What Force Lifts a Bicycle Wheel Over a Step?

[leezak]

The figure shows a bicycle wheel resting against a small step whose height is h = 0.110 m. The weight and radius of the wheel are W = 29.0 N and r = 0.350 m. A horizontal force F is applied to the axle of the wheel. As the magnitude of F increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. What is the magnitude of the force when this happens?

I drew a diagram, yet I'm not even sure how to start this problem... help please! thanks!

 

[Andrew Mason]

The figure shows a bicycle wheel resting against a small step whose height is h = 0.110 m. The weight and radius of the wheel are W = 29.0 N and r = 0.350 m. A horizontal force F is applied to the axle of the wheel. As the magnitude of F increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. What is the magnitude of the force when this happens?
I drew a diagram, yet I'm not even sure how to start this problem... help please! thanks!

The step provides an axis (a line about which the wheel rotates) which divides the wheel into two parts. You have to determine how that axis divides the mass of the wheel. Then is a matter of determining the centre of mass of each part of the wheel to find the torques.

AM

 

[quicknote]

The bicycle torque problem can be solved using basic principles of mechanics and torque. To begin, we can set up the equation for torque, which is T = F * r * sin(theta), where T is torque, F is the applied force, r is the distance between the point of rotation and the point where the force is applied, and theta is the angle between the force and the lever arm.

In this case, the lever arm is the radius of the wheel (0.350 m) and the angle between the force and the lever arm is 90 degrees, since the force is being applied horizontally. Therefore, the equation becomes T = F * 0.350 m * sin(90 degrees). Since sin(90 degrees) = 1, the equation simplifies to T = 0.350 F.

Next, we need to consider the torque caused by the weight of the wheel. This torque is equal to the weight (W) multiplied by the lever arm (0.110 m). Therefore, the torque due to the weight is T = W * 0.110 m = 29.0 N * 0.110 m = 3.19 Nm.

In order for the wheel to start to rise up and lose contact with the ground, the applied torque (T) must be equal to or greater than the torque due to the weight (3.19 Nm). Therefore, we can set up the following equation:

0.350 F = 3.19 Nm

Solving for F, we get F = 3.19 Nm / 0.350 m = 9.11 N.

Therefore, the magnitude of the force when the wheel just begins to rise up and loses contact with the ground is 9.11 N.