At a railway station a traveller pulls a suitcase by its long handle. The handle points upward at an angle, not along the floor. Another traveller pushes a luggage trolley with the arms sloping downward. Both bodies speed up along the platform. Yet a porter will tell you that the pulled suitcase feels lighter than the pushed trolley, even when the two loads are the same.
Two equations instead of one
In the last two chapters the forces on a body balanced, and the body stayed at rest or moved steadily. This chapter is about bodies whose forces do not balance. The net force then gives the body an acceleration, and Newton's second law says how much.
\[ \sum \vec F = m\vec a \]
You met this law in Forces and Newton's Laws as a vector law. A vector equation in a plane is two ordinary equations, one for each axis. Choose two perpendicular axes \(x\) and \(y\), and split every force into its components along them.
\[ \sum F_x = m a_x, \qquad \sum F_y = m a_y \]
The acceleration first, then the forces
The two equations hold for any pair of axes, but one choice makes them easy. Ask first where the body will accelerate. A box pulled along the floor can only move along the floor. A block on a slope can only move along the slope. Put the \(x\) axis along that direction. Then \(a_y = 0\), and the \(y\) equation says that the forces across the motion balance.
Before writing any equation, decide from the situation the direction in which the body can accelerate. Put the \(x\) axis along that direction and the \(y\) axis across it. Along \(y\) the forces balance, \(\sum F_y = 0\), and this equation usually gives the normal force. Along \(x\), \(\sum F_x = m a\) gives the acceleration or an unknown force.
The quantity \(ma\) is not a force. Never draw an arrow labelled \(ma\) among the forces on a body. The forces are pushes and pulls by other bodies; \(ma\) is what their sum produces.
A crate is dragged along a level floor by a rope. Which choice of axes makes the second law easiest to use?
A pull at an angle
A block of mass \(m\) rests on a smooth, level floor. A force \(F\) pulls it at an angle \(\theta\) above the horizontal. The block cannot leave the floor or sink into it, so its acceleration is along the floor. Take \(x\) along the floor and \(y\) upward. The force has two components: \(F\cos\theta\) along the floor and \(F\sin\theta\) upward.
\[ y:\; N + F\sin\theta - mg = 0 \quad\Rightarrow\quad N = mg - F\sin\theta \] \[ x:\; F\cos\theta = ma \quad\Rightarrow\quad a = \dfrac{F\cos\theta}{m} \]
The upward part of the pull carries some of the weight. So the floor has to push up less, and the normal force is smaller than \(mg\).
A push at an angle
Now let the same force push the block at an angle \(\theta\) below the horizontal. The component along the floor is still \(F\cos\theta\). The other component now points downward and presses the block into the floor.
\[ N = mg + F\sin\theta, \qquad a = \dfrac{F\cos\theta}{m} \]
On a smooth floor the acceleration is the same in both cases. The difference is only in the normal force. This difference will matter a great deal in the chapters on friction, because friction grows with the normal force.
This pays off the promise of the railway station. The pulled suitcase presses on the floor with less than its weight. The pushed trolley presses with more than its weight. On a real floor the smaller normal force means less friction, and that is why a pull at a small upward angle feels easier.
A porter pushes a \(30\,\text{kg}\) luggage trolley along a smooth platform. The push is \(150\,\text{N}\), directed \(37°\) below the horizontal. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). Find the acceleration of the trolley and the normal force of the platform.
Forces on the trolley: weight \(300\,\text{N}\) down, normal force \(N\) up, push \(150\,\text{N}\) at \(37°\) below the horizontal. The trolley accelerates along the platform, so \(x\) is along the platform and \(y\) is upward.
\(y: N - 300 - 150 \times 0.6 = 0\). \(x: 150 \times 0.8 = 30a\).
\(N = 300 + 90 = 390\,\text{N}\). \(a = 120 / 30 = 4\,\text{m/s}^2\).
The trolley speeds up at \(4\,\text{m/s}^2\) and presses on the platform with \(390\,\text{N}\), more than its weight of \(300\,\text{N}\).
A \(4\,\text{kg}\) block on a smooth floor is pulled with a force of \(20\,\text{N}\) directed \(37°\) above the horizontal. Take \(g = 10\,\text{m/s}^2\) and \(\sin 37° = 0.6\). Find the normal force in newtons.
The same force pulls a block on a smooth floor, first horizontally and then at \(60°\) above the horizontal. How does the acceleration compare?
A block on a smooth incline
An incline is a plane surface tilted at an angle \(\theta\) to the horizontal. Let a block of mass \(m\) rest on a smooth surface tilted this way, and let it go. Two forces act on the block: its weight \(mg\), straight down, and the normal force \(N\), perpendicular to the surface. The block can only move along the slope, so choose \(x\) down the slope and \(y\) perpendicular to it.
The normal force already lies along \(y\). Only the weight must be split. The angle between the weight and the \(y\) axis equals the angle of the incline, \(\theta\). So the weight has a part \(mg\sin\theta\) down the slope and a part \(mg\cos\theta\) into the slope.
\[ y:\; N - mg\cos\theta = 0, \qquad x:\; mg\sin\theta = ma \]
The mass has cancelled from the acceleration. A heavy block and a light block on the same smooth incline slide down together. The check at the two ends is useful. At \(\theta = 0\) the floor is level and \(a = 0\). At \(\theta = 90°\) the block falls freely and \(a = g\).
A \(1\,\text{kg}\) block and a \(5\,\text{kg}\) block are released together from the same height on two identical smooth inclines. Which reaches the foot first?
Galileo used exactly this cancellation. A ball on a gentle slope has the acceleration of free fall made smaller by the factor \(\sin\theta\). The motion is slow enough to time, and it keeps the property that mass does not matter.
A \(5\,\text{kg}\) box is pulled up a smooth ramp of angle \(30°\) by a rope parallel to the ramp. The tension is \(40\,\text{N}\). Take \(g = 10\,\text{m/s}^2\). Find the acceleration of the box and the normal force.
The box moves along the ramp, so take \(x\) up the slope. The forces along \(x\) are the tension \(40\,\text{N}\) up the slope and \(mg\sin 30° = 25\,\text{N}\) down it.
\(\sum F_x = 40 - 25 = 15\,\text{N}\), so \(a = 15 / 5 = 3\,\text{m/s}^2\) up the slope.
Across the slope nothing moves, so \(N = mg\cos 30° = 50 \times 0.866 = 43.3\,\text{N}\).
A block slides down a smooth incline of angle \(37°\). Take \(g = 10\,\text{m/s}^2\) and \(\sin 37° = 0.6\). Find its acceleration in m/s².
A block rests on a smooth incline that is slowly made steeper. What happens to the normal force?
A student writes \(mg\cos\theta\) for the component of the weight along the slope. How can the student check the choice without drawing anything?
Test the two ends. At \(\theta = 0\) the incline is level, and the weight can have no part along it. \(mg\sin 0 = 0\) passes this test; \(mg\cos 0 = mg\) fails it. At \(\theta = 90°\) the whole weight acts along the vertical face, and \(mg\sin 90° = mg\) passes. So the component along the slope is \(mg\sin\theta\).
- Writing \(N = mg\) in every problem.\(N\) comes from the \(y\) equation. A pull at an angle, a push at an angle or a slope all change it: \(mg - F\sin\theta\), \(mg + F\sin\theta\), \(mg\cos\theta\).
- Drawing \(ma\) as one of the forces.Draw only pushes and pulls by other bodies. \(ma\) belongs on the other side of the equation.
- Swapping sine and cosine on an incline.Check at \(\theta = 0\): the part along a level floor must vanish, so it is \(mg\sin\theta\).