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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.

m1

FA

m2

y

x

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Qualitative analysis

1. Which of the following statements is true?

- 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.

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: 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

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.

Qualitative analysis

- 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.

- 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: A

Incorrect

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).

Choice: B

Correct

Choice: C

Incorrect

If objects move together, they will have the same acceleration.

a1

a2

a

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 aforce 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|

Choice: A

This is not the only correct choice.

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: 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.

This pair of forces, as with all action-reaction pairs, act in opposite directions, and have the same magnitude.

Choice: C

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: 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.

Planning

N1

N2

F2on1

FA

F2 on 1

F1 on 2

- A:
- B:
- C:
- D:

m1g

m2g

N1

N2

F2on1

FA

FA

Bold type denotes vector quantities in all free-body diagrams.

m1g

m2g

N1

N2

FA

F1 on 2

F2on1

FA

m1g

m2g

N1

N2

F2on1

FA

F1 on 2

m1g

m2g

Choice: A

Incorrect

There is no force that pushes the smaller red block from the left.

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: C

Incorrect

The applied force does not act on the 2nd smaller 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

FA

(m1+m2)g

A: None of the vertical forces are relevant for solving this problem.

Planning

B: Vertical and horizontal motions are completely decoupled in this problem since the surface is frictionless.

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.

D: Gravitational force is important for solving the problem.

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

This statement is correct.We can solve this problem by only analyzing the forces in the x-directions.

Choice: B

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: C

Incorrect

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.

Implementation

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

We get this equation be applying Newton’s 2nd Law to the larger block:

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:

N1

F2 on 1

FA

m1g

Choice: A

Incorrect

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

We 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

F1 on 2

m2g

Choice: C

Incorrect

No motion in the y-direction

Choice: D

Incorrect

We get this equation be applying Newton’s 2nd Law to the system of blocks moving together:

- Make a free-body diagram for the 2 blocks together:
- Use Newton’s 2nd Law to sum up the forces in each direction:

N

FA

(m1+m2)g

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

8. Recall from using Newton’s 3rd Law in question 3 that |F2 on 1| = |F1 on 2|, so the magnitude of F2on 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?

- 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: