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?

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To find the magnitude of the force when the wheel just begins to rise up and loses contact with the ground, we can analyze the forces acting on the wheel.

The weight of the wheel is acting downward with a magnitude of W = 23.1 N.

There are two forces acting on the wheel at the point of contact with the ground: the normal force (N) and the friction force (f). The normal force is perpendicular to the surface and acts upward, balancing the downward force due to the weight. The friction force opposes any tendency of motion and acts parallel to the surface.

When the wheel is about to break contact with the ground, the normal force becomes zero because it cannot balance the downward force anymore. At this point, the friction force is also at its maximum value.

To analyze the forces, we can consider the torque equation about the point where the wheel contacts the ground:

τ = r * F * sin(θ)

where τ is the torque, r is the radius of the wheel, F is the applied force, and θ is the angle between the force and the line connecting the point of contact and the axle.

Since the wheel is just beginning to rise up, the point of contact is about to leave the ground, and the angle between the force and the line connecting the point of contact and the axle is 90 degrees (θ = 90 degrees).

Now, we can rewrite the torque equation:

τ = r * F

The torque due to the weight is given by:

τ_weight = r * W * sin(90)

Since sin(90) = 1, the torque due to the weight simplifies to:

τ_weight = r * W

At the point where the wheel is about to lose contact with the ground, the torque due to the applied force is equal to the torque due to the weight:

τ = τ_weight

Substituting the expressions for the torques:

r * F = r * W

The radius cancels out, and we are left with:

F = W

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