/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 The drawing shows a bicycle whee... [FREE SOLUTION] | 91Ó°ÊÓ

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The drawing shows a bicycle wheel resting against a small step whose height is \(h=0.120 \mathrm{~m}\). The weight and radius of the wheel are \(W=25.0 \mathrm{~N}\) and \(r=0.340 \mathrm{~m}\). A horizontal force \(\overrightarrow{\mathbf{F}}\) is applied to the axle of the wheel. As the magnitude of \(\overrightarrow{\mathbf{F}}\) 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?

Short Answer

Expert verified
The force magnitude is 67.2 N.

Step by step solution

01

Identify the critical point

The critical point occurs when the wheel starts to rise over the step and loses contact with the ground. At this moment, the wheel's contact point with the step becomes the pivot for rotation.
02

Use the geometry of the problem

Since the wheel's radius is \(r = 0.340 \text{ m}\) and the height of the step is \(h = 0.120 \text{ m}\), the geometry of the contact point tells us that the horizontal distance from the wheel's center to the edge of the step is \(\sqrt{r^2 - h^2}\).
03

Center of mass gravitational torque

The gravitational force \(W = 25.0 \text{ N}\) acts through the center of the wheel. The torque due to this weight around the pivot is \(W \times (r - \sqrt{r^2 - h^2})\).
04

Torque due to the applied force

The horizontal force \(\overrightarrow{F}\) causes a clockwise torque around the same pivot, calculated as \(F \times h\), where \(h = 0.120 \text{ m}\) is the lever arm perpendicular to \(\overrightarrow{F}\).
05

Set up the equilibrium equation

For the wheel to begin rising, the torques must be balanced: \(W \times (r - \sqrt{r^2 - h^2}) = F \times h\).
06

Solve for \(F\)

Rearrange the equation to solve for \(F\): \[ F = \frac{W \times (r - \sqrt{r^2 - h^2})}{h} \]. Plug in the values: \(W = 25.0 \text{ N}\), \(r = 0.340 \text{ m}\), \(\sqrt{0.340^2 - 0.120^2} = 0.320\) and \(h = 0.120 \text{ m}\). Substitute and calculate to find \(F = 67.2 \text{ N}\).

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Key Concepts

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

Bicycle Wheel Mechanics
A bicycle wheel is essentially a circular frame with spokes, which can rotate around its central axle. Understanding this involves considering the wheel's balance, the forces it experiences, and its motion. The axle is the point where forces are usually applied to propel a bicycle forward. In our scenario, when a horizontal force is applied to the axle, it results in rotational motion due to the wheel's interaction with a step.

The wheel's contact point with the step acts as a pivot, a crucial consideration when analyzing the mechanics of the wheel moving over the step. At this pivot point, the wheel's center of mass, its radius, and the dimensions of external forces and objects, such as the step's height, all play a role in determining how it will behave when force is applied.
Rotational Motion
Rotational motion occurs when an object spins around a pivot or axis. In our example, that pivot is the contact point of the wheel and the step. It's essential to understand that rotational motion is different from linear motion. While linear motion just considers distance and speed in a straight line, rotational motion involves angular displacement, velocity, and acceleration.

One way to visualize rotational motion is to imagine the wheel rotating around the pivot point, with each point on the wheel tracing a circular path. The angle of rotation, typically measured in radians, is a core element here. Importantly, the distance from the pivot affects the rotational effect of any force applied. Larger distances (or radii) result in more substantial rotational effects. Therefore, knowing the radius of the wheel and the pivot’s location helps us determine how much force is needed to overcome resistance and initiate rotation, leading to the wheel rising over the step.
Force Calculation
To calculate the force necessary for the wheel to rise over the step, we employ the concept of torque. Torque, often defined as a measure of the force that can cause an object to rotate about an axis, is calculated by multiplying the force by the perpendicular distance to the pivot. In mathematical terms, torque (\(\tau\)) is given by \(\tau = r \times F \times \sin(\theta)\), where \(r\) is the lever arm distance, \(F\) is the force, and \(\theta\) is the angle between the force and the lever arm.

In the problem at hand:
  • The gravitational force acts at the center of the wheel, creating a counterclockwise torque about the contact point.
  • The horizontal force \(F\) applied to the axle generates a clockwise torque.
Setting these torques equal allows us to solve for \(F\). This equilibrium condition ensures the wheel is on the verge of rising, without yet having started to move upward.
Physics Problem Solving
Physics problem solving often involves breaking down complex scenarios into simpler parts. In this case, we focused on the interaction between forces, pivot points, and rotational motion to understand equilibrium conditions. Solving these problems usually involves:
  • Identifying key points, such as the pivot or points of force application.
  • Using known values and physics principles, like torque and equilibrium.
  • Organizing them into equations that can be solved mathematically.

Real-world physics problems, like our bicycle wheel example, require logical reasoning and methodical calculation. Practicing these steps sharpens problem-solving skills, preparing one for more complex scenarios.

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Most popular questions from this chapter

A wrecking ball (weight \(=4800 \mathrm{~N}\) ) is supported by a boom, which may be assumed to be uniform and has a weight of \(3600 \mathrm{~N}\). As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is \(32^{\circ}\), and the angle between the boom and the horizontal is \(48^{\circ} .\) Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by the hinge at point \(P\).

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of \(0.80 \mathrm{~m}\) and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed \(v_{0}\), so that the rod begins to rotate upward about the pivot. What must be the value of \(v_{0}\), such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?

A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is \(82.5 \mathrm{~m}\). The moment of inertia of the station without people is \(3.00 \times 10^{9} \mathrm{~kg} \cdot \mathrm{m}^{2}\). Suppose 500 people, with an average mass of \(70.0 \mathrm{~kg}\) each, live on this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station's angular speed due to the radial movement of the people?

Interactive Solution \(9.47\) at offers a model for solving problems of this type. A solid sphere is rolling on a surface. What fraction of its total kinetic energy is in the form of rotational kinetic energy about the center of mass?

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{~kg} \cdot \mathrm{m}^{2}\). From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{~m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{~kg}\). After all the sand is in place, what is the angular velocity of the disk?

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