Splitting weight on a slope

Mathematics · Mechanics · ages 17-18

Name ______________________ Date ____________

A block weighing 60 N remains at rest on a fixed rough slope with sin theta = 0.6 and cos theta = 0.8. Its static coefficient is mu_s = 1. Only weight and contact forces act. What is the actual static friction force, in N?
Answer: ______________
A block weighs 50 N. Its slope makes angle theta with the horizontal, and sin theta = 0.6. What is the downhill component of its weight, in N?
Answer: ______________
A 5 kg block rests on a fixed slope with cos theta = 0.8. Only weight and contact forces act. Take g = 10 m/s2. What is the normal reaction?
- 40 N
- 30 N
- 50 N
A block slides downhill relative to a fixed rough plane. Which way does friction on the block act?
- Down the slope, with the sliding
- Up the slope, opposite the sliding
- Perpendicular to the slope, outward
A 4 kg block is sliding uphill on a fixed rough slope with sin theta = 0.6 and cos theta = 0.8. Its kinetic coefficient is mu_k = 0.5. Only weight and contact forces act. Take g = 10 m/s2. Find the magnitude of its acceleration in m/s2.
Answer: ______________
A resting block has static coefficient mu_s = 0.5 on a fixed plane. Only weight and contact forces act. At what limiting slope angle is it on the point of slipping downhill? Give the angle to one decimal place.
- 30.0 degrees
- 60.0 degrees
- 26.6 degrees
A 2 kg block slides downhill on a fixed slope with sin theta = 0.6 and cos theta = 0.8. Its kinetic coefficient is mu_k = 0.25. Only weight and contact forces act. Take g = 10 m/s2. Find its downhill acceleration in m/s2.
Answer: ______________
A 4 kg block slides downhill on a fixed slope with sin theta = 0.8 and cos theta = 0.6. Only weight and contact forces act. Take g = 10 m/s2. Its measured downhill acceleration is 5 m/s2. Find its kinetic coefficient mu_k.
Answer: ______________
A resting block is on the point of slipping downhill on a fixed slope with sin theta = 0.6 and cos theta = 0.8. Only weight and contact forces act. Find its static coefficient mu_s as a decimal.
Answer: ______________
Two blocks of masses m and 2m slide downhill on the same fixed rough plane. They have the same mu_k, and only weight and contact forces act. What follows about their accelerations while both are sliding?
- They have equal accelerations because mass cancels from F = ma.
- The heavier block has twice the acceleration because its weight is doubled.
- The heavier block has half the acceleration because its inertia is doubled.
- The heavier block has no acceleration because its friction is doubled.

Answer key

For grown-ups. Fold this page away before handing over the rest.

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36 · Rest needs 60 × 0.6 = 36 N of uphill friction. The 48 N static limit is a bound, not the actual force.
30 · The downhill weight component is W sin theta = 50 × 0.6 = 30 N.
40 N · Perpendicular forces balance, so R = mg cos theta = 5 × 10 × 0.8 = 40 N.
Up the slope, opposite the sliding · Kinetic friction opposes sliding relative to the plane. Here the block slides downhill, so friction acts uphill.
10 · Weight contributes 24 N downhill and friction contributes 16 N downhill. Their total is 40 N, giving acceleration 40/4 = 10 m/s2 downhill.
26.6 degrees · At the limit, tan theta = 0.5. In degree mode, arctan(0.5) is about 26.565 degrees, which rounds to 26.6 degrees.
4 · Weight contributes 12 N downhill and friction contributes 4 N uphill. The net force is 8 N, giving a = 8/2 = 4 m/s2 downhill.
0.5 · The downhill pull is 32 N and the net force is 20 N. Friction is 12 N and R = 24 N, so mu_k = 12/24 = 0.5.
0.75 · At impending downhill slip, mu_s = sin theta / cos theta = 0.6/0.8 = 0.75.
They have equal accelerations because mass cancels from F = ma. · Both the downhill weight component and kinetic friction scale with mass. Dividing the net force by mass gives the same acceleration for both blocks.