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Chapter 9: Problem 29

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

Expert verified
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

01

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.
02

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.
03

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 \(F = \frac{W*(r-h)}{r}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) Show that a solid cylinder of mass \(M\) and radius \(R\) is equivalent to a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\), for rotation about a central axis. ( \(b\) ) The radial distance from a given axis at which the mass of a body could be concentrated without altering the rotational inertia of the body about that axis is called the radius of gyration. Let \(k\) represent the radius of gyration and show that $$ k=\sqrt{I / M} $$ This gives the radius of the "equivalent hoop" in the general case.

Two identical blocks, each of mass \(M\), are connected by a light string over a frictionless pulley of radius \(R\) and rotational inertia \(I\) (Fig. 9-55). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and the sliding block. When this system is released, it is found that the pulley turns through an angle \(\theta\) in time \(t\) and the acceleration of the blocks is constant. (a) What is the angular acceleration of the pulley? ( \(b\) ) What is the acceleration of the two blocks? ( \(c\) ) What are the tensions in the upper and lower sections of the string? All answers are to be expressed in terms of \(M, I, R, \theta, g\), and \(t\).

What is the torque about the origin on a particle located at \(x=1.5 \mathrm{~m}, \quad y=-2.0 \mathrm{~m}, z=1.6 \mathrm{~m}\) due to a force \(\overrightarrow{\mathbf{F}}=\) \((3.5 \mathrm{~N}) \hat{\mathbf{i}}-(2.4 \mathrm{~N}) \hat{\mathbf{j}}+(4.3 \mathrm{~N}) \hat{\mathbf{k}} ?\) Express your result in unit vector notation.

Three particles are attached to a thin rod of length \(1.00 \mathrm{~m}\) and negligible mass that pivots about the origin in the \(x y\) plane. Particle 1 (mass \(52 \mathrm{~g}\) ) is attached a distance of \(27 \mathrm{~cm}\) from the origin, particle \(2(35 \mathrm{~g})\) is at \(45 \mathrm{~cm}\), and particle \(3(24 \mathrm{~g})\) at \(65 \mathrm{~cm} .(a)\) What is the rotational inertia of the assembly? (b) If the rod were instead pivoted about the center of mass of the assembly, what would be the rotational inertia?

A small lead sphere of mass \(25 \mathrm{~g}\) is attached to the origin by a thin rod of length \(74 \mathrm{~cm}\) and negligible mass. The rod pivots about the \(z\) axis in the \(x y\) plane. A constant force of \(22 \mathrm{~N}\) in the \(y\) direction acts on the sphere. (a) Considering the sphere to be a particle, what is the rotational inertia about the origin? (b) If the rod makes an angle of \(40^{\circ}\) with the positive \(x\) axis, find the angular acceleration of the rod.
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