Maximizing Friction: Solving the 10kg Block and 5kg Bracket Question

[p_o_o_k_i]

hey guys! I've been stuck on the question for 2 days now, and it's really frustrating! I'm really hoping that someone can help me out! =] thanks in advance!

a 10kg block rests on a 5kg bracket (img below). the 5kg bracket sits on a frictionless surface. The coefficents of friction between the 10kg block and the bracket on which it rests are $$\u_s$$=.40 and $$\u_k$$=.3.
(a) what is the maximum force F that can be applied if the 10kg block is not to slide on the bracket?
(b)what is the corresponding acceleration of the 5kg bracket?

http://img114.imageshack.us/my.php?image=56qe9.jpg

my work:
is it safe to say, because the 10kg block does not slide on the bracket, that it is static friction, and also an acceleration to the left?

$$\vec{F}_{net} = \Sigma \vec{F}x = F_s - F = - m a$$

=$$\u_s m g + m a = F$$

but then, my equation has two unknow variables, F and a. So then, I'm guessing that there is no acceleration?

And I don't get, how the bracket would have an acceleration...

Thanks in advance!

 

[chroot]

When you push on the static block, the bracket pushes back on the block with a force equal and opposite the force you applied. Thus, you're pushing in one direction, but the friction force is pushing back in the opposite direction -- resulting in no net force, and no acceleration or movement.

When you push hard enough, however, you reach a point where the friction cannot push back with enough force to counter yours, and the block begins to move.

Both forms of friction are modeled by a proportionality between two forces, namely the force due to friction and the normal force, which is what the block exerts in pressing down on the bracket. If the coefficient of static friction is 0.4, that means that static friction is capable of 40% as much force as the normal force.

A block of mass m exerts a normal force of mg on the bracket, and static friction between the block and bracket is capable of 40% as much force as the normal force.

- Warren

 

[p_o_o_k_i]

thanks so much warren! your reply increased my understanding of friction. thanks!

would the acceleration of the bracket refer to the pully part? and would it have a kinetic friction instead of static friction?

 

[chroot]

I'm glad that my explanation helped you understand friction a little better.

We know that a force of 10kg * 0.4 * g will overcome the static friction between the block and the bracket.

Now, we just need to figure out how hard you'd have to pull on the rope to get such a force between the block and bracket. Note that my previous discussion necessarily involved the bracket pushing back on the block. If the bracket is to push back on the block, then the bracket has to be affixed to some immovable structure.

But we're told the bracket rests on a frictionless surface... which means that any force applied to the bracket will cause motion.

Can you draw a free-body diagram of the bracket, showing the forces due to the rope acting on the pulley?

- Warren