Quote:
Originally Posted by John
the forward friction would be F= tau/r
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Correct.
Quote:
Originally Posted by John
sideways friction = tau/r * cotan(theta)
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Correct.
Quote:
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)
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Correct so far...
Quote:
Originally Posted by John
F = tau/r * sqrt(1+cotan(theta)^2) = tau/r * sqrt(1/sin(theta))^2= tau/r * cosec(theta)^2
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Oops. You forgot the "sqrt". So the answer is wrong.
Quote:
Originally Posted by John
The angle from the horizontal is
arctan((tau/r)/(tau/r * cotan(theta))) = arctan (tan(theta)) =theta
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Correct. (You have a typo, shown in blue)
Quote:
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.
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Correct!