Friction problem about 3 boxes on top of one another?

Question

I just need someone to tell me what to do. I want to know how to solve this. Here is an overview of the problem.

Find the least horizontal force Q to start motion of any part of the system of 3 blocks resting upon one another. The weight of block A is 300lbs, B is 100lbs and C is 200lbs. Between A and B, and B and C, C and the ground there is a coefficient of friction.

The diagram of the system looks like this:

A --->Q

B

C

A, B, and C are the boxes and Q is directed away from A.

Answer 1

You have to exert the force on the bottom block if they all have the same amount of friction between them otherwise you would only be moving the block you are pushing and anything above it.  This is because friction depends on the normal force acting on an object which is dependent on mass.  So the route to go here is treat the system as if it is one block with mass equal to A+B+C and find the minimum force (that is the force that will just overcome the static friction) to move it.  That will be your answer.

Answer 2

Driving Force - Resistive Force = Net f\orce

DF - RF = ma

Q - ( coefficient of friction X reaction force ) = ma

For the least value of Q , ma must be somehow 0 (achieved by giving a jerk to maintain a constant velocity)

Where , Reaction force = mg = mass of A + mass of B + mass of C (Consider it as one system)

Coefficient of friction is given , what's the deal put these up in the formula and go ahead.

Answer 3

draw a free body diagram (FBD)of each of the boxes sum the forces

so for box A

sum the forces in the vertical direction

F= N - w = 0

N=w

sum the forces in the horizontal direction

F= Q- f = 0

Q = f = u(N)

when Q > f the box moves

f= frictional force opposing motion

u=coefficient of friction

N= normal force

w= weight

for box B

sum the forces in the vertical direction

F= N -w_a -w_b = 0

N1= w_a+ w_b

sum the forces in the horizontal direction

F= Q- f = 0

Q = f = u(N1)

same method for box C