At a building site, a truck tips sand onto the ground. The sand forms a cone. The workers pour more sand, and the cone grows taller and wider, but its sides keep the same slope. Grains poured on top roll down until the slope is back to that same angle. A heap of dry gravel keeps its own angle too, steeper than the sand.
Weight on a slope
Place a block of mass \(m\) on a plank tilted at an angle \(\theta\). Split the weight \(mg\) into two components. One lies along the slope, \(mg\sin\theta\), pointing down the slope. The other is at right angles to the slope, \(mg\cos\theta\), pressing the block into the plank. Nothing moves the block off the plank or into it, so the normal reaction balances the second component.
\[ N = mg\cos\theta \]
The first component tries to slide the block down. Friction acts along the slope and resists it.
At rest: friction balances the pull down the slope
If the block stays at rest, the forces along the slope balance. Static friction points up the slope and equals \(mg\sin\theta\). It must stay within its limit:
\[ mg\sin\theta \le \mu_s\, mg\cos\theta \]
Tilt the plank a little more and \(mg\sin\theta\) grows while the limit \(\mu_s mg\cos\theta\) shrinks. At some angle the two become equal, and the block is on the point of sliding.
The angle of repose
The largest angle at which a body can rest on a slope without sliding is the angle of repose. At this angle static friction has reached its limit, and the block is on the point of sliding.
\[ mg\sin\theta_r = f_s \]
Why. on the point of sliding the block is still at rest, so friction balances the pull down the slope.
\[ f_s = \mu_s N = \mu_s\, mg\cos\theta_r \]
Why. on the point of sliding, static friction has reached its limit, and \(N = mg\cos\theta\) on a slope.
\[ mg\sin\theta_r = \mu_s\, mg\cos\theta_r \]
Why. put the two expressions for friction equal.
\[ \tan\theta_r = \mu_s \]
Read it. divide both sides by \(mg\cos\theta_r\). The mass cancels.
The mass cancels. A heavy block and a light block of the same material start to slide at the same angle. The angle of repose depends only on the two surfaces. This gives a simple way to measure \(\mu_s\). Tilt a plank slowly until a block just starts to slide, read the angle, and take its tangent.
Now return to the sand heap. Each grain on the surface of the cone is a small body resting on a slope of other grains. If the slope is steeper than the angle of repose of sand on sand, the surface grains slide down. If the slope is gentler, grains poured on top stay where they land and steepen it. So the heap settles at exactly the angle of repose, whatever its size. Rough, angular gravel has a larger \(\mu_s\), so its heap is steeper.
A block just starts to slide when a plank is tilted to the angle \(\theta\) with \(\tan\theta = 0.75\). What is the angle of repose, in degrees?
A \(1\,\text{kg}\) wooden block starts to slide down a plank at \(22°\). A \(3\,\text{kg}\) block of the same wood is placed on the same plank. At what angle does it start to slide?
Sliding down
Beyond the angle of repose, the block slides down. Friction now is kinetic, \(\mu_k mg\cos\theta\), and it still points up the slope, against the motion. The net force down the slope is \(mg\sin\theta - \mu_k mg\cos\theta\), so the acceleration is
\[ a = g(\sin\theta - \mu_k\cos\theta) \]
Once again the mass cancels. The acceleration is smaller than \(g\sin\theta\), the value on a smooth slope.
Pushed up the slope
Now push the block up the slope. As it moves up, it slides over the plank in the upward direction, so kinetic friction points down the slope. Friction and \(mg\sin\theta\) now point the same way, and both resist the push. To move the block up at a steady speed, the push must be
\[ F = mg\sin\theta + \mu_k mg\cos\theta \]
This is the key idea of the whole lesson: friction has no fixed direction on a slope. It points against the sliding, up the slope for a block moving down and down the slope for a block moving up.
A \(2\,\text{kg}\) block lies on a plank inclined at \(37°\), with \(\mu_s = 0.8\) and \(\mu_k = 0.5\). Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). (a) Does the block stay at rest? If so, find the friction. (b) The plank is made slippery with oil so that \(\mu_s = 0.6\) and \(\mu_k = 0.5\), and the block is given a small tap. Find its acceleration. (c) On the oiled plank, what push along the slope moves the block up at a steady speed?
\(mg\sin 37° = 20 \times 0.6 = 12\,\text{N}\) down the slope. \(N = mg\cos 37° = 20 \times 0.8 = 16\,\text{N}\).
(a) Limit \(\mu_s N = 0.8 \times 16 = 12.8\,\text{N}\). (b) Sliding down: \(a = (mg\sin\theta - \mu_k N)/m\). (c) Moving up steadily: \(F = mg\sin\theta + \mu_k N\).
(a) \(12\,\text{N} < 12.8\,\text{N}\), so the block stays at rest, with friction \(12\,\text{N}\) up the slope. (b) \(\tan 37° = 0.75 > 0.6\), so the block slides once tapped: \(a = (12 - 0.5 \times 16)/2 = 2\,\text{m/s}^2\) down the slope. (c) \(F = 12 + 8 = 20\,\text{N}\) up the slope.
On the dry plank the block rests, and its friction of \(12\,\text{N}\) is only \(0.8\,\text{N}\) below the limit. Oiled, it slides down at \(2\,\text{m/s}^2\), and a push of \(20\,\text{N}\) is needed to move it up steadily. Pushing it up needs \(12 + 8\) newtons, while letting it slide down at a steady speed needs a push of only \(12 - 8 = 4\,\text{N}\) up the slope.
A \(5\,\text{kg}\) box rests on a ramp inclined at \(30°\), with \(\mu_s = 0.7\). Take \(g = 10\,\text{m/s}^2\). What is the friction on the box?
A block slides down a slope inclined at \(37°\) with \(\mu_k = 0.25\). Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). Find its acceleration down the slope, in \(\text{m/s}^2\).
A worker pushes a sack up a rough ramp into a truck. The sack moves up the ramp. Which way does friction on the sack point?
The angle of repose of a block on an incline depends on which of these?
Engineers cutting a road into a Himalayan hillside cannot leave the cut face steeper than the angle of repose of the loose rock and soil. A steeper face sheds stones onto the road, and in the monsoon, water lowers \(\mu_s\) and the angle of repose, so whole slopes slide. This is why hill roads have walls, nets and gentle, stepped cuttings. The same idea sets the slope of the sides of a canal, a railway embankment and a heap of coal at a power station.
- Writing \(N = mg\) on a slope.On a slope \(N = mg\cos\theta\). Only the component of the weight at right angles to the slope presses the surfaces together.
- Always drawing friction up the slope.Friction opposes the sliding. For a body moving up the slope, friction points down the slope.
- Writing friction as \(\mu_s mg\cos\theta\) for a body at rest.At rest, friction equals \(mg\sin\theta\). It equals \(\mu_s mg\cos\theta\) only at the angle of repose.