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## 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.