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Chapter 12: Problem 47

What horizontal force applied at its highest point is necessary to keep a wheel of mass \(M\) from rolling down a slope inclined at angle \(\theta\) to the horizontal?

Short Answer

Expert verified
The horizontal force necessary to keep the wheel from rolling down a slope is \( F = Mg \sin(\theta) \).

Step by step solution

01

Setup the force diagram

First, draw the force diagram of the wheel. include: (1) The weight of the wheel, \( Mg \), acting vertically downward from the center. (2) The horizontal force \( F \) acting from the highest point of the wheel. (3) The frictional force \( f \), acting from its bottom point, opposite to the direction of potential motion. (4) The Normal force \( N \), acting perpendicular to the surface from the bottom point of the wheel. The wheel also has a radius \( r \).
02

Resolve Forces

As it is an inclined plane, resolve the weight into two components: one parallel to the slope, \( Mg \sin(\theta) \), acting downwards along the slope, and one perpendicular to the slope, \( Mg \cos(\theta) \), acting into the slope.
03

Apply Equilibrium Conditions

Since the wheel is not moving, all the forces and moments should be balanced. It means that the system of the wheel is in equilibrium. So, use equilibrium conditions for the wheel, which are that the sum of vertical forces is 0, the sum of horizontal forces is 0, and the sum of moments is 0. From the conditions sum of vertical forces equals 0, we get: \( N = Mg \cos(\theta) \). The condition of sum of horizontal forces equals to 0 gives: \( F = f + Mg \sin(\theta) \).
04

Determine the Torque

According to the nature of the problem, the moment due to the horizontal force \( F \) at the highest point of the wheel is also considered. Only the forces \(\ F\) and \(\ f \) will cause a moment about the center of the wheel. As the wheel is not rotating, the sum of moments about the center of the wheel equals to 0. The moment equation (torques balanced) will be: \( rF = rf \), hence \( F=f \).
05

Substitute f back into the equation

Substitute \( f \) from equation in step 4 into equation in step 3, we get \( F=F+Mg \sin(\theta) \). Therefore, the horizontal force necessary to keep the wheel from rolling down a slope is \( F = Mg \sin(\theta) \).

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