Recent content by Osh

Hi, Arin. It's Gareth. :) I was looking for help on the same question, and this was the first page on Google. Anyway, if you still want help, I found this other page that has the step-by-step answers: http://www.csupomona.edu/~skboddeker/132/assign/ch22h.htm

Well, that's confusing. In the professor's approach, there was no reaction force on the centre of mass acting backwards. I don't see how they would yield the same answer, considering that. Let me see how it works: For the torque from the reaction force on the bottom of the back wheel...

Apparently my professor used a method in which he used the reaction force from the ground on the back wheel to create a torque around the centre of the back wheel, causing the bike to rotate at a certain acceleration. I think I'll just start over that way. Thanks for your help, though!

Oh, I think I see it now: because the back wheel is interacting with the body of the bike through the chain, an equal and opposite force to the force of acceleration on the back wheel is applied to the centre of mass. Essentially, Fa is in the opposite direction as in the picture. So, I can set...

I don't see how it would do that, though. I envision something like this: Where Fg is the force of gravity and Fa is the force from the acceleration. So in order for the wheel to lift, there would have to be a torque acting counterclockwise from the pivot of the back wheel, right? But...

Homework Statement The distance between the centers of the wheels of a motorcycle is 155 cm. The center of mass of the motorcycle, including the biker, is 88.0 cm above the ground and halfway between the wheels. Assume the mass of each wheel is small compared with the body of the motorcycle...