Question

What minimum force \(F\) applied horizontally at the axle of the wheel in Fig. \(9-49\) is necessary to raise the wheel over an obstacle of height \(h\) ? Take \(r\) as the radius of the wheel and \(W\) as its weight.

Short answer

The minimum force \(F\) required to lift the wheel over the obstacle of height \(h\) must be \(F = \frac{W*(r-h)}{r}\), where \(W\) is the weight of the wheel and \(r\) is the radius of the wheel.

Step-by-step solution

Understand the problem

A force \(F\) is to be applied at the axle of a wheel to lift it over an obstacle of height \(h\). The wheel has a weight \(W\) and a radius \(r\). The force must overcome the gravitational pull of the weight and hence create a torque sufficient to lift the wheel over the obstacle.

Apply the principle of Torque

Torque is the rotational equivalent of linear force. The equilibrium condition for rotations is that total torque about any point must be zero. The pivotal point is the edge of the wheel touching the obstacle. As the direction of the torque due to \(W\) (weight) and \(F\) (force applied) is opposite, we can set them equal to each other.

Calculate the Force

According to the principle of lever arm, Torque = Force * Distance. For \(W\), lever arm is \(r-h\). For \(F\), lever arm is \(r\). Hence we have, Torque due to \(W\) = Torque due to \(F\), which simplifies to \(W*(r-h) = F*r\). Solving it for \(F\), we get minimum force \(F = \frac{W*(r-h)}{r}\)

Key concepts

Torque and Lever Arm

Torque can be thought of as the rotational equivalent of a linear force and plays a critical role in problems involving rotational motion. When a force is applied at a certain distance from the pivot point, this distance is known as the lever arm. The torque produced by a force is calculated by multiplying the force by the length of the lever arm, following the equation \( Torque = Force \times Lever \, Arm \).

In the case of a wheel trying to overcome an obstacle, the force applied at the axle generates a torque around the edge of the wheel that is in contact with the obstacle. The lever arm in this scenario is the radius of the wheel, and the generated torque must be strong enough to oppose and overcome the torque due to the wheel's weight, which acts downward through the wheel's center.

Equilibrium Condition for Rotations

The equilibrium condition for rotations is fundamental in solving many physics problems involving torque. It states that for an object to be in rotational equilibrium, the net torque acting on the object must be zero. This means the sum of all clockwise torques about any pivot point must be equal to the sum of all counterclockwise torques.

In the exercise, the wheel is on the verge of rotating over the obstacle, meaning the sum of torques around the contact point must balance out. The weight of the wheel creates a clockwise torque, and the force applied at the axle attempts to generate a counterclockwise torque. When these two are equal, the wheel is perfectly balanced on the obstacle and can be moved over it with the application of the minimum necessary force.

Calculating Force in Rotational Dynamics

In rotational dynamics, the force required to produce a certain motion can be deduced by understanding the relationship between torque and lever arms. From the equation \( Torque = Force \times Lever \, Arm \) we can solve for the force if the torque and lever arm are known. In this context, 'minimum force' is the least amount of force needed to generate a torque that matches the torque due to the object's weight trying to rotate in the opposite direction.

To calculate the force needed to raise the wheel over an obstacle, as shown in the exercise, we can rearrange the equation to \( F = \frac{Torque}{Lever \, Arm} \). By substituting the torque created by the weight of the wheel and the effective lever arms, we can find the minimum force. This process illustrates the practical application of torque and lever arm concepts to solve real-world problems involving rotational motion.