Calculate Torque on Unicycle for Homework Statement

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To calculate the maximum torque a woman can generate on a unicycle, her weight (mass times gravity) is used as the force applied on the pedal. The torque is determined by multiplying the crank length by this force. There is some confusion regarding the angle of force application, but it is clarified that the angle between the radius and the force is 90 degrees, making the torque calculation valid. The pedal serves as a means to transfer force to the crank, not affecting the torque formula itself. Understanding the relationship between force, crank length, and torque is crucial for solving this problem.
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Homework Statement


A woman is riding a unicycle. The mass of the woman is: m(w)=66.0 kg,
the mass of the unicycle is: m(u)=5.40 kg, the diameter of the unicycle
wheel is: d = 55.0 cm, and the length of the unicycle crank is L =
11.0 cm

What is the maximum torque that the woman is able to
generate just by using her weight to push on the unicycle
pedal?


Homework Equations


τ = r x F

The Attempt at a Solution


So, the maximum force that she can exert on the unicycle is her weight i.e. her mass times g. To find the torque all you have to do is multiply the length of the crank by the weight. I'm just wondering though why this is correct. I mean her weight acts vertically downwards, directly above the centre of the unicycle wheel. By the cross product formula r x F = |r||F|Sin(θ). But isn't theta in this case 0? Implying that the torque is 0, which can't be correct. I may just be visualizing this incorrectly though. If someone could explain this it'd be great. I'll attach a diagram of what I'm visualizing. Thanks.
 
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Am I visualising this incorrectly?
 
Your diagram did not show the force acting on the pedal but other than that I think you are correct. I think the angle is between th radius and the force and this would be 90 deg and sin 90 = 1 so T = F X r
 
Yeah, I was thinking that initially but our lecturer seemed pretty sure that it was the length of the crank times the applied force. This makes I sense to me as the rider is applying the force to the pedals and not the crank itself.
 
It is the length of the crank times the force on the pedal. There is nothing magic about the pedal other than a way to get the force from the foot to the end of the crank.
 
Ohhh, ok. I see. Thanks man.
 
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