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Two blocks (m1 = 5kg, m2 = 2.5kg) are in contact on a frictionless table. A constant horizontal force FA = 3N is applied to the larger block as shown.
Find the magnitude of the force F1 on 2 exerted by the larger block on the smaller block.
Find the magnitude of the force F2 on 1 exerted by the smaller block on the larger block.
FA
m1
m2y
x
• A : This is an equilibrium problem, the two blocks remain at rest.
• B : This is an equilibrium problem, the two blocks move at a constant velocity.
• C : This is a non-equilibrium problem, the two blocks move at a constant velocity.
• D : This is a non-equilibrium problem, the two blocks accelerate.
1. Which of the following statements is true?
Qualitative analysis
Choice: A
Incorrect
Since our blocks are on a frictionless table, there is no friction force resisting the force that is applied to the larger block. Only if there was a friction force equal to the applied force would
the system by in equilibrium.
-Equilibrium requires that no net force acts on object
- Otherwise, the object is not in equilibrium.
- If velocity is constant (or zero) then object is in equilibrium.
Choice: B
Incorrect
Since our blocks are on a frictionless table, there is no friction force resisting the force that is applied to the larger block. Only if there was a friction force equal to the applied force would
the system by in equilibrium.
-Equilibrium requires that no net force acts on object
- Otherwise, the object is not in equilibrium.
- If velocity is constant (or zero) then object is in equilibrium.
Choice: C
Incorrect
Since our blocks are on a frictionless table, there is no friction force resisting the force that is applied to the larger block. Only if there was a friction force equal to the applied force would
the system by in equilibrium.
-Equilibrium requires that no net force acts on object
- Otherwise, the object is not in equilibrium.
- If velocity is constant (or zero) then object is in equilibrium.
Choice: D
Correct
This is a true statement. Since our blocks are on a frictionless table, there is no friction force resisting the force that is applied to the larger block. Only if there was a friction force equal
to the applied force would the system by in equilibrium.
The blocks will accelerate, and our system is not in equilibrium.
2. Which of the following statements is true?
• A : The applied force will act on both blocks because the smaller block is in the way of the larger block.
• B : The applied force will act only on the larger block. The smaller block will feel a smaller horizontal force exerted by the larger block.
• C : The acceleration of the two blocks will be different because they have different masses.
Qualitative analysis
Choice: A
Incorrect
The applied force to our blocks, perhaps by someone’s hand, is a type of contact force.
A contact force acts only at the place of contact (it does not act on objects it is not touching).
Choice: B
Correct
This is true because the force applied to our blocks, perhaps by someone’s hand, is a type of contact force.
A contact force acts only at the place of contact (it does not act on objects it is not touching).
3. Newton’s 3rd Law, in general, states:
“If body A exerts a force on body B (action), then body B exerts a force on body A (reaction) that is equal in magnitude and opposite in
direction.”
Considering this law, what do we know is true about our situation?
• A: There is a force exerted on the larger block by the smaller block (F2 on 1), and there is also a force exerted on the
smaller block by the larger block(F1 on 2).
• B: The direction of F2on1 should be opposite of F1 on 2.
• C:
• D: All of the above are true.
|F2 on 1| = |F1 on 2|
This is true according to Newton’s third law. Since block 1 is pushed into the block 2 there is a force (action) on block 1 by block 2. Block two must exert a force (reaction) on block 1.
Choice: A
This is not the only correct choice.
This is true according to Newton’s third law. Since block 1 exerts a force (action) on
block 2 by being pushed into it, block two must exert a force (reaction) on block 1.
This pair of forces, as with all action-reaction pairs, act in opposite directions, and have the same
magnitude.
Choice: B
This is not the only correct choice.
This is true according to Newton’s third law. Since block 1 exerts a force (action) on block 2 by being pushed into it, block two must exert a
force (reaction) on block 1.
These two forces, as with all action-reaction pairs, have equal magnitudes.
Choice: C
This is not the only correct choice.
Choice: D
Correct
Since block 1 exerts a force (action) on block 2 by being pushed into it, block two must
exert a force (reaction) on block 1.
This pair of forces, as with all action-reaction pairs, act in opposite directions, and have the same
magnitude.
4. Which pair of free-body diagrams are correct ?• A:
• B:
• C:
• D:
Planning
m1g
m1g
m1g
m1g
FA
FA
FA
FA
N1
N1
N1
N1
F2on1
F2on1
F2on1
F2on1
m2g
m2g
m2g
m2g
N2
N2
N2
N2
FA F1 on 2
F1 on 2
FA
F1 on 2
F2 on 1
Bold type denotes vector
quantities in all free-body
diagrams.
Bold type denotes vector
quantities in all free-body
diagrams.
Choice: B
Incorrect
The applied force does not act on the 2nd smaller red block, but there is a force from the
1st block pushing on the 2nd block.
Choice: D
Correct
These diagrams represent all of the forces on each block correctly.
Note that we can also make a free-body diagram for the two blocks together as a system:
N
(m1+m2)g
FA
5. Which one of the following statements is incorrect about the vertical motion or vertical forces acting on the blocks?
A: None of the vertical forces are relevant for solving this problem.
C: The solution to this problem will be the same on the earth and on the moon because the gravitational force is not important for this problem.
B: Vertical and horizontal motions are completely decoupled in this problem since the surface is frictionless.
D: Gravitational force is important for solving the problem.
Planning
This statement is correct.
We are only concerned with the forces and motion in the positive and negative x-directions
in this problem, because the vertical forces cancel out. The force of gravity on the blocks is
canceled by the normal force.
Choice: A
Incorrect
Choice: B
Incorrect
This statement is correct.
We can solve this problem by only analyzing the forces in the x-directions.
Choice: C
Incorrect
This statement is correct.
We can solve this problem by only analyzing the forces in the x-directions.
The force of gravity act in the negative y-direction and is cancelled by the normal
force.
Choice: D
Correct
This statement is incorrect, because gravity pulls in the negative y-direction and is exactly
canceled by the normal force acting in the positive y-directon.
Gravitational force is not important for solving
the problem.
6. Apply Newton’s 2nd Law to each block individually, and then to the two blocks together as a system.
Which of the following equations is not a result of applying Newton’s 2nd Law? (a is acceleration)
• A: FA – F 2 on 1 = m1a
• B: FA – F1 on 2 = (m1 + m2)a
• C: F1 on 2 = m2a
• D : FA = (m1+m2)a
Implementation
Choice: A
Incorrect
Recall the free body diagram for the first block from question 3:
Use Newton’s 2nd Law to sum up the forces in each direction:
Fx m1ax
FA F2 on 1 m1ax
Fy m1ay 0
m1g N1 0
m1gFA
N1
F2 on 1
We get this equation be applying Newton’s 2nd Law to the larger block:
No motion in the y-direction
No motion in the y-direction
Applying Newton’s 2nd Law to either block or both
together will not lead to this equation.
Choice: B
Correct
Choice: C
IncorrectWe get this equation be applying Newton’s 2nd Law to the smaller block:
Recall the free body diagram for the second block from question 3:
Use Newton’s 2nd Law to sum up the forces in each direction:
N2
m2g
F1 on 2
Fx m2ax
F1 on 2 m2ax
Fy m2ay 0
m2g N2 0
No motion in the y-direction
No motion in the y-direction
Choice: D
IncorrectWe get this equation be applying Newton’s 2nd Law to the system of blocks moving together:
1. Make a free-body diagram for the 2 blocks together:
2. Use Newton’s 2nd Law to sum up the forces in each direction:
N
FA
(m1+m2)g
Fx max
m m1 m2
FA m1 m2 ax
Fy may 0
m1 m2 g N0
No motion in the y-direction
No motion in the y-direction
7. Now use the correct equations that we found in the previous question to solve for the magnitude of the force (F1 on 2) exerted on the smaller block by the larger block.
Answer
1st: Solve for acceleration 2nd: Substitute to find F1 on 2
F1 on 2 m2ax
FA m1 m2 ax
FA
m1 m2 ax
3N5kg 2.5kg
ax
0.4m/s2 ax
F1 on 2 2.5kg 0.4m/s2 F1 on 2 1N
8. Recall from using Newton’s 3rd Law in question 3 that |F2 on 1| = |F1 on 2|, so the magnitude of F2 on 1 is also 1 N.
Let’s check that this is true using the correct equations that we found in the previous questions to solve for the magnitude of the normal force (F2 on 1) exerted on the larger block by the smaller block.
Answer
Assessment
• Do units look correct?• Is |F2 on 1| = |F1 on 2|?
• Is the direction of F2 on 1 opposite FA?
Reflection• Newton’s 3rd law gave an important relation for the forces
between blocks. This law explains the normal forces that the blocks exert on each other.
• Blocks moving together have the same acceleration.
• The applied force (FA) only acts on larger block.
• If the force FA were applied to the smaller block in the opposite direction, the solution can be obtained by swapping m2 and m1:
F1 on 2 m1
FA
m1 m2
FA
1m2
m1
F2 on 1 m2
FA
m2 m1
FA
1m1
m2