Physics Quiz

[BRAVESaj25bd8]

I'll use x,y,z to denote right direction, up direction, out of the page direction respectively, based on your drawing.

Sum of moments about the center of the wheel (z axis) shows that the y component of the friction force is tau/r.

Knowing that the axle is a two force member (cannot translate a moment), the friction force must be in the direction theta away from the x axis pointing up and to the right.

The magnitude of the force can be found by finding the resultant of the friction force from the y component we have already. Thus, the friction force is acting at a magnitude of tau/(rsin(theta)) at an angle theta above the horizontal (pointing up and to the right).

Solid question.

[Ether]

Originally Posted by brndn

F = τ/(rsinθ)

Correct!

Originally Posted by brndn

...opposite the direction of the wheel turning?

Not clear what you mean by the above. But if you mean that F lies in the plane of the wheel, then no.

[Ether]

Originally Posted by John

the forward friction would be F= tau/r

Correct.

Originally Posted by John

sideways friction = tau/r * cotan(theta)

Correct.

Originally Posted by John

The total magnitude of the total friction force is just
F = sqrt ((tau/r)^2 + (tau/r * cotan(theta))^2)
F = tau/r * sqrt(1+cotan(theta)^2)

Correct so far...

Originally Posted by John

F = tau/r * sqrt(1+cotan(theta)^2) = tau/r * sqrt(1/sin(theta))^2= tau/r * cosec(theta)^2

Oops. You forgot the "sqrt". So the answer is wrong.

Originally Posted by John

The angle from the horizontal is
arctan((tau/r)/(tau/r * cotan(theta))) = arctan (tan(theta)) =theta

Correct. (You have a typo, shown in blue)

Originally Posted by John

This makes sense because the friction force has to act exactly opposite to the tension force, or the wheel won't be in static equilibrium.

Correct!

[Ether]

Originally Posted by BRAVESaj25bd8

the y component of the friction force is tau/r.

Correct.

the friction force must be in the direction theta away from the x axis pointing up and to the right.

Correct.

the friction force is acting at a magnitude of tau/(rsin(theta)) at an angle theta above the horizontal (pointing up and to the right)

Correct. Nicely done.

[Ninja_Bait]

I figured it out in the middle of a driving class (about 11:15, EST) but I didn't have internet access when I got it. I was double counting the tau/r which led to the erroneous assumption about the direction of friction I had before. The friction is what drives the wheel forward, not tau/r.

I got the right answer all by myself, though, after I figured that bit out. I learned a little physics, too.

[James Critchley]

The "correct" solution is singular about theta = N*pi and is incomplete.

The puzzle contains enough information to solve this. Anybody... ?