Checking My Work: Solving for Applied Force in Friction Problem

[mohlam12]

hey everybody, i have an easy problem in friction, i just want to make sure if what i did is right; here is the problem:
A 22kg mass is slid up an incline of 71degree to the horizontal at constant velocity. If mu=.29, what is the applied force?

here is what i did:

Fgravity = 22*9.8*cos(71) = 70.19
Fparallel = 22.9.8*sin(71) = 203.85
Ffriction = 70.19/.26 = 269.96
Fapplied = 269.96+203.85 = 473.81 N

do u think that is right (not the calculations, but the formulas i used) ?

 

[Chi Meson]

Sorry. No.

What is it that you are calling "Fgravity"?

The force of gravity is "weight" which is "mg." The parallel component of weight that you have is correct, but the frictional force is wrong.

Friction is "mu" times the normal force. Here, the normal force balances the perpendicular compnent of the weight, so friction is
f = (0.29)(22kg)(9.8N/kg)cos71 = 20.3 N

since the velocity is constant, then the pushing force "up" the incline is balanced by the two forces "down" the incline. It appears that this part of the concept you have correct.

 

[HallsofIvy]

To second Chi Meson, the frictional force is mu times the normal force, not normal force divided by mu!