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

Work through these on your own. Draw each body on its own first. Let the two bodies share the same size of acceleration, and use one symbol for the tension. Take $g = 9.8$ m s $^{-2}$ unless a question says otherwise.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Masses of $3$ kg and $5$ kg hang from the ends of a light inextensible string that passes over a smooth pulley. The system is released from rest. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
For the $3$ kg and $5$ kg masses of the previous question, find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .
Masses of $4$ kg and $6$ kg hang over a smooth pulley on a light inextensible string and are released. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
Masses of $4$ kg and $6$ kg hang over a smooth pulley on a light inextensible string and are released. Find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $4$ 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 $1$ kg. The system is released from rest. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
For the $4$ kg block on the table and the $1$ kg hanging mass, find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $5$ kg on a smooth table is joined over a smooth pulley to a hanging mass of $2$ kg. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
For the $5$ kg block on the table and the $2$ kg hanging mass, find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $8$ kg on a smooth table is joined over a smooth pulley to a hanging mass of $2$ kg. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
For the $8$ kg block on the table and the $2$ kg hanging mass, find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $m$ kg on a smooth table is joined over a smooth pulley to a hanging mass of $2$ kg. The system is released and the acceleration is $1.96$ m s $^{-2}$ . Find $m$ . Take $g = 9.8$ m s $^{-2}$ .
Masses of $2$ kg and $3$ kg hang over a smooth pulley on a light inextensible string and are released. Find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .

Answers and concise solutions

1. The acceleration

Answer: The driving force is the difference in the weights, $(5 - 3) \times 9.8 = 19.6$ N, and the mass moved is $3 + 5 = 8$ kg. So $a = \dfrac{19.6}{8} = 2.45$ m s $^{-2}$ . The $5$ kg mass moves down and the $3$ kg mass moves up at the same rate.

2. The tension

Answer: For the $3$ kg mass, moving up: $T - 3g = 3a$ . With $a = 2.45$ m s $^{-2}$ , $T = 3(9.8 + 2.45) = 36.75$ N. The same tension acts on the $5$ kg mass, because the string is light and the pulley is smooth.

3. The acceleration

Answer: The driving force is $(6 - 4) \times 9.8 = 19.6$ N and the mass moved is $10$ kg, so $a = \dfrac{19.6}{10} = 1.96$ m s $^{-2}$ .

4. The tension

Answer: For the $4$ kg mass moving up, $T - 4g = 4a$ with $a = 1.96$ m s $^{-2}$ . So $T = 4(9.8 + 1.96) = 47.04$ N. Check with the $6$ kg mass: $6g - T = 6 \times 1.96 = 11.76$ , and $58.8 - 47.04 = 11.76$ , which agrees.

5. The acceleration

Answer: The hanging weight, $1 \times 9.8 = 9.8$ N, drives the whole system. The mass moved is $4 + 1 = 5$ kg, so $a = \dfrac{9.8}{5} = 1.96$ m s $^{-2}$ . The table is smooth, so nothing opposes the block.

6. The tension

Answer: For the block on the table, the only horizontal force is the tension, so $T = 4 \times 1.96 = 7.84$ N. Check with the hanging mass: $9.8 - T = 1 \times 1.96$ , which also gives $T = 7.84$ N.

7. The acceleration

Answer: The hanging weight is $2 \times 9.8 = 19.6$ N and the mass moved is $7$ kg, so $a = \dfrac{19.6}{7} = 2.8$ m s $^{-2}$ .

8. The tension

Answer: The block on the table has only the tension along the table, so $T = 5 \times 2.8 = 14$ N. Check with the hanging mass: $19.6 - 14 = 5.6 = 2 \times 2.8$ , which agrees.

9. The acceleration

Answer: The hanging weight is $19.6$ N and the mass moved is $10$ kg, so $a = \dfrac{19.6}{10} = 1.96$ m s $^{-2}$ .

10. The tension

Answer: The block on the table has only the tension along the table, so $T = 8 \times 1.96 = 15.68$ N. The tension is larger than in the $4$ kg case because a larger mass needs a larger force for the same acceleration.

11. The mass

Answer: The hanging weight drives the total mass: $19.6 = (m + 2) \times 1.96$ . Divide by $1.96$ : $m + 2 = 10$ , so $m = 8$ kg.

12. The tension

Answer: First find the acceleration: $a = \dfrac{(3 - 2) \times 9.8}{5} = 1.96$ m s $^{-2}$ . Then for the $2$ kg mass moving up, $T - 2g = 2a$ , so $T = 2(9.8 + 1.96) = 23.52$ N.