16.8 Moments in static contexts — Practise

Work through these on your own. Draw the diagram and mark every force with its distance from the pivot. Remember the weight of a uniform beam acts at its midpoint, and use the perpendicular distance for a force at an angle. Take $g = 9.8$ m s $^{-2}$ .

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

A force of $20$ N acts at right angles to a spanner at a distance of $0.5$ m from the nut. Find the moment of the force about the nut.
A force of $15$ N acts at the end of a rod of length $2$ m. The force makes an angle of $30^\circ$ with the rod. Find the moment of the force about the other end of the rod.
A child of mass $30$ kg sits $2$ m from the pivot of a seesaw. A second child of mass $60$ kg sits on the other side. How far from the pivot must the second child sit to balance the seesaw? Take $g = 9.8$ m s $^{-2}$ .
A uniform plank $AB$ has length $4.0$ m and mass $20$ kg. It rests on two supports at $A$ and $B$ . A crate of mass $45$ kg rests on the plank $1.5$ m from $A$ . Find the reaction at each support, to $1$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A uniform beam $AB$ has length $3.0$ m and mass $12$ kg. It rests on supports at $A$ and $B$ . A load of mass $20$ kg rests on the beam $2.0$ m from $A$ . Find the reaction at each support, to $1$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A uniform metre rule is pivoted at its centre, the $50$ cm mark. A mass of $0.2$ kg hangs at the $20$ cm mark. Where must a mass of $0.3$ kg hang to balance the rule?
A non-uniform rod $AB$ has length $4$ m and mass $6$ kg. It is pivoted at its centre and balances when a mass of $2$ kg is placed at $B$ . Find the distance of the centre of mass of the rod from $A$ , to $2$ d.p.
A uniform beam $AB$ has length $5$ m and mass $30$ kg. It rests on supports at $A$ and $B$ . A load of mass $50$ kg rests at the midpoint. Find the reaction at each support. Take $g = 9.8$ m s $^{-2}$ .
A uniform beam $AB$ has length $4$ m and mass $25$ kg. It rests on a support at $A$ and a support at $C$ , where $AC = 3$ m. A load of mass $40$ kg rests on the beam $2$ m from $A$ . Find the reaction at each support, to $1$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A uniform plank $AB$ has length $6$ m and mass $40$ kg. It rests on supports at $A$ and $B$ . A person of mass $80$ kg stands $2$ m from $A$ . Find the reaction at each support, to $1$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A force of $50$ N acts at the end of a rod of length $1.5$ m. The force makes an angle of $40^\circ$ with the rod. Find the moment of the force about the other end, to $2$ d.p.
A uniform rod $AB$ has mass $8$ kg and length $2.5$ m. It is hinged at $A$ and held horizontal by a vertical string at $B$ . Find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .

Answers and concise solutions

1. The moment

Answer: The moment is force times perpendicular distance: $20 \times 0.5 = 10$ N m. The force is at right angles to the spanner, so the distance given is already the perpendicular distance.

2. The moment

Answer: The perpendicular distance from the pivot to the line of the force is $2\sin 30^\circ = 1$ m. The moment is $15 \times 1 = 15$ N m. Only the component of the force at right angles to the rod makes a moment.

3. The distance

Answer: For balance, the clockwise and anticlockwise moments at the pivot are equal: $30g \times 2 = 60g \times d$ . The $g$ cancels, so $d = \dfrac{30 \times 2}{60} = 1$ m.

4. The two reactions

Answer: The plank is uniform, so its weight of $196$ N acts at its midpoint, $2$ m from $A$ . The crate weighs $441$ N and acts $1.5$ m from $A$ . Taking moments about $A$ : $4R_B = 196 \times 2 + 441 \times 1.5 = 1053.5$ , so $R_B = 263.4$ N. The total load is $637$ N, so $R_A = 637 - 263.4 = 373.6$ N. The support nearer the crate carries more.

5. The two reactions

Answer: The beam weight is $117.6$ N at $1.5$ m from $A$ , and the load is $196$ N at $2.0$ m. Taking moments about $A$ : $3R_B = 117.6 \times 1.5 + 196 \times 2 = 568.4$ , so $R_B = 189.5$ N. The total is $313.6$ N, so $R_A = 313.6 - 189.5 = 124.1$ N.

6. The balance distance

Answer: The $0.2$ kg mass is $0.30$ m from the pivot, so its moment is $0.2g \times 0.30$ . Let the $0.3$ kg mass be $d$ m from the pivot on the other side: $0.3g\,d = 0.2g \times 0.30$ . The $g$ cancels, so $d = 0.20$ m. The mass hangs at the $70$ cm mark.

7. The distance of the centre of mass from A

Answer: The $2$ kg mass at $B$ is $2$ m from the pivot, giving a moment of $2g \times 2 = 4g$ . The weight of the rod, $6g$ , must act on the other side of the pivot at distance $x$ : $6gx = 4g$ , so $x = \dfrac{2}{3}$ m. The centre of mass is $2 - \dfrac{2}{3} \approx 1.33$ m from $A$ . A non-uniform rod's centre of mass is not at its midpoint.

8. The two reactions

Answer: The load is at the midpoint, and the beam is uniform, so every weight acts at the centre. The total weight is $(30 + 50) \times 9.8 = 784$ N. By symmetry each support carries half: $392$ N.

9. The two reactions

Answer: The beam weight is $245$ N at $2$ m from $A$ and the load is $392$ N at $2$ m from $A$ . Taking moments about $A$ : $3R_C = 245 \times 2 + 392 \times 2 = 1274$ , so $R_C = 424.7$ N. The total is $637$ N, so $R_A = 637 - 424.7 = 212.3$ N.

10. The two reactions

Answer: The plank weight is $392$ N at $3$ m from $A$ ; the person weighs $784$ N at $2$ m from $A$ . Taking moments about $A$ : $6R_B = 392 \times 3 + 784 \times 2 = 2744$ , so $R_B = 457.3$ N. The total is $1176$ N, so $R_A = 1176 - 457.3 = 718.7$ N.

11. The moment

Answer: The perpendicular distance is $1.5\sin 40^\circ = 0.964$ m. The moment is $50 \times 0.964 \approx 48.21$ N m. Only the component of the force at right angles to the rod contributes.

12. The tension

Answer: The weight is $78.4$ N and acts at the midpoint, $1.25$ m from $A$ . Taking moments about $A$ : $2.5T = 78.4 \times 1.25 = 98$ , so $T = 39.2$ N. The hinge carries the rest of the load.