The drawing shows a bicycle wheel resting against a small step whose height is h = 0.125 m. The weight and radius of the wheel are W = 23.1 N and r = 0.330 m, respectively. A horizontal force vector F is applied to the axle of the wheel. As the magnitude of vector 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?

N

Distance axle above step edge = .33 - .125 = .205

Angle T is angle between straight down from Axle and the top edge of the step
so
cos T = .205/.33
T = 10.58 deg
sin T = .184

Take moments about the top edge of the step.
F * .205 clockwise (my step is on the right)

m g * .33 sin T counterclockwise
distance of wheel axle above step
but m g =23.1 N
so
23.1 * .33 * .184 = .205 F
F = 6.83 N

Thank you so much for your time and help!

To solve this problem, we can use the principle of torques. The torque exerted by a force about a pivot point is given by the formula:

Torque = Force * Lever arm

In this case, the pivot point is the point of contact between the wheel and the ground, and the force is the weight of the wheel. Since the wheel is just about to lift off the ground, the torque exerted by the weight is equal and opposite to the torque exerted by the applied force.

We can calculate the torque exerted by the weight using the formula:

Torque_weight = Weight * Lever arm_weight

The weight of the wheel acts at its center, so the lever arm will be the radius of the wheel, r.

Thus, Torque_weight = Weight * r

The torque exerted by the applied force, Torque_force, is given by:

Torque_force = Force * Lever arm

In this case, the lever arm will be the distance between the point of contact between the wheel and the ground and the axle of the wheel. Let's call this distance 'd'.

To solve for the magnitude of the force when the wheel just begins to rise up and loses contact with the ground, we set Torque_weight equal to Torque_force and solve for Force.

Torque_weight = Torque_force

Weight * r = Force * d

Substituting the known values:

23.1 N * 0.330 m = Force * d

Simplifying:

7.623 N*m = Force * d

Since d is not given in the problem, we cannot determine the exact magnitude of the force without additional information.

To determine the magnitude of the force when the wheel just begins to rise up and loses contact with the ground, we need to consider the balance of forces acting on the wheel.

When the wheel is resting on the ground, there are two main forces acting on it: the weight (W) acting downward and the normal force (N) exerted by the ground upward. These two forces are balanced, resulting in a net force of zero and keeping the wheel stationary.

When the force vector F is applied horizontally to the axle, it creates a torque that tends to rotate the wheel in the clockwise direction. This torque must be balanced by an equal and opposite torque to keep the wheel stationary.

The torque created by the applied force F can be calculated as the product of the applied force and the perpendicular distance between the point of application and the axis of rotation. In this case, the axis of rotation is at the point of contact between the wheel and the ground.

The perpendicular distance between the point of application of the force and the axis of rotation is equal to the radius of the wheel (r).

Therefore, the torque created by the applied force F is given by:

τ = F * r

To keep the wheel stationary, the torque created by the applied force F must be balanced by an equal and opposite torque created by the weight W. The torque due to weight can be calculated as the product of the weight and the perpendicular distance between the center of mass of the wheel and the axis of rotation. In this case, the center of mass of the wheel is at the center of the wheel.

The perpendicular distance between the center of mass of the wheel and the axis of rotation is also equal to the radius of the wheel (r).

Therefore, the torque due to weight is given by:

τ = W * r

For the wheel to just begin to rise up and lose contact with the ground, the torque created by the applied force F must be equal to the torque due to weight.

Therefore, we have the equation:

F * r = W * r

Simplifying and solving for the magnitude of force F:

F = W

Substituting the given value for the weight W:

F = 23.1 N

Therefore, the magnitude of the force when the wheel just begins to rise up and loses contact with the ground is 23.1 N.