16.6 Resultant forces and dynamics in a plane — Practise with help

Work each one first, closed book. Choose one axis along the acceleration and one perpendicular to it. Resolve in both directions, then use the perpendicular equation for the normal reaction and the other equation for the acceleration. Take $g = 9.8$ m s $^{-2}$ .

Warm-up: remember

Find the horizontal and vertical components of a force of $50$ N acting at $40^\circ$ above the horizontal.
Add the vectors $(3, 4)$ and $(-1, 2)$ .
Solve $5a = 12.5$ .

Answers

Horizontal $= 50\cos 40^\circ \approx 38.30$ N. Vertical $= 50\sin 40^\circ \approx 32.14$ N.
$(3 - 1, 4 + 2) = (2, 6)$ .
$a = 2.5$ .

If question 1 was hard, revise resolving forces in Section 16.5 before you continue.

Practise with help

1. A force of $40$ N acts due east and a force of $30$ N acts due north. Find the size of the resultant and its direction from the east direction.

Hint

The two forces are perpendicular, so add them as the sides of a right-angled triangle.

Next step

The size is $\sqrt{40^2 + 30^2}$ . The direction uses the tangent of the north force over the east force.

Full working

R = \sqrt{40^2 + 30^2} = 50 \text{ N},

\theta = \tan^{-1}\left(\frac{30}{40}\right) = 36.87^\circ

north of east.

2. A body of mass $20$ kg is pulled along a horizontal floor by a rope at $30^\circ$ to the horizontal with a tension of $150$ N. A resistance of $50$ N opposes the motion. Find the normal reaction and the acceleration.

Hint

Resolve vertically first. The rope has an upward component, so the normal reaction is less than the weight.

Next step

Vertically: $R + 150\sin 30^\circ = 20 \times 9.8$ . Horizontally, the resultant is $150\cos 30^\circ - 50$ .

Full working

R = 196 - 75 = 121 \text{ N}.

a = \frac{150\cos 30^\circ - 50}{20} = \frac{129.90 - 50}{20} \approx 4.00 \text{ m s}^{-2}.

3. A block of mass $6$ kg is released on a smooth slope of $25^\circ$ . Find the acceleration and the normal reaction.

Hint

On a smooth slope the only force along the slope is the component of the weight.

Next step

The weight is $6 \times 9.8 = 58.8$ N. Down the slope the resultant is $58.8\sin 25^\circ$ . Perpendicular to the slope the reaction is $58.8\cos 25^\circ$ .

Full working

a = \frac{58.8\sin 25^\circ}{6} = 9.8\sin 25^\circ \approx 4.14 \text{ m s}^{-2},

R = 58.8\cos 25^\circ \approx 53.29 \text{ N}.

4. A block of mass $5$ kg is pulled up a smooth slope of $35^\circ$ by a rope parallel to the slope with a tension of $60$ N. Find the acceleration.

Hint

Along the slope the tension acts up and the weight component acts down.

Next step

The weight component down the slope is $5 \times 9.8\sin 35^\circ = 28.10$ N. Subtract it from the tension, then divide by the mass.

Full working

a = \frac{60 - 49\sin 35^\circ}{5} = \frac{60 - 28.10}{5} \approx 6.38 \text{ m s}^{-2}

up the slope.

5. A block of mass $4$ kg rests on a horizontal table. It is joined by a light inextensible string over a smooth pulley at the table edge to a hanging mass of $3$ kg. A resistance of $8$ N acts on the block. Find the acceleration and the tension.

Hint

The hanging weight drives the system. The resistance opposes the motion, so it subtracts from the driving force.

Next step

The driving force is $3 \times 9.8 = 29.4$ N and the total mass is $7$ kg. Then use $T - 8 = 4a$ for the block.

Full working

a = \frac{29.4 - 8}{7} = \frac{21.4}{7} \approx 3.06 \text{ m s}^{-2}.

T = 4a + 8 = 12.23 + 8 \approx 20.23 \text{ N}.

6. Forces of $8$ N due east, $6$ N due north and $3$ N due west act at a point. Find the resultant and its size.

Hint

Add the east and west forces first, since they act along the same line.

Next step

The east-west resultant is $8 - 3 = 5$ N east. Now combine $5$ N and $6$ N at right angles.

Full working

The resultant is $(5, 6)$ N. Its size is

\sqrt{5^2 + 6^2} = \sqrt{61} \approx 7.81 \text{ N}.

7. Two forces of $200$ N and $150$ N act at a point with an angle of $60^\circ$ between them. Find the size of the resultant.

Hint

Use the cosine rule form for two forces at an angle: $R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$ .

Next step

Substitute $F_1 = 200$ , $F_2 = 150$ and $\cos 60^\circ = 0.5$ .

Full working

R^2 = 200^2 + 150^2 + 2(200)(150)(0.5) = 40000 + 22500 + 30000 = 92500,

R = \sqrt{92500} \approx 304.14 \text{ N}.

8. A block of mass $12$ kg slides down a slope of $20^\circ$ . A friction force of $15$ N acts up the slope. Find the acceleration.

Hint

The friction opposes the sliding, so it acts up the slope and reduces the downhill resultant.

Next step

The weight component down the slope is $12 \times 9.8\sin 20^\circ = 40.22$ N. Subtract the friction, then divide by the mass.

Full working

a = \frac{12 \times 9.8\sin 20^\circ - 15}{12} = \frac{40.22 - 15}{12} \approx 2.10 \text{ m s}^{-2}

down the slope.