Newton's laws of motion

4.4 Newton’s laws of motion

The three laws

Weight and mass. Mass is the amount of matter, in kg. Weight is the force of gravity on it, in newtons: $W = mg$.

$3000 - 600 = 1200a \quad\Rightarrow\quad a = 2 \text{ m s}^{-2}$

Lifts and vertical motion

Taking up as positive: $R - 700 = 70 \times 2$, so $R = 840$ N.

The person feels heavier when the lift accelerates upwards, and lighter when it accelerates downwards. This also applies when a lift going up slows down, since the acceleration is then downwards.

Inclined planes

Resolve along the slope for $F = ma$, and perpendicular to the slope for $R$, where there is no acceleration.

Perpendicular: $R = 50\cos 30^\circ = 43.3$ N. So $F = 0.2 \times 43.3 = 8.66$ N.

Along the slope: $50\sin 30^\circ - 8.66 = 5a$, so $a = 3.27$ m s⁻².

Connected particles

For two particles joined by a light inextensible string or a rod:

• They have the same size of acceleration while the string is taut.
• The tension is the same along a string over a smooth pulley.

Particles over a pulley

For the 3 kg particle, moving down: $30 - T = 3a$. For the 2 kg particle, moving up: $T - 20 = 2a$.

Adding gives $10 = 5a$, so $a = 2$ m s⁻². Then $T = 24$ N.

On the table, $R = 40$ N, so $F = 10$ N and $T - 10 = 4a$.

The hanging particle has $20 - T = 2a$.

Adding gives $10 = 6a$, so $a = \tfrac{5}{3}$ m s⁻² and $T = 16.7$ N.

Vehicles and tow bars

A car and trailer are connected particles with the same acceleration. The tow bar pulls the trailer forward and the car backward.

The whole system: $3000 - 600 = 1200a$, so $a = 2$ m s⁻².

The trailer alone: $T - 100 = 200 \times 2$, so $T = 500$ N.

After the string goes slack

If a hanging particle hits the ground, the string goes slack and the two parts move separately:

1. Find the common acceleration and the speed when it hits the ground.
2. The other particle continues with this speed, with a new acceleration: only friction on a table, or only gravity if it is hanging.
3. Find the further distance it moves, and add this if a total is asked for.

Speed at impact: $v^2 = 2 \times \tfrac{5}{3} \times 1 = \tfrac{10}{3}$.

After this, friction alone acts on the table: $-10 = 4a$, so $a = -2.5$ m s⁻².

$0 = \frac{10}{3} + 2(-2.5)s \quad\Rightarrow\quad s = \frac{2}{3} \text{ m}$