.0 +y +x a

53. Example Step 1: Free body Diagram FT FT m1 + FG=9 kg +
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object.
Find the acceleration of the two objects
Find the tension in the string.
Step 1: Free body Diagram
FG=9 kg FT FT m1 + +
The easiest way to choose the signs for the forces is to logical choose what you believe will be correct direction and follow that direction from one object to the other. If your final answer is negative, it just means your initial direction choice was wrong.

54. Example (Solution) FT FT m1 + FG=9 kg +
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object.
Find the acceleration of the two objects
FG=9 kg FT FT m1 + +
Adding these equations will remove the force of tension, thus giving allowing us to solve for acceleration.
Horizontal
Vertical
Since the answer is positive our initial direction choice was correct.
Note: don’t just say that a force of (9.8x9) is pulling on the 5 kg mass. This will give an acceleration of 17.6, which is greater than the acceleration due to gravity of the falling 9 kg mass, and you would not have tension in the rope. Ugly!

55. Example (Solution) FT FT m1 + FG +
A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object.
Find the tension in the string. FG FT FT m1 + +
We need only substitute the acceleration value into either the horizontal or vertical equation.
Horizontal
Vertical

56. Example
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block C descends with constant velocity.
Determine the tension in the rope connecting Block A and B.
What is the weight of Block C?
If the rope connecting A and B were cut, what would be the acceleration of C?

57. Example Normal Block A Block B Normal Tension from C Object Tension
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity.
Determine the tension in the rope connecting Block A and B.
What is the weight of Block C?
If the rope connecting A and B were cut, what would be the acceleration of C?
Normal
Block A
Block B
Normal
Tension from C
Object +x +y
Tension
Object
Friction
Tension from A
Gravity
Gravity
Friction

58. Example (Force Vectors)
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity.
Determine the tension in the rope connecting Block A and B.
What is the weight of Block C?
If the rope connecting A and B were cut, what would be the acceleration of C?
Block A
Block C
Block B

59. Example (Force Vectors)
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity.
Determine the tension in the rope connecting Block A and B.
What is the weight of Block C?
If the rope connecting A and B were cut, what would be the acceleration of C?
Block A
Block C
Block B
Solving for the tension between block A and B
Applying Block C’s equation to Block B
The weight of Block C

60. Example (Force Vectors)
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is Block c descends with constant velocity.
Determine the tension in the rope connecting Block A and B.
What is the weight of Block C?
If the rope connecting A and B were cut, what would be the acceleration of C?
When the rope is cut:

61. Example
A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.

62. Example (Free Body Diagram)
A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.
We will look at this from outside the truck (ie the ground) because we would prefer an inertial frame of reference.
Tension broken into components
Tension
Object
Gravity

63. Example (Force Vectors)
A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.
Vertical Forces
Horizontal Forces

64. Example
Calculate the acceleration of a box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.
5.0 kg
25N

65. Example (Free Body Diagram)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.
Since the gravity force down (5x9.8) is greater than force up (25sin(30), the box slides down, so friction is up.
Applied
Friction
5.0 kg
25N
Object
Magnetic
Normal
Gravity

66. Example (Vector Forces)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.
5.0 kg
25N +y +x
Vertical
Horizontal

67. Example (Insert Numbers)
Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.
5.0 kg
25N +y +x

68. Example
Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.

69. Example (Free Body Diagram)
Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.
Treat objects 300 thru 121 as a single mass of 180*0.5kg
For Masses 121 to 300
Normal
Big Mass
Tension from above
Gravity
Friction

70. Example (Vector Forces)
Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.
For Combined Masses 121 to 300 a +y +x
y-axis
x-axis

71. Example (Insert Values)
Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.
For Combined Masses 1 to 120 a +y +x

72. Example
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,
Find the normal force on the crate
Find the force of friction on the crate
Find the acceleration of the crate.
300 a

73. Example (Free Body Diagram)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,
Find the normal force on the crate
Find the force of friction on the crate
Find the acceleration of the crate.
300 FN a Ff
300 +y +x Fa Fg

74. Example (Vector Forces)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,
Find the normal force on the crate
Find the force of friction on the crate
Find the acceleration of the crate. +y +x Fg FN Ff Fa
300 a
y-axis
Insert Values

75. Example (Vector Forces)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,
Find the normal force on the crate
Find the force of friction on the crate
Find the acceleration of the crate. Fg FN Ff Fa
300 +y +x a
Friction

76. Example (Vector Forces)
A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,
Find the normal force on the crate
Find the force of friction on the crate
Find the acceleration of the crate. Fg FN Ff Fa
300 +y +x a
Acceleration

77. Example Calculate the unknowns for each accelerated block. b) a) a c)
18 kg a) F
6 kg b) a m c)

78. Example (Solution) Calculate the unknowns for each accelerated block.
18 kg a) F

79. Example (Solution) Calculate the unknowns for each accelerated block.
6 kg b) a

80. Example (Solution) Calculate the unknowns for each accelerated block.

81. Example
Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs

82. The only horizontal force is the Normal Force .Therefore F=N=ma
Example
Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs
If I do not fall, then the friction force, Ff, must balance my weight mg, that is Ff = mg
The only horizontal force is the Normal Force .Therefore F=N=ma
Putting this together we obtain:

83. Attached bodies on two inclined planes
smooth peg m2 m1 1 2
all surfaces frictionless
peg is frictionless

84. How will the bodies move?
From the free body diagrams for each body, and the chosen coordinate system for each block, we can apply Newton’s Second Law:
Taking “x” components:
1) T1 - m1g sin 1 = m1 a1X
2) T2 - m2g sin 2 = m2 a2X
But T1 = T2 = T
and -a1X = a2X = a
(constraints) x y x y
m2g T2 N 2 T1 N
m1g 1 m2 m1

85. Solving the equations Using the constraints, we get 2 eqn and 2 unks,
solve the equations.
T - m1gsin 1 = -m1 a (a)
T - m2gsin 2 = m2 a (b)
Subtracting (a) from (b) gives:
m1gsin 1 - m2gsin 2 = (m1+m2 )a
So: - a m g = + 1 2
sin q

86. Two-body dynamics Case (1) Case (2)
In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. m
10kg a a m
F = 98.1 N
Case (1)
Case (2)
(a) Case (1) (b) Case (2) (c) same

87. Solution
For case (1) draw FBD and write FNET = ma for each block:
(a)
T = ma (a)
mWg -T = mWa (b) m
10kg a
Add (a) and (b):
mWg = (m + mW)a
mW=10kg
(b)
Note:

88. Solution T = 98.1 N = ma Case (1) Case (2) For case (2) m m a a 10kg
F = 98.1 N
Case (2)
The answer is (b) Case (2) In this case the block experiences a larger acceleratioin

89. Problem: Accelerometer
A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle  with respect to the vertical (y) axis. Solve for  in terms of a and g. a  i

90. Accelerometer... Draw a free body diagram for the mass:
What are all of the forces acting? m
T (string tension)
mg (gravitational force)  i

91. Accelerometer... Using components (recommended):
i: FX = TX = T sin  = ma
j: FY = TY - mg
= T cos - mg = 0 TX  TY T  j i m ma mg

92. Accelerometer... Using components : i: T sin  = ma
j: T cos - mg = 0
Eliminate T : TX TY T  j i m ma mg
T sin = ma
T cos = mg

93. Accelerometer...
Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
T (string tension) T  mg  m
FTOT
mg (gravitational force)

94. Accelerometer...
Alternative solution using vectors (elegant but not as systematic):
Find the total vector force FNET:
Recall that FNET = ma: So
T (string tension)  T mg  m ma
mg (gravitational force)

95. Accelerometer... Let’s put in some numbers:
Say the car goes from 0 to 60 mph in 10 seconds:
60 mph = 60 x 0.45 m/s = 27 m/s.
Acceleration a = Δv/Δt = 2.7 m/s2.
So a/g = 2.7 / 9.8 =
 = arctan (a/g) = 15.6 deg a 

96. Understanding
A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. These two forces
Have equal magnitudes and form an action/reaction pair
Have equal magnitudes but do not form an action/reaction pair
Have unequal magnitudes and form an action/reaction pair
Have unequal magnitudes and do not form an action/reaction pair
None of the above
Because the person is not accelerating, the net force they feel is zero. Therefore the magnitudes must be the same (opposite directions. These are not action/reaction forces because they act of the same object (the person). Action/Reaction pairs always act on different objects.