Calculating the Force Required to Move a Crate at Constant Velocity

[moy13]

Homework Statement

A factory worker pushes a 28.4 kg crate a distance of 4.6 m along a level floor at constant velocity by pushing downward at an angle of 31 degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

What magnitude of force (Fpush) must the worker apply to move the crate at constant velocity?

Homework Equations

sum of Fx = Fpush*cos(31) - (kinetic friction) = 0
sum of Fy = n - Fpush*sin(31) - mg = 0

total work = K2 - K1 = 0
work = F dot s = Fscos(31)

kinetic friction = mu-sub-k * n

The Attempt at a Solution

Fpush = (kinetic friction) / cos(31)

and

n = Fpush*sin(31) + mg

I need n, but I can't find it this way, what else can I do?

Thanks for any help.

 

[joseg707]

Solve for Fpush and plug that into the other equation. I'm having trouble with forces myself, but that's what I would do.

 

[joseg707]

Actually F_N is equal to mass times gravity. You already have everything else so I think all you have to do is solve.

 

[moy13]

Actually F_N is equal to mass times gravity. You already have everything else so I think all you have to do is solve.

I tried keeping n as the mass times gravity, but I get the wrong answer since there is a downward force exerted on the crate by the factory worker.