# Two blocks on frictionless table, friction between blocks (Blocks side by side)

2 really tricky ones for me on this homework assignment. I think I got the other one settled (Thanks to help here) -- now I just need to finish this last one.

## Homework Statement

Two blocks sit on this frictionless table, but there is friction between the blocks.

Static = .589
Kinetic = .332

A = 100 kg
B = 10 kg

What minimum force F must be exerted on A in order for B not to fall?

## Homework Equations

F=m•a

R - the static resistive force ---> fs*F=R

## The Attempt at a Solution

Attempt 1:

Since B isn't moving, I'll use the static friction force. I know that the force that A applies on B, times the static force, will be my upward force -- which I need to counter the downward force of gravity.

Since B is 10kg, I need 98 kg of up force.

98/589 = 166.4 N

So

Force to move block A = 980 N
Force to move block B = 98 N

980+98+166.4=1244 N

The only problem is I am given a range for the answer, and I need to be between 1700 and 2200, so I know that answer is wrong.

Once I find the force, then I'm supposed to find the overall acceleration...I think that should be easy, once I know the force.

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ehild
Homework Helper
F is a horizontal force, all the others you used are vertical, so there is a problem with your method.

Draw the free body diagram for the blocks: there is a horizontal force F acting on A and there is the normal force N acting between the blocks. The blocks move with the same acceleration "a" on the (horizontal) table.

The resultant horizontal force on A is F-N. F-N=100a.
The horizontal force on B is N: N=10a.
In the vertical direction, you have the downward force of gravity and the upward static friction, they should cancel.
The maximum static friction is N*fs, N*fs=10g.
You have to eliminate a and N from the three equations to get F.

ehild

Ok I was able to figure it out with that explanation. I'm not completely catching onto why it worked out that way, but I'm starting to get it. We have some more to do with this topic in the class, and I'm pretty sure it will click the next time it's explained.

Thanks very much for your help.