Put a bag of rice on the kitchen scale at home. The scale reads \(5\,\text{kg}\). Now press down on the bag with one finger, and the reading rises. Tie a string to the bag and lift gently without raising the bag, and the reading falls. The rice has not changed. What has changed is the push of the scale, and that push is the normal force.
A surface pushes only as hard as it must
A table is not a spring with a fixed push. The surface of the table deforms very slightly under a load, and it pushes back just hard enough to stop the load from sinking in. So the normal force has no formula of its own. The normal force is found, every time, from the equilibrium of the body. Only when the weight is the one force pressing the body into a level surface does the normal force equal \(mg\).
The normal force on a body is perpendicular to the surface of contact and has whatever size keeps the body from moving into the surface. Its value comes from the equilibrium equation across the surface. It can never be negative, because a surface can push but cannot pull.
Four cases, one method
Take a block of weight \(20\,\text{N}\) and look across the surface each time. On a level table, nothing else acts vertically, so \(N = 20\,\text{N}\). Press down with \(P = 10\,\text{N}\), and the table must balance both, so \(N = W + P = 30\,\text{N}\). Pull up with \(F = 8\,\text{N}\), and the string takes part of the weight, so \(N = W - F = 12\,\text{N}\). Put the block on a smooth \(37°\) incline, held by a string along the slope, and only \(W\cos 37°\) presses into the surface, so \(N = 16\,\text{N}\).
When the upward pull reaches the weight, the normal force reaches zero, and the block is just about to leave the surface. If the pull grows further, the formula \(N = W - F\) would give a negative number. A negative normal force is impossible: the surface cannot pull the block down. Instead, the block lifts off and is no longer in equilibrium.
A lift at rest
Stand on a bathroom scale inside a lift that is at rest. Two forces act on you: your weight down and the push of the scale up. Nothing else acts, so the push of the scale equals your weight, and the scale reads your true mass. A lift at rest is just a room. The reading changes only when the lift speeds up or slows down, and that case needs the second law, in the next chapter.
A box of weight \(W\) rests on a level floor. In which situations is the normal force on the box less than \(W\)?
A block of weight \(20\,\text{N}\) rests on a table. A string pulls it straight up with \(25\,\text{N}\). What is the normal force on the block?
Choose the axis that holds one unknown
The fastest route to the normal force is an axis along which only one force is unknown. On a level table that axis is the vertical. On an incline held by a string along the slope, it is the line across the slope. When a push is horizontal on an incline, the vertical axis often works best. The push then drops out, because the push has no vertical component, and the normal force is the only unknown left.
Block A of mass \(5\,\text{kg}\) rests on a table. A light string runs from the top of A up over a light pulley on the ceiling. It then runs down to block B of mass \(3\,\text{kg}\). B hangs at rest. Take \(g = 10\,\text{m/s}^2\). Find the normal force of the table on A.
Two bodies. On B: \(W_B = 30\,\text{N}\) down and \(T\) up. On A: \(W_A = 50\,\text{N}\) down, \(T\) up and \(N\) up. The pulley is light, so the tension is the same on both sides.
For B: \(T - 30 = 0\). For A, vertically: \(T + N - 50 = 0\).
\(T = 30\,\text{N}\). Then \(N = 50 - 30 = 20\,\text{N}\).
The table pushes with only \(20\,\text{N}\), although A weighs \(50\,\text{N}\). The string, through the pulley, carries the other \(30\,\text{N}\). If B had a mass of more than \(5\,\text{kg}\), N would become zero and A would lift off.
A block of mass \(4\,\text{kg}\) is held at rest on a smooth incline of \(37°\) by a horizontal push \(F\). Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find \(F\) and the normal force \(N\).
The body is the block. The forces are \(W = 40\,\text{N}\) down, \(F\) horizontal towards the incline, and \(N\) perpendicular to the slope. The normal force leans away from the vertical by \(37°\), the angle of the incline.
Vertical: \(N\cos 37° - 40 = 0\). Horizontal: \(F - N\sin 37° = 0\).
\(N = 40 / 0.8 = 50\,\text{N}\). Then \(F = 50 \times 0.6 = 30\,\text{N}\).
The normal force is \(50\,\text{N}\), larger than the weight of \(40\,\text{N}\). The horizontal push presses the block into the slope as well as holding it up the slope. Compare the string along the slope in Lesson 2, where \(N = mg\cos\theta\) was smaller than the weight.
A carton of mass \(20\,\text{kg}\) sits on a platform scale. A worker presses straight down on the carton with a force of \(50\,\text{N}\). Take \(g = 10\,\text{m/s}^2\). What normal force does the scale exert on the carton, in newtons?
The same block is held at rest on the same smooth incline in two ways. First a string along the slope holds it, then a horizontal push. In which case is the normal force larger?
A student says, "The normal force is the reaction to the weight, so it always equals mg." Explain the mistake.
The weight and the normal force act on the same body, so they are not an action-reaction pair. The reaction to the weight is the pull of the body on the Earth. The normal force is a separate contact force whose size is fixed by equilibrium across the surface. It equals \(mg\) only when the surface is level and no other force acts across it. Pressing, pulling or tilting changes it.
A bathroom scale measures the normal force of its platform on your feet, then divides by \(g\). Lean with one hand on the washbasin, and the basin carries part of your weight: the reading drops. Press down on the basin instead, and the basin pushes you down in return: the reading rises. Only when you stand still with nothing else touching you does the scale read your true weight.
- Writing \(N = mg\) in every problem.Find \(N\) from the equilibrium equation across the surface, every time. List every force with a component across the surface.
- Getting a negative normal force and keeping it.A negative \(N\) means the body has left the surface. Set \(N = 0\); the body is no longer in equilibrium.
- Calling the normal force the reaction to the weight.Both act on the same body, so they cannot be a third-law pair. The reaction to the weight acts on the Earth.