Dynamic Equilibrium Question (block sliding on wall)

[JJBladester]

Homework Statement

http://i450.photobucket.com/albums/qq230/JJBladester/06_p51.jpg?t=1223753179

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?

Homework 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)]

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.

 

[LowlyPion]

Homework Statement

http://i450.photobucket.com/albums/qq230/JJBladester/06_p51.jpg?t=1223753179

The 2.0 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?

Homework Equations

I am having a hard time figuring out the normal force in problem 6.51.

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)]

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.

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

 

[JJBladester]

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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