# Homework Help: Dynamic Equilibrium Question (block sliding on wall)

1. Oct 11, 2008

1. The problem statement, all variables and given/known data

The 2.0kg wood box in the figure slides down a vertical wood wall while you push on it at a 45 degree angle.

What magnitude of force should you apply to cause the box to slide down at a constant speed?

2. Relevant equations

I am having a hard time figuring out the normal force in this problem.

I have set the problem up with four forces (f_k (friction), F_g (gravity), n (normal), F_push (push)). I got the coefficient of friction (f_k) as 0.20 from my textbook.

Fnet(x) = F_push(x) + F_g(x) + n + f_k(x)
Fnet(x) = F_push*cos(45) + 0 + n + 0

Fnet(y) = F_push(y) + F_g(y) + f_k(y)
Fnet(y) = F_push*sin(45) + (2.0)(9.8) -[0.20*2.0*(sin(45)]

3. The attempt at a solution
I know that the normal force acts perpendicular to the surface, so their should be no n in the y-direction. I'm stuck in finding the normal force in the x-direction and I believe this is the missing link in me figuring out this problem.

Last edited: Oct 11, 2008
2. Oct 11, 2008

### LowlyPion

Welcome to PF.

Since the angle is 45 degrees, don't you know what the normal force is in terms of the force?

And isn't that really a part of the equation I highlighted and not the 2.0 as it relates to friction?

As in 2.0*(9.8) needs to balance with .2 * F * cos45 as well as the vertical component F * Sin45

Last edited: Oct 11, 2008
3. Oct 11, 2008

LowlyPion,

Thanks for welcoming me to this site and thanks for your guidance.
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The normal force in terms of the push force is F_push*cos45. Using that to find f_k, I think I have the solution.

Fnet(y) = f_k + F_g + F_push(y) = 0

Fnet(y) = 0.20*F_push*cos45 - (2.0)(9.8) + F_push(sin45) = 0
19.6 = 0.1414*F_push + 0.7071*F_push

Therefore, F_push is roughly 23 N.
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