Two students move a heavy steel trunk across a hostel floor with a rope. One pulls the rope flat along the floor. The other lifts the rope and pulls at an angle upward, and the trunk moves more easily. Lifting the rope too high makes it hard again. Somewhere between flat and steep there is a best angle.
Two forces from one surface are one force
A surface pushes on a block in two ways at once. The normal force \(N\) pushes straight out of the surface. Friction \(f\) acts along the surface. Both come from the same contact, so we may add them into a single contact force \(R\). Its size is \(R = \sqrt{N^2 + f^2}\). Its direction leans away from the normal, towards the friction side, by an angle whose tangent is \(f/N\).
As the pull on a block grows, static friction grows and the contact force leans further back. The lean stops growing when friction reaches its limit \(\mu_s N\). The largest lean is the angle of friction \(\lambda\), with \(\tan\lambda = \mu_s N / N = \mu_s\). The contact force can lean less than \(\lambda\), but never more, while the block is at rest.
The angle of friction \(\lambda\) is the angle between the normal force and the resultant contact force when friction is limiting. It satisfies \(\tan\lambda = \mu_s\).
The angle of friction is the angle of repose
In Part 1 you tilted an incline until a block just began to slide. That angle was the angle of repose, with \(\tan\theta = \mu_s\). The two angles are the same angle. On the incline at the angle of repose, only two forces act on the block: its weight and the contact force. So the contact force must be vertical. The normal is tilted by \(\theta\) from the vertical, so the contact force leans from the normal by exactly \(\theta\), and friction is limiting.
A block rests on a floor with \(\mu_s = 1/\sqrt{3}\). What is the angle of friction?
Why a raised rope helps
Pull a block with a rope at an angle \(\theta\) above the horizontal. The pull has two jobs now. Its horizontal part \(P\cos\theta\) fights friction. Its vertical part \(P\sin\theta\) lifts a little of the weight. That lift lowers the normal force to \(N = mg - P\sin\theta\), and a smaller normal force means a smaller limiting friction. So raising the rope loses some horizontal pull but also weakens the friction it must beat.
\[ N = mg - P\sin\theta \]
Why. vertically the block does not move, so \(N + P\sin\theta = mg\).
\[ P\cos\theta = \mu_s\,(mg - P\sin\theta) \]
Why. just about to slide, the horizontal part of the pull equals limiting friction \(\mu_s N\).
\[ P = \dfrac{\mu_s mg}{\cos\theta + \mu_s\sin\theta} \]
Why. collect the terms in \(P\) on one side.
\[ P = \dfrac{\tan\lambda\, mg}{\cos\theta + \tan\lambda\sin\theta} = \dfrac{mg\sin\lambda}{\cos\lambda\cos\theta + \sin\lambda\sin\theta} \]
Why. write \(\mu_s = \tan\lambda\) and multiply the top and the bottom by \(\cos\lambda\).
\[ P = \dfrac{mg\sin\lambda}{\cos(\theta - \lambda)} \]
Read it. the bottom is the expansion of \(\cos(\theta - \lambda)\).
The last line of the derivation answers the question at once. The top, \(mg\sin\lambda\), does not depend on \(\theta\). The bottom, \(\cos(\theta - \lambda)\), is largest, equal to one, when \(\theta = \lambda\). So the pull is least when the rope makes the angle of friction with the floor.
Look at the shape of the graph in Fig 3.2. It is flat near the bottom. At \(20°\) the pull is \(62.7\,\text{N}\), and at \(60°\) it is \(65.2\,\text{N}\). So the exact angle matters less than raising the rope at all. A pull a little off the best angle costs only a few newtons. A pull far off, flat or steep, costs much more.
A crate of mass \(20\,\text{kg}\) stands on a floor with \(\mu_s = 0.75\). Take \(g = 10\,\text{m/s}^2\). At what angle should a rope pull to start it with the least force, and what is that force? Compare it with a horizontal pull.
\(\tan\lambda = 0.75\), so \(\lambda = 37°\), with \(\sin\lambda = 0.6\) and \(\cos\lambda = 0.8\). The weight is \(200\,\text{N}\).
Best angle \(\theta = \lambda\). \(P_{\min} = mg\sin\lambda\). Horizontal pull: \(P_0 = \mu_s mg\).
\(P_{\min} = 200 \times 0.6 = 120\,\text{N}\). \(P_0 = 0.75 \times 200 = 150\,\text{N}\).
Pull at \(37°\) with \(120\,\text{N}\). That is one fifth less than the \(150\,\text{N}\) needed with a flat rope.
A \(4\,\text{kg}\) box rests on a floor with \(\mu_s = 1/\sqrt{3}\). Take \(g = 10\,\text{m/s}^2\). What is the least pull, in newtons, that can start it moving?
A lawn roller can be pulled by its handle at \(37°\) above the ground, or pushed by the same handle at \(37°\) below the horizontal. Which is easier, and why?
Why does the needed pull grow again when the rope is raised far above the angle of friction?
A steep rope puts most of its force into lifting and little into dragging. The horizontal part \(P\cos\theta\) becomes small, so a large \(P\) is needed to match even a reduced friction. The best balance between lifting and dragging is at \(\theta = \lambda\). Beyond it, the loss in horizontal pull outweighs the gain from a smaller normal force.
In 1699 Guillaume Amontons reported to the Paris Academy that friction is proportional to the load pressing two surfaces together. He also said it does not depend on the area of contact. Many members doubted the second claim, since it seemed to contradict common sense. Eighty years later Charles-Augustin de Coulomb tested the laws with careful experiments on wood and metals. He confirmed them, and he separated the friction needed to start sliding from the smaller friction during sliding. His memoir won the Academy's prize in 1781. The coefficient \(\mu\), and with it the angle of friction, comes from their work.
A steel trunk of mass \(80\,\text{kg}\) stands on a floor with \(\mu_s = 0.4\). Estimate the least pull, in newtons, with a rope at the best angle. Take \(g = 10\,\text{m/s}^2\).
- Taking \(\tan\lambda = 1/\mu_s\).The angle of friction is measured from the normal, and \(\tan\lambda = f/N = \mu_s\).
- Using \(N = mg\) when the pull is at an angle.A pull at \(\theta\) above the horizontal lifts part of the weight, so \(N = mg - P\sin\theta\). A push at \(\theta\) below the horizontal gives \(N = mg + P\sin\theta\).
- Believing the steepest rope is the best rope.The best angle is the angle of friction. Beyond it, too little of the pull drags.