1. Consider the torque (moment) applied to the wheel, as measured about the point where the wheel touches the curb. To make the wheel rise above the curb, the torque due to the force must exceed the torque due to the weight, which equals m g sqrt(R^2-h^2) and is in the opposite direction. There will be no force where the wheel touches the road since the wheel will be rising.
The torque due to the applied force is
F (r-h) when it is applied at the center of the wheel.
I will use m instead of x to represent the mass.
F (r-h) = m g sqrt(r^2-h^2)
sqrt(r^2-h^2) is the horizontal distance from the center of the wheel to the point where the wheel touches the curb. It is the "lever arm" for the torque due to the weight
2. Do this case similarly, but the torque due to the applied force becomes
F(2r-h). This will make the required F less.
how would you go about doing this type of problem thank you
You are trying to raise a bicycle wheel of mass x and radius r up over a curb of height h . To do this, you apply a horizontal force .
1. What is the least magnitude of the force that will succeed in raising the wheel onto
the curb when the force is applied at the center of the wheel?
2. What is the least magnitude of the force that will succeed in raising the wheel onto the curb when the force is applied at the top of the wheel?
3. In which case is less force required?
case (a)
case (b)
2 answers
It's sqrt(r^2-(r^2-h^2) not sqrt(r^2-h^2)
2 answers