Motion on a Rough Inclined Plane
Resolve weight along and perpendicular to a slope, include friction in the correct direction, and find the acceleration or the angle at which sliding starts.
What a learner can do afterwards
- Identify the forces on a block, resolve weight along and perpendicular to a fixed slope, and find the normal reaction and friction direction
- Apply Newton's second law to uphill or downhill sliding on a rough fixed plane to find acceleration or the kinetic coefficient, and explain why mass cancels
- Find actual static friction, decide whether rest is possible, and derive or calculate the limiting slope angle using the static coefficient of friction
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Let theta be a fixed plane's angle above the horizontal, between 0 and 90 degrees. Weight W = mg acts vertically downward. Resolve it as mg sin theta downhill and mg cos theta into the plane. These components replace the weight arrow in calculations; they are not extra forces. With only weight and contact forces, perpendicular acceleration is zero, so the outward normal reaction is R = mg cos theta.
For a block at rest, static friction supplies the force needed for balance, up to mu_s R. At limiting equilibrium, f_s = mu_s R. During sliding, use f_k = mu_k R. The symbols mu_s and mu_k are the static and kinetic coefficients for the contacting surfaces. Friction opposes sliding relative to the fixed plane, or the tendency to slide when at rest. Identify that motion or tendency before drawing friction.
For downhill sliding with only weight and contact forces, choose downhill as positive: ma = mg sin theta - mu_k mg cos theta. Thus a = g(sin theta - mu_k cos theta), and mu_k = (g sin theta - a)/(g cos theta). Mass cancels, so changing mass alone leaves a unchanged. At constant speed a = 0, so friction balances mg sin theta. For uphill sliding, both along-plane forces act downhill: a = g(sin theta + mu_k cos theta).
With only weight and contact forces, rest requires mg sin theta <= mu_s mg cos theta. The left side is the friction needed; the right side is its limit. Since cos theta > 0, rest requires tan theta <= mu_s. At impending downhill slip, equality gives mu_s = tan theta = sin theta / cos theta. This is the limiting slope angle, also called the angle of repose. Equality describes limiting equilibrium, not motion already underway.
For a block acted on only by weight and contact forces, the limiting angle is theta = arctan(mu_s), between 0 and 90 degrees. If mu_s = 0.2, degree mode gives about 11.3 degrees. Below the limiting angle the block can remain at rest. Above it, static friction cannot hold it. Once it slides, use mu_k to find acceleration.
Resolve weight along and across the fixed plane. Use mu_k during sliding and the static limit mu_s R to find the limiting slope angle.
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The readings, extra questions and flashcards were written for this dataset. The linked videos are by Khan Academy.
More to read
Choosing the friction model
A block at rest does not automatically experience the largest possible friction force. First find the force needed to balance everything acting along the plane. Static friction can supply that force only if its magnitude is no greater than mu_s R. Equality describes limiting equilibrium. If the required force exceeds the limit, the block cannot stay at rest.
For a block already sliding, use the kinetic model f_k = mu_k R. Its friction direction is opposite the sliding relative to the plane. On a fixed plane, downhill sliding gives uphill friction, while uphill sliding gives downhill friction. The coefficients describe the contacting surfaces. The static coefficient is usually larger than the kinetic coefficient, but an actual static friction force can be smaller than either friction limit or a sliding friction force.
Forces and components on a plane
For a block on a rough fixed plane with no rope or extra applied force, list three physical forces: weight, normal reaction and friction. Weight points vertically downward. The reaction is perpendicular to the contact surface. Friction acts along that surface.
Let theta be the plane's angle above the horizontal. In calculations, replace weight by its two components: mg sin theta downhill and mg cos theta into the plane. Do not count those components as two additional forces while also counting weight.
A block that remains in contact has no acceleration perpendicular to this fixed plane. The normal reaction therefore balances mg cos theta when no other force has a perpendicular component. An angled rope or push would change the balance. Check the forces before using R = mg cos theta.
Sliding and the limiting angle
Take downhill as positive on a fixed plane. During downhill sliding, weight contributes mg sin theta and friction contributes -mu_k mg cos theta. Divide their sum by mass to obtain a = g(sin theta - mu_k cos theta). Mass cancels from both terms. This means that changing mass alone does not change the acceleration when the plane and coefficient stay the same.
For uphill sliding, the downhill weight component and friction point the same way. Add their magnitudes to obtain the downhill acceleration g(sin theta + mu_k cos theta).
The start of sliding uses the static coefficient instead. With only weight and contact forces, a resting block needs mg sin theta <= mu_s mg cos theta. At the limiting angle, equality gives tan theta = mu_s, so theta = arctan(mu_s). A block already sliding downhill at constant speed instead satisfies tan theta = mu_k. The similar equations describe different friction states.
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