General Physics 1: Calculating Force for Constant Speed

[Kp0684]

General Physics 1...

A crate of mass 30.0kg rests on level surface. The coefficient of kinetic friction between the crate and surface is .400 . What force applied at an angle of 30.0 degrees below the horizontal (i.e. pushing down) is required to keep the mass moving at constant speed?......on this problem i came up with the equation Fn= Uk[mgcos(30)]...with Uk=.400...Fn=102N...as the answer...iam i doing this right...need help...please.

 

[scholzie]

You made some erroneous assumtions. The normal force of the crate is not just mgcos\theta, if the angle is that of an applied force. This method assumes that the box is on an incline (in which case that would be true). This is a level surface, so the normal force is caused by the box's full weight (mg) and the force you apply (F_y=F\cos\theta) by pushing or pulling.

Force of friction: f=\mu N with N=mg+F\sin30 and so f=\mu\left(mg + F\sin30\right)

To move at a constant speed, F\cos30 must equal the force of the friction trying to keep it from moving:
F_{x net} = F\cos30 - F_{friction} = ma =0

That should help.

PS, as an answer, I got close to 180N - I'll hold off on the exact answer I got until you've done the problem. :)

 

[thepatientmental]

Yes, you are correct! The equation you used, Fn = Uk[mgcos(30)], is the correct formula for calculating the force needed to keep the crate moving at a constant speed. The normal force (Fn) is equal to the coefficient of kinetic friction (Uk) multiplied by the weight of the crate (mg) and the cosine of the angle of the applied force (30 degrees).

Plugging in the given values, we get:
Fn = (0.400)(30.0kg)(9.8m/s^2)(cos30) = 102N

So, the force required to keep the crate moving at a constant speed is 102N, which is the same answer you got. Great job!