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K^2
K^2 is offline
#2
Dec16-12, 05:14 AM
Sci Advisor
P: 2,470
Quote Quote by peanutaxis View Post
I figure that the weight of the bicycle is probably acting through the center of the wheel because the COM is at the same height (would this change if the COM was in a different place?). And because the wheel is not moving laterally I figure the friction force is opposing the weight, but acting at ground level.
First of all, force from the frame is acting on center, right. That's property of the axis, so it will not change even if CoM is higher/lower.

You are also correct that the friction will be equal to weight's projection along the incline. No problem there.

What you are forgetting about is Newton's 3rd Law for the brake. If brake is applying force on wheel, wheel is applying force back on the brake. That force would accelerate the bike backwards if it weren't attached to the wheel. So rather than just bike's weight, front wheel experiences bike's weight plus the reaction force of the brake through it's center.

Now you can balance the tangential forces. Keep in mind that you also have normal forces on the wheel, and that normal force on the two wheels together adds up to component of weight perpendicular to the incline.