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xstetsonx
Oct9-09, 05:32 PM
1. A m = 3.0 kg wood box slides down a vertical wood wall while you push on it at a θ = 49° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

2. Fk-Mk*n


3. i know there are 4 forces acting on the object. gravity, normal force from the wall, kinetic friction, and the force that i applied. but i do not know how to set up the equation to solve this problem

Vykan12
Oct9-09, 05:43 PM
Do you know how to draw a free body diagram?

xstetsonx
Oct9-09, 05:47 PM
yea i did one but the n is on the x axis so that means i need to know the F to get the normal force right?

Vykan12
Oct9-09, 05:53 PM
First off, you have to neglect friction since you do not know the kinetic friction coefficient of the surface. Second, the normal force doesn't cause the block to move in what you defined to be the x direction. Are you using newton's second law for net force in the x and y directions?

xstetsonx
Oct9-09, 05:54 PM
First off, you have to neglect friction since you do not know the kinetic friction coefficient of the surface. Second, the normal force doesn't cause the block to move in what you defined to be the x direction. Are you using newton's second law for net force in the x and y directions?

o i am sorry this question some how expected us to find the coefficient of kinetic friction on wood which is 0.2. and yes

tiny-tim
Oct9-09, 05:55 PM
Hi xstetsonx! Welcome to PF! :wink:
i need to know the F to get the normal force right?

No, you don't need to know F, you can just call it F.

For constant speed, you need zero acceleration, which means zero net force in both the x and y direction.

So call the force F, find N, and then find ľN. :smile:

xstetsonx
Oct9-09, 05:58 PM
so the total force will be
mK*Fcos49-mg+Fsin49?

tiny-tim
Oct9-09, 06:02 PM
so the total force will be
mK*Fcos49-mg+Fsin49?

Yes, except I'm not sure about that m at the beginning.

Vykan12
Oct9-09, 06:02 PM
You should get

F_{app} - F_{k} - mg\sin(\theta) = 0
n - mg\cos(\theta) = 0

The algebra's up to you.

Edit:

so the total force will be
mK*Fcos48-mg+Fsin48?

Yes, that's correct, assuming F is the force of gravity.