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Sliding down a ramp

Forces and motion · level 6, lesson 4 of 6

A question to start

A child on Mars rides a sledge down a 25° dune. The same child, same sledge and same slope on Earth. Which reaches the bottom faster, and by how much?

The idea

Take the direction down the slope as positive. Once the object slides, two forces act along the slope:

  • the part of the weight down the slope, \(mg\sin\theta\);
  • kinetic friction up the slope, \(\mu_k N = \mu_k mg\cos\theta\).

The normal force and the part of the weight into the slope cancel. So the second law gives

\[ma = mg\sin\theta - \mu_k mg\cos\theta \qquad\Rightarrow\qquad a = g(\sin\theta - \mu_k\cos\theta).\]

The mass cancels. And the whole answer is proportional to \(g\): on Mars, anything sliding down a slope accelerates at 0.38 times its Earth rate, for the same slope and surfaces.

mg sin θfriction μk mg cos θa = g(sin θ − μk cos θ)
A sledge on a 25° slope, \(\mu_k = 0.15\), 30 m long Mars Earth
Acceleration \(a\) 1.06 m/s² 2.81 m/s²
Time to the bottom, from rest 7.5 s 4.6 s
Speed at the bottom 8.0 m/s 13.0 m/s

The time comes from \(s = \tfrac{1}{2}at^2\), so \(t = \sqrt{2s/a}\), and the speed from \(v = at\).

If \(\sin\theta < \mu_k\cos\theta\), the formula gives a negative \(a\): friction wins, and a sliding object slows down and stops.

Worked examples

1. The Mars sledge. \(\sin 25° = 0.423\), \(\cos 25° = 0.906\). \(a = 3.71 \times (0.423 - 0.15 \times 0.906) = 3.71 \times 0.287 = 1.06\) m/s². \(t = \sqrt{2 \times 30 / 1.06} = 7.5\) s. \(v = 1.06 \times 7.5 = 8.0\) m/s.

2. A crate slid down a ramp. \(\theta = 15°\), \(\mu_k = 0.35\). \(\sin 15° - 0.35\cos 15° = 0.259 - 0.338 < 0\). It slows down and stops: the ramp is too gentle for it to keep sliding.

Try it

  1. Find the acceleration down a 30° frictionless slope on Mars.
  2. Find it on a 30° slope with \(\mu_k = 0.2\) on Mars.
  3. On that slope, how long does it take to slide 20 m from rest?
  4. Why does the mass not matter?
Answers
  1. \(3.71 \times 0.5 = 1.86\) m/s².
  2. \(3.71 \times (0.5 - 0.2 \times 0.866) = 3.71 \times 0.327 = 1.21\) m/s².
  3. \(t = \sqrt{40 / 1.21} = 5.7\) s.
  4. Both the pull down the slope and the friction are proportional to \(m\), and so is the inertia.

Common mistake

Using \(N = mg\) on a slope. On a slope, \(N = mg\cos\theta\). Using \(mg\) makes the friction too big.

Where it leads