16.3 Weight and motion under gravity — Learn

Prerequisites: Newton's second law, $F = ma$ (Section 16.2); rearranging a formula.

By the end of this section you can: explain the difference between mass and weight; use $W = mg$; describe free fall and terminal velocity; and find the normal reaction on a body in a lift that accelerates.

1. Mass and weight

Mass is the amount of matter in a body. It is measured in kilograms and it does not change when the body moves. Mass is a scalar.

Weight is the force of gravity on a body. It is measured in newtons, and it acts vertically downwards. Weight is a vector.

The link between them is

where $m$ is the mass in kilograms and $g$ is the acceleration due to gravity. Near the Earth's surface, $g = 9.8$ m s$^{-2}$. The value of $g$ is smaller on the Moon and on other planets, so a body's weight changes from place to place while its mass stays the same.

A body of mass $1$ kg has a weight of $9.8$ N on the Earth. That is why a kilogram of sugar feels heavy but has no weight at all in deep space.

2. Motion under gravity

When the only force on a falling body is its weight, the body is in free fall. The resultant force is $mg$, so by $F = ma$:

So every freely falling body accelerates at $g$ downwards, whatever its mass.

In real motion, air resistance $R$ acts upwards and opposes the fall. Newton's second law along the vertical then gives

As the body speeds up, the air resistance grows. When $R$ reaches $mg$, the resultant force is zero and the acceleration is zero. The body now falls at a constant velocity called the terminal velocity.

Example 1. Find the weight of a crate of mass $45$ kg. Take $g = 9.8$ m s$^{-2}$.

Use $W = mg$:

The weight acts downwards. The mass is still $45$ kg.

Example 2. A body has a weight of $588$ N. Find its mass. Take $g = 9.8$ m s$^{-2}$.

Rearrange $W = mg$ to make the mass the subject:

Example 3. A skydiver of mass $60$ kg falls with an air resistance of $200$ N. Find the acceleration. Take $g = 9.8$ m s$^{-2}$.

Take downwards as positive. The weight is

The resultant force downwards is

Then

The air resistance makes the acceleration less than $g$.

Example 4. The same skydiver continues to fall. Explain what happens to the acceleration and to the velocity.

As the skydiver speeds up, the air resistance increases. The resultant force $588 - R$ therefore decreases, so the acceleration decreases. When $R$ reaches $588$ N the resultant force is zero and the acceleration is zero. The skydiver then falls at a constant terminal velocity.

3. Apparent weight in a lift

A person standing in a lift feels the normal reaction $R$ from the floor, not the weight. That reaction is the apparent weight.

Take upwards as positive. The resultant force is $R - mg$, so

• If the lift accelerates upwards, $R$ is larger than $mg$, and the person feels heavier.
• If the lift accelerates downwards, $a$ is negative, so $R$ is smaller than $mg$, and the person feels lighter.
• If the lift moves at a constant velocity, $a = 0$ and $R = mg$.

Example 5. A person of mass $70$ kg stands in a lift that accelerates upwards at $1.5$ m s$^{-2}$. Find the normal reaction from the floor. Take $g = 9.8$ m s$^{-2}$.

Use $R = m(g + a)$:

The reaction is $791$ N, which is larger than the weight of $686$ N. The person feels heavier during the upward acceleration.

Watch out. Three mistakes are common.

Calling weight a mass in kilograms. Weight is a force in newtons. In Example 1 the crate has mass $45$ kg and weight $441$ N. Writing "$45$ kg" where a force is wanted breaks every later step.

Assuming $R = mg$ in a lift. That is true only when the acceleration is zero. While the lift accelerates, use $R = m(g + a)$ with the sign of $a$. Re-check: if the lift accelerates downwards, the person should feel lighter, so $R$ must come out less than $mg$.

Using $g$ far from the Earth. The value $9.8$ m s$^{-2}$ belongs to the Earth's surface. A question set on the Moon gives its own value of $g$, and the mass stays the same.

You should now be able to

• say what mass is and what weight is, and give the unit of each;
• use $W = mg$ in both directions;
• describe free fall and explain why all bodies fall with the same acceleration when air resistance is negligible;
• explain terminal velocity;
• find the normal reaction on a body in a lift that accelerates up, accelerates down, or moves at a constant velocity.