What is the frictional force on a wheel-disk system?

[jake010]

Homework Statement

Find the magnitude and direction of the frictional force applied at point P if the system is to be in equilibrium.

Relevant Equations

m= 0.2 kg
M= 1.5 kg
r= 0.1 m
R= 0.3 m
g = 10 m/s^2

friction in axle and pulley, and mass of spokes and pulley are negligible

Correct answer: Fr = 5N

The figure is in the attached image.

My attempt:

I(wheel-disk system) = 0.5mr^2 + MR ^2
= 0.5(0.2kg)(0.1m)^2 + (1.5kg)(0.3m)^2
= 0.136(kg)(m^2)

Fnet(of object) = Ma
Mg - T = Ma
T = Mg - Ma

Fnet (wheel-disk system) = (M+m)a
Mg- Ma - Fr = (M+m)a
a = 0 because system in equilibrium
Mg - Fr = 0
Fr = Mg
Fr = 15N

This answer is incorrect, the correct answer is 5N and I'm not sure what I'm doing wrong.

 

[phinds]

I think you are way over complicating the problem. The weight of the disk and hoop are irrelvant, since at equilibruim they are totally in balance. The only thing that matters is the ration of the radii and the force pulling on the inner rim. I get an answer of 5N

 

[jake010]

I think you are way over complicating the problem. The weight of the disk and hoop are irrelvant, since at equilibruim they are totally in balance. The only thing that matters is the ration of the radii and the force pulling on the inner rim. I get an answer of 5N

Sorry if I'm not getting this but isn't the force pulling on the inner rim Mg (=15N) making frictional force 15N to cancel out the force pulling on the disk?

 

[phinds]

Sorry if I'm not getting this but isn't the force pulling on the inner rim Mg (=15N) making frictional force 15N to cancel out the force pulling on the disk?

Yes, that is exactly what it would do if the frictional force were on the same rim as the weight, but it's 3 times farther out. 15/3 = 5

You may now smack yourself in the forehead

 

[jake010]

Yes, that is exactly what it would do if the frictional force were on the same rim as the weight, but it's 3 times farther out. 15/3 = 5

You may now smack yourself in the forehead

Wait I think I get it now. So you divide force by 3 because angular acceleration stays the same but radius is 3 times larger, and linear acceleration = (angular acceleration)(radius), right? Thanks for the help!

 

[DEvens]

Wait I think I get it now. So you divide force by 3 because angular acceleration stays the same but radius is 3 times larger, and linear acceleration = (angular acceleration)(radius), right? Thanks for the help!

Hmmm... Equilibrium means what? So how much acceleration?

Forces, not acceleration.

In this case, the wheel-hoop thing is just a slightly unusually shaped lever.

Also, the question explicitly says to indicate the direction.