At a building site, a sack of cement lies on a sloping plank. A rope ties it over a pulley at the top to a hanging bucket. Add a little sand to the bucket and nothing happens. Add more and still nothing. Only when the bucket is heavy enough does the sack start up the plank. Empty the bucket and the sack may slide down instead.
Find the tendency first
Let a block of mass \(m_1\) rest on a rough incline of angle \(\theta\). A light string runs up the slope, over a smooth pulley at the top, to a hanging mass \(m_2\). Two forces try to move the pair along the string. The hanging weight \(m_2 g\) pulls the block up the slope. The weight component \(m_1 g\sin\theta\) pulls it down. Ignore friction for a moment and compare them. The larger one sets the tendency of motion.
Friction on the block then points against the tendency. Its limit is \(\mu_s m_1 g\cos\theta\). If the difference between the two driving forces is within this limit, static friction absorbs it and nothing moves. If the difference is larger, the pair accelerates the way of the tendency, and kinetic friction \(\mu_k m_1 g\cos\theta\) acts against it.
(1) Find the tendency by comparing the driving forces with friction ignored. (2) Point friction against the tendency. (3) If the net driving force is within the limit \(\mu_s N\), the system stays at rest, and friction equals the net driving force. (4) Otherwise the system moves the way of the tendency, with acceleration equal to (net driving force \(- \mu_k N\)) divided by the total mass.
A \(5\,\text{kg}\) block rests on a \(37°\) incline with \(\mu = 0.25\), tied over a pulley to a hanging mass of \(2.5\,\text{kg}\). The weight component down the slope is \(30\,\text{N}\), and the hanging weight is \(25\,\text{N}\). What happens?
Step through the four chips of Fig 5.1. With \(m_1 = 5\,\text{kg}\), \(m_1 g\sin 37° = 30\,\text{N}\) and the friction limit is \(10\,\text{N}\). So the pair stays at rest for every hanging weight from \(20\,\text{N}\) to \(40\,\text{N}\), that is for \(m_2\) from \(2\,\text{kg}\) to \(4\,\text{kg}\). At \(3\,\text{kg}\) the two driving forces balance and friction is zero. Outside the band the pair moves, and the friction arrow points against the motion.
In Fig 5.1, the hanging mass is \(6\,\text{kg}\). Take \(\mu_s = \mu_k = 0.25\), \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). Find the acceleration and the tension in the string.
Driving forces: \(m_2 g = 60\,\text{N}\) down on the hanging side, \(m_1 g\sin 37° = 30\,\text{N}\) down the slope. Tendency: the block moves up the slope. \(N = 40\,\text{N}\), limit \(10\,\text{N}\).
Net driving force \(60 - 30 = 30\,\text{N}\), more than \(10\,\text{N}\), so the pair moves. \(a = (60 - 30 - 10)/(5 + 6)\).
\(a = 20/11 \approx 1.82\,\text{m/s}^2\). For the hanging mass: \(60 - T = 6a\), so \(T = 60 - 10.9 = 49.1\,\text{N}\).
The block slides up the incline and the bucket falls, both at \(1.82\,\text{m/s}^2\). The tension is about \(49\,\text{N}\), less than the bucket's weight, because the bucket accelerates downward.
The same system, with a \(1\,\text{kg}\) hanging mass. The block slides down the slope. What is its acceleration, in m/s²? Take \(\mu_k = 0.25\) and \(g = 10\,\text{m/s}^2\).
A \(10\,\text{kg}\) block lies on a \(37°\) incline with \(\mu_s = 0.5\), tied over a pulley at the top to a hanging mass \(m_2\). For which \(m_2\) does the block stay at rest? Take \(g = 10\,\text{m/s}^2\).
A box dropped on a moving belt
At an airport, bags are placed at rest on a belt that is already moving. At the moment a bag lands, the belt slides under it. Relative to the belt, the bag moves backward. So kinetic friction on the bag points forward, the direction of the belt. The bag gains speed at \(a = \mu_k g\). It slides on the belt until its speed equals the belt speed \(V\). After that there is no relative motion, and on a level belt at steady speed friction is zero.
Read the graph of Fig 5.2 as a relative motion. Relative to the belt, the box starts with velocity \(-V\) and has acceleration \(+\mu_k g\). So it is a body thrown backward and brought to rest by a steady deceleration. It stops, relative to the belt, after \(t = V/(\mu_k g)\), having covered \(V^2/(2\mu_k g)\) backward along the belt. That distance is the slip distance.
In Fig 5.2 the box needs \(t = 2/2 = 1\,\text{s}\). In that second the box covers \(\tfrac{1}{2} \times 2 \times 1^2 = 1\,\text{m}\), while the paint mark covers \(2 \times 1 = 2\,\text{m}\). So the box ends \(1\,\text{m}\) behind the mark, which matches \(V^2/(2\mu_k g) = 4/4 = 1\,\text{m}\). A worker who wants the bags not to smear the belt must make the slip short: a slow belt or a grippy surface.
A belt runs at \(3\,\text{m/s}\). A parcel is placed on it at rest, with \(\mu_k = 0.3\). Take \(g = 10\,\text{m/s}^2\). How far, in metres, does the parcel slide along the belt before it moves with it?
While a box slips on a belt that moves to the right, which way does friction act on the belt?
Friction and work, named for later
Friction in this chapter has pushed bodies forward as often as it has held them back. On the belt, friction speeds the box up, so it does positive work on the box. At the same time, the equal friction on the belt does negative work on the belt, over a longer distance. The difference is the heat made in the slip. The unit on Work, Energy and Power returns to these three numbers with the tools to count them.
For the \(5\,\text{kg}\) box of Fig 5.2, friction is \(\mu_k mg = 10\,\text{N}\). It acts on the box over \(1\,\text{m}\) and on the belt over \(2\,\text{m}\). So friction does \(+10\,\text{J}\) on the box, exactly the kinetic energy it gains, \(\tfrac{1}{2} \times 5 \times 2^2\). It does \(-20\,\text{J}\) on the belt, which the motor must supply. The missing \(10\,\text{J}\), friction times the slip, becomes heat.
An airport baggage belt runs at about \(0.5\,\text{m/s}\), and the rubber belt has \(\mu_k \approx 0.4\) with a suitcase. Estimate how far, in centimetres, a suitcase slides on the belt after it is put down at rest. Take \(g = 10\,\text{m/s}^2\).
The belts that carry bags up into an aircraft hold are tilted. There, friction must not only match the belt speed, it must also hold the bag against \(mg\sin\theta\) for the whole climb. The belt surface is ribbed rubber, with a high \(\mu_s\), and the angle is kept below the angle of friction of a bag on rubber. On a belt steeper than that angle, the bags would slide back down however slowly the belt moved.
- Writing the equation of motion before deciding the tendency.Compare the driving forces with friction ignored. Friction points against the tendency, and it may hold the system at rest.
- Using the kinetic formula when the system is at rest.If the net driving force is within \(\mu_s N\), the acceleration is zero and friction equals the net driving force, not \(\mu_k N\).
- Pointing friction on a belt-carried box backward.The box slips backward relative to the belt, so friction on it points forward, in the direction of the belt.