16.4 Newton's third law, equilibrium and connected particles — Practise with help

Work each one first, closed book. Draw each body on its own before you write any equation. Give the two bodies the same size of acceleration, and use one symbol for the tension. Take $g = 9.8$ m s $^{-2}$ .

Warm-up: remember

Solve the pair of equations $2x + y = 11$ and $3x - y = 4$ together.
Work out $5 \times 9.8$ .
A resultant force of $20$ N acts on a mass of $5$ kg. Find the acceleration.

Answers

Adding gives $5x = 15$ , so $x = 3$ . Then $y = 11 - 6 = 5$ .
$5 \times 9.8 = 49$ .
$a = \dfrac{20}{5} = 4$ m s $^{-2}$ .

If question 1 was hard, revise solving simultaneous equations in Section 2.5 before you continue.

Practise with help

1. Masses of $2$ kg and $4$ kg hang over a smooth pulley on a light inextensible string and are released. Find the acceleration and the tension.

Hint

The heavier mass moves down and the lighter mass moves up. The difference in the weights drives both masses.

Next step

The driving force is $(4 - 2) \times 9.8 = 19.6$ N and the mass moved is $2 + 4 = 6$ kg. Then use $T - 2g = 2a$ for the lighter mass.

Full working

a = \frac{(4 - 2) \times 9.8}{2 + 4} = \frac{19.6}{6} = \frac{49}{15} \approx 3.27 \text{ m s}^{-2}.

For the $2$ kg mass, moving up:

T = 2(9.8 + 3.27) = 26.13 \text{ N}.

2. A block of mass $3$ kg rests on a smooth horizontal table. It is joined by a light inextensible string over a smooth pulley at the table edge to a hanging mass of $2$ kg. Find the acceleration and the tension.

Hint

Only the hanging weight drives the system, and the total mass moved is the sum of the two masses.

Next step

The hanging weight is $2 \times 9.8 = 19.6$ N and the total mass is $5$ kg. Then use $T = 3a$ for the block on the table.

Full working

a = \frac{2 \times 9.8}{3 + 2} = \frac{19.6}{5} = 3.92 \text{ m s}^{-2}.

T = 3a = 3 \times 3.92 = 11.76 \text{ N}.

3. A book rests on a table. A student says the weight of the book and the normal reaction on the book are a Newton's third law pair. Explain why the student is wrong.

Hint

A third-law pair acts on two different bodies. Ask which body each force acts on here.

Next step

Both the weight and the normal reaction act on the book. Find the real pair: the force the book exerts on the table, and the force the table exerts on the book.

Full working

The weight and the normal reaction both act on the book, so they are forces on one body, not a third-law pair. They happen to be equal in size because the book is in equilibrium.

The third-law pair is the contact force between the book and the table. The book pushes down on the table, and the table pushes up on the book with the same size of force.

4. Masses of $6$ kg and $4$ kg hang over a smooth pulley on a light inextensible string and are released. Find the acceleration and the tension.

Hint

This is the same pattern as question 1, with different masses.

Next step

The driving force is $(6 - 4) \times 9.8 = 19.6$ N and the mass moved is $10$ kg. Then use the lighter mass for the tension.

Full working

a = \frac{(6 - 4) \times 9.8}{6 + 4} = \frac{19.6}{10} = 1.96 \text{ m s}^{-2}.

For the $4$ kg mass, moving up:

T = 4(9.8 + 1.96) = 47.04 \text{ N}.

5. A block of mass $5$ kg rests on a smooth table and is joined over a smooth pulley to a hanging mass of $1$ kg. Find the acceleration and the tension.

Hint

The hanging weight is the only driving force.

Next step

The total mass is $6$ kg and the driving force is $9.8$ N. Then use $T = 5a$ .

Full working

a = \frac{1 \times 9.8}{5 + 1} = \frac{9.8}{6} = \frac{49}{30} \approx 1.63 \text{ m s}^{-2}.

T = 5a = 5 \times 1.633 \approx 8.17 \text{ N}.

6. A block on a smooth table is joined over a smooth pulley to a hanging mass of $3$ kg. The acceleration is $1.96$ m s $^{-2}$ . Find the mass of the block.

Hint

Write $F = ma$ for the whole system, with the block's mass as the unknown.

Next step

The hanging weight is $3 \times 9.8 = 29.4$ N, so $29.4 = (m + 3) \times 1.96$ .

Full working

29.4 = (m + 3) \times 1.96,

so $m + 3 = \dfrac{29.4}{1.96} = 15$ , giving $m = 12$ kg.

7. Masses of $7$ kg and $3$ kg hang over a smooth pulley on a light inextensible string and are released. Find the acceleration and the tension.

Hint

Follow the four steps: draw each mass, choose the direction, write two equations, then solve.

Next step

The driving force is $(7 - 3) \times 9.8 = 39.2$ N and the mass moved is $10$ kg. Then use the lighter mass for the tension.

Full working

a = \frac{(7 - 3) \times 9.8}{7 + 3} = \frac{39.2}{10} = 3.92 \text{ m s}^{-2}.

For the $3$ kg mass, moving up:

T = 3(9.8 + 3.92) = 41.16 \text{ N}.

8. A particle is in equilibrium under three forces. Explain what this says about the forces, and what a triangle of forces looks like.

Hint

Equilibrium means the resultant force is zero.

Next step

If the three force arrows are placed head to tail, their result is the zero vector. What shape does that make?

Full working

In equilibrium the resultant force is zero. Placing the three arrows head to tail returns to the starting point, so the arrows form a closed triangle. Each force is balanced by the sum of the other two.