Solving 2 Simple Force Problems: Understanding Force Magnitude and Application

[musicfairy]

There's two force problems I couldn't get on my study guide.

1. Two blocks (A and B) are in contact on a horizontal frictionless surface. A 36-N constant force is applied to A as shown. The magnitude of the force of A on B is:

There is a picture. Block B is on the right of Block A. 36N is applied to block A. Block A = 4.0kg and block B = 20kg.

A. 1.5 N
B. 6.0 N
C. 29 N
D. 30 N
E. 36 N

What I did is:

F = (m_A + m_B)a
a = 36 / (4.0 + 20)
a = 3/2 m/s²

FAB = 4.0(3/2)
= 6.0 N

The correct answer is D. 30N. I can see where it comes from. I thought "the magnitude of the force of A on B" means I should multiply m_A and a to get the force, but to get 30 N they multiplied m_B and a. Why?

2. Two blocks with masses m and M are pushed along a horizontal frictionless surface by a horizontal applied force F as shown. The magnitude of the force of either of these blocks on the other is:

In this picture. M and m are in contact. m is on the right of M. Force F is applied to M.

A. mF/(m + M)
B. mF/M
C. mF/(M - m)
D. MF/(m + M)
E. MF/m


The correct answer is A. I don't even know how to start this problem. I tried to solve for a and substitute but that failed. Any ideas?



These came from a study guide I was given in class. I thought I knew this topic well but these problems really scare me. Can someone please explain these to me? I have a test coming up pretty soon. I read this section in the book, but sometimes I can't apply what I read to every problem.

 

[Dschumanji]

There's two force problems I couldn't get on my study guide.

1. Two blocks (A and B) are in contact on a horizontal frictionless surface. A 36-N constant force is applied to A as shown. The magnitude of the force of A on B is:

There is a picture. Block B is on the right of Block A. 36N is applied to block A. Block A = 4.0kg and block B = 20kg.

A. 1.5 N
B. 6.0 N
C. 29 N
D. 30 N
E. 36 N

What I did is:

F = (m_A + m_B)a
a = 36 / (4.0 + 20)
a = 3/2 m/s²

FAB = 4.0(3/2)
= 6.0 N

The correct answer is D. 30N. I can see where it comes from. I thought "the magnitude of the force of A on B" means I should multiply m_A and a to get the force, but to get 30 N they multiplied m_B and a. Why?

The blocks are in contact with each other so when the force is applied to the composite system, both parts must have the same acceleration. The force being applied to the composite system happens on block A, but block A also exerts a force on block B to make it accelerate. We know that block B must have the same acceleration as block A so the force acting block B must be equal to m_ba.

 

[Mattowander]

For the second problem, it can be solved the same as the first one.

The acceleration of both blocks together can be found using F = ma, where m is (M + m).

Once you find the acceleration, you can multiply by m to get mF/M+m

You multiply by m and not M for the same reason the above poster explained.

Hope this helps