Force necessary to create enough friction.

[Elmnt]

Homework Statement

The two blocks (m = 18 kg and M = 101 kg) in the figure below are not attached to each other. The coefficient of static friction between the blocks is µs = 0.59, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block?


http://www.webassign.net/hrw/6-36.gif

Homework Equations

The Attempt at a Solution

My thought process involved me finding the force of friction that would be large enough to cancel out the the force due to gravity and hold the block in place. The friction force would have to have a magnitude of 176.4N (18kg*9.8m/s$$^{2}$$) since the static friction coefficient is .59 I just solved: friction force = static coefficient * Normal force , for the normal force, which = 298.98 N. Now that I know what the normal force needs to be, I am confused on how to solve for the force necessary to produce that normal force.

 

[rock.freak667]

I think the minimum force would be the normal force.

 

[Elmnt]

I was thinking along those lines, but it is wrong. I think it has something to do with the frictionless plane.

 

[Doc Al]

My thought process involved me finding the force of friction that would be large enough to cancel out the the force due to gravity and hold the block in place. The friction force would have to have a magnitude of 176.4N (18kg*9.8m/s$$^{2}$$) since the static friction coefficient is .59 I just solved: friction force = static coefficient * Normal force , for the normal force, which = 298.98 N. Now that I know what the normal force needs to be, I am confused on how to solve for the force necessary to produce that normal force.

So far, so good. Hint: What's the acceleration of the blocks?