Force needed to push a 2.0 kg box down a wall at constant speed

[tiffany09]

Hello everyone..so I am new to this forum. I missed the first couple weeks of my AP physics class due to injury and have so much work to catch up on. The problem is, I need a lot of help even though I'm still at a pretty basic level. I keep mixing up a lot of things.

Homework Statement

"A 2.0 kg wood box 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?"

So we know that there are 4 forces working on this object.
Applied force, Normal Force, Force due to Gravity, and Force of Friction.

We also the angle at which we're pushing the object against the wall: 45 degrees.

Homework Equations

F = ma
then what?

The Attempt at a Solution

Well I drew the diagram, and I know that because the object is moving down at a constant speed, the acceleration is 0. Therefore due to Newton's 2nd law, F=ma, the net force is 0. I don't know how this helps me though. However, I don't know what it means to find the magnitude, and I don't know how to find the values of any of the 4 forces. And what does the angle of his push have to do with anything?

 

[Rake-MC]

You cannot solve the problem with only this information. You must also know the coefficient of kinetic friction of wood on wood. This should either be listed in the question or at the end of your book.

 

[tiffany09]

I now understand what you mean by finding the kinetic friction of wood on wood. I found that the coeffeicient is .20 . what can i do with that information? if i know the normal force i can figure out with the net force is as well right?

 

[ejeong87]

by using your co-efficient of friction of 0.2 you can solve this problem.

what you have to realize is that the force that makes 45degrees on the boss actually causes a normal force from the wall and the box therefore causing friction. which can be found by using mu*r = friction [ which is pointing up]. r being reaction force which is Force * cos(45) [or sin(45) but doesn't make a difference].

So for the forces acting in y - axis is actually the following.

F*cos(45) + Fr - mg = 0

*note u = mu*

Fr = u*F*sin(45)

Fcos(45) + u*F*sin(45) = mg

F(cos(45)+u*sin(45)) = 2*9.8
F = 19.6 / (cos(45) + u*sin(45))

which should give you the correct answer.

I had the same question for one of my homework problems.

I had all that but I couldn't figure out what u was.

Could you explain how you got the value for u ?