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 prevent the wheel from rolling down the slope is given by \(F = Mg \sin \theta\).

Step by step solution

01

Identify Forces

\nFirst, identify the forces acting on the wheel. These are:\n\n1. The gravitational force \(Mg\) which acts vertically downward at the center of the wheel.\n2. The normal force \(N\) which is the force exerted by the inclined plane on the wheel, acting perpendicularly to the plane.\n3. The frictional force \(f\) which prevents the wheel from sliding down the slope. It acts along the incline but upwards.\n4. The applied horizontal force \(F\), which also prevents the wheel from rolling down the slope. It acts horizontally on the highest point of the wheel.
02

Resolve Forces

Next, resolve the forces into components along the slope (parallel to the slope) and perpendicular to the slope. \nThe gravitational force \(Mg\) can be resolved into two components. The component \(Mg \cos \theta\) acts perpendicularly to the slope (counterbalancing the normal force) and the component \(Mg \sin \theta\) acts along the slope (counterbalanced by friction and the horizontal force \(F\)).
03

Use Newton's Second Law

The next step is the application of Newton's second law of motion in the horizontal and vertical directions. As the wheel is in equilibrium (not in motion), the resultant forces in both directions must be equal to zero. \nTherefore we obtain the equations: \n- \(F - Mg \sin \theta = 0 \) (along the slope) \n- \(N - Mg \cos \theta = 0\) (Perpendicular to the slope)
04

Solve for Applied Force

Finally, solving the first equation for \(F\) gives the horizontal force necessary to prevent the wheel from rolling down the slope, which is \(F = Mg \sin \theta\).

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