An auto-rickshaw driver keeps a steel tiffin box on the flat shelf in front of the steering handle. When the auto starts from a signal, the box stays with the shelf. On a smooth road at steady speed, it stays too. When the driver brakes hard, the box slides forward and hits the windscreen. Something held the box in the first two cases and failed in the third.
Friction opposes relative motion, not motion
In Part 1 you met friction as the force that opposes sliding. The word that matters is relative. Friction on a body opposes the motion of that body relative to the surface it touches. It does not oppose the motion of the body relative to the ground. The tiffin box moves forward with the auto all the time. Yet the friction on it may point forward, backward or nowhere.
When the two surfaces do not slide, there is still a direction in which they would slide if there were no friction. This is the tendency of relative motion. Static friction points opposite to this tendency. To find the tendency, imagine the contact made perfectly smooth and ask what would happen next.
The friction on a body points opposite to the motion of the body relative to the surface in contact. When there is no sliding, it points opposite to the tendency of relative motion, the slip that would happen if the surface were smooth.
The box on a speeding trolley
Put a box on a flat trolley and let the trolley speed up to the right. Imagine the deck smooth. The deck would move forward under the box, and the box would stay behind. Relative to the deck, the box would slip backward. So static friction on the box points forward. That forward friction is the only horizontal force on the box, and it is what makes the box accelerate with the trolley.
Read the three chips of Fig 1.1 against each other. The velocity is the same in all three, so the velocity cannot decide the direction of friction. The acceleration decides it. The box must have the same acceleration as the trolley, and friction is the only force that can give it one. So the friction on the box is \(ma\), in the direction of the acceleration. At steady speed \(a = 0\), and no friction is needed at all.
This also settles the tiffin box. When the auto brakes, the box needs a backward friction \(ma\). If the braking is hard, \(ma\) is larger than the limiting friction \(\mu_s mg\). The shelf cannot supply it, and the box slides forward relative to the shelf, towards the windscreen.
A box rests on the floor of a delivery van. The coefficient of static friction between them is \(0.4\). Take \(g = 10\,\text{m/s}^2\). What is the largest acceleration of the van, in m/s², for which the box does not slip?
A truck carries a crate on its flat bed and is braking while moving forward. The crate does not slip. Which way does friction from the bed push the crate?
A block held against a wall
A harder case: press a block flat against a vertical wall with a horizontal push \(F\). Nothing pushes the block up or down except gravity. If the wall were smooth, the block would slide down. So friction from the wall points up. The normal reaction from the wall balances the push, \(N = F\), and friction balances the weight, \(f = mg\).
The push does not hold the block up directly. The push is horizontal and gravity is vertical. The push only sets the normal force, and the normal force sets the most friction the wall can give. The block stays if \(mg \le \mu_s F\). So the least push that holds it is \(F = mg/\mu_s\).
The \(2\,\text{kg}\) block of Fig 1.2 has \(\mu_s = 0.5\) with the wall. Take \(g = 10\,\text{m/s}^2\). What is the least horizontal push, in newtons, that keeps it from sliding down?
When the tendency is not obvious: assume, then test
Sometimes two forces compete and you cannot see the tendency at a glance. A block on a rough incline pulled up the slope is the standard case. The pull tends to drag the block up, and gravity tends to drag it down. Which wins? You do not need to guess correctly. Assume a direction for friction, write \(\Sigma F = 0\), and solve for \(f\).
The sign of the answer settles the direction. A positive \(f\) means your assumed direction is right. A negative \(f\) means friction points the other way, with the same size. One more test remains. Static friction can hold the body only if its size does not exceed the limit \(\mu_s N\). If \(|f| > \mu_s N\), the body is not at rest after all. It slides, and the friction on it becomes kinetic friction, \(\mu_k N\), opposite to the sliding.
To find static friction on a body believed to be at rest, take four steps. (1) Draw friction in any direction along the surface. (2) Write the equilibrium equations and solve for \(f\). (3) If \(f < 0\), reverse the arrow. (4) Check \(|f| \le \mu_s N\). If the check fails, the body moves and kinetic friction \(\mu_k N\) opposes the sliding.
A \(10\,\text{kg}\) crate rests on a ramp at \(37°\) to the horizontal. The coefficient of static friction is \(0.5\). A rope pulls it up the ramp with \(70\,\text{N}\), parallel to the ramp. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). Find the friction on the crate.
Along the ramp: the pull \(70\,\text{N}\) up, the weight component \(mg\sin 37° = 60\,\text{N}\) down, and friction \(f\), assumed up the ramp. Perpendicular: \(N = mg\cos 37° = 80\,\text{N}\).
At rest: \(70 + f - 60 = 0\).
\(f = -10\,\text{N}\). The minus sign says friction points down the ramp, with size \(10\,\text{N}\). The limit is \(\mu_s N = 0.5 \times 80 = 40\,\text{N}\), and \(10 \le 40\).
The crate stays at rest. Friction is \(10\,\text{N}\) down the ramp. The pull is \(10\,\text{N}\) more than the weight component, so the crate tends to move up, and friction opposes that tendency.
The same crate is now pulled up the ramp with \(110\,\text{N}\). Solving \(110 + f - 60 = 0\) gives \(f = -50\,\text{N}\). What happens?
A student says: "Friction always acts opposite to the velocity of a body." Give one example from this lesson where this is false. Then state the correct rule.
A box on a trolley that is speeding up moves forward, and the friction on it also points forward. The rule is that friction opposes the motion of the body relative to the surface it touches, or the tendency of that relative motion. The box tends to slip backward relative to the deck, so friction on it points forward.
The words "opposite to the motion" are the most common mistake about friction. Always ask two questions: relative to which surface, and would the body slip that way if the contact were smooth? The ground is almost never the right surface.
- Pointing friction opposite to the velocity of the body.Point it opposite to the slip relative to the surface in contact. On a speeding trolley, the box moves forward and friction on it points forward.
- Taking \(f = \mu_s N\) for a body at rest.Static friction is only as large as needed. Solve for it from the equilibrium equations. \(\mu_s N\) is its limit, reached only when the body is about to slip.
- Stopping at a negative answer for \(f\).Reverse the arrow, then compare the size with \(\mu_s N\). If the size is larger, the body is not at rest.