16.8 Moments in static contexts — Practise with help

Work each one first, closed book. Draw the beam or rod and mark every force with its distance from the pivot. Take moments about a point that removes an unknown force. Take $g = 9.8$ m s $^{-2}$ .

Warm-up: remember

Work out $196 \times 2$ .
Solve $4x = 1053.5$ .
A uniform rod has length $4.0$ m. How far from one end is its centre of mass?

Answers

$196 \times 2 = 392$ .
$x = \dfrac{1053.5}{4} = 263.375$ .
The centre of mass is at the midpoint, $2.0$ m from either end.

If question 3 was hard, remember that a uniform rod balances at its midpoint.

Practise with help

1. A force of $25$ N acts at right angles to a lever at a distance of $0.4$ m from the pivot. Find the moment of the force about the pivot.

Hint

The force is at right angles, so the distance given is the perpendicular distance.

Next step

Multiply the force by the distance.

Full working

\text{moment} = 25 \times 0.4 = 10 \text{ N m}.

2. A force of $40$ N acts at the end of a rod of length $1.2$ m. The force makes an angle of $60^\circ$ with the rod. Find the moment of the force about the other end.

Hint

The perpendicular distance is $1.2\sin 60^\circ$ , not the length of the rod.

Next step

Work out $1.2\sin 60^\circ$ first, then multiply by $40$ .

Full working

\text{moment} = 40 \times 1.2 \times \sin 60^\circ \approx 40 \times 1.039 = 41.57 \text{ N m}.

3. A child of mass $40$ kg sits $1.5$ 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?

Hint

For balance, the two moments about the pivot are equal.

Next step

Set $40g \times 1.5 = 60g \times d$ . The $g$ cancels.

Full working

40 \times 1.5 = 60d, \qquad 60 = 60d, \qquad d = 1 \text{ m}.

4. A uniform beam $AB$ has length $5$ m and mass $20$ kg. It rests on supports at $A$ and $B$ . A load of mass $30$ kg rests $2$ m from $A$ . Find the reaction at each support.

Hint

The beam's weight acts at its midpoint, $2.5$ m from $A$ .

Next step

Take moments about $A$ : $5R_B = 196 \times 2.5 + 294 \times 2$ . Then the reactions add to the total load.

Full working

5R_B = 490 + 588 = 1078, \qquad R_B = 215.6 \text{ N}.

R_A = (196 + 294) - 215.6 = 274.4 \text{ N}.

5. A uniform beam $AB$ has length $3$ m and mass $15$ kg. It rests on supports at $A$ and $B$ . A load of mass $25$ kg rests $1$ m from $A$ . Find the reaction at each support.

Hint

The beam weight is $147$ N at $1.5$ m from $A$ , and the load is $245$ N at $1$ m.

Next step

Take moments about $A$ : $3R_B = 147 \times 1.5 + 245 \times 1$ .

Full working

3R_B = 220.5 + 245 = 465.5, \qquad R_B \approx 155.17 \text{ N}.

R_A = (147 + 245) - 155.17 \approx 236.83 \text{ N}.

6. A uniform rod $AB$ has mass $10$ kg and length $3$ m. It is hinged at $A$ and held horizontal by a vertical string at $B$ . Find the tension in the string.

Hint

Take moments about the hinge, so the hinge force does not appear in the equation.

Next step

The weight is $98$ N at $1.5$ m from $A$ . Set $3T = 98 \times 1.5$ .

Full working

T = \frac{98 \times 1.5}{3} = \frac{147}{3} = 49 \text{ N}.

7. A non-uniform rod $AB$ has length $2$ m and mass $5$ kg. It is pivoted at its centre and balances when a mass of $1$ kg is placed at $B$ . Find the distance of the centre of mass of the rod from $A$ .

Hint

The $1$ kg mass is $1$ m from the pivot. Its moment must be balanced by the rod's weight acting on the other side of the pivot.

Next step

Set $5g\,x = 1g \times 1$ , so $x = 0.2$ m from the pivot. Then subtract from $1$ m.

Full working

5x = 1, \qquad x = 0.2 \text{ m}.

The centre of mass is $1 - 0.2 = 0.8$ m from $A$ .

8. A uniform plank $AB$ has length $6$ m and mass $30$ kg. It rests on supports at $A$ and $B$ . A person of mass $60$ kg stands $4$ m from $A$ . Find the reaction at each support.

Hint

The plank's weight acts $3$ m from $A$ . The person's weight acts $4$ m from $A$ .

Next step

Take moments about $A$ : $6R_B = 294 \times 3 + 588 \times 4$ .

Full working

6R_B = 882 + 2352 = 3234, \qquad R_B = 539 \text{ N}.

R_A = (294 + 588) - 539 = 343 \text{ N}.