16.5 Resolving forces and coplanar equilibrium — Practise with help

Work each one first, closed book. Choose the two axes before you write anything. Resolve every force, then put the sum in each direction equal to zero. Take $g = 9.8$ m s $^{-2}$ .

Warm-up: remember

Write down $\cos 30^\circ$ and $\sin 30^\circ$ exactly.
In a right-angled triangle, the hypotenuse is $10$ cm and one angle is $40^\circ$ . Find the side adjacent to that angle, to $2$ d.p.
Work out $\sqrt{3^2 + 4^2}$ .

Answers

$\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ , $\sin 30^\circ = \dfrac{1}{2}$ .
$10\cos 40^\circ \approx 7.66$ cm.
$\sqrt{25} = 5$ .

If question 2 was hard, revise the sine and cosine ratios before you continue.

Practise with help

1. Resolve a force of $70$ N acting at $25^\circ$ above the horizontal.

Hint

The angle is measured from the horizontal, so the horizontal component uses the cosine.

Next step

The horizontal component is $70\cos 25^\circ$ and the vertical component is $70\sin 25^\circ$ .

Full working

\text{horizontal} = 70\cos 25^\circ \approx 63.44 \text{ N},

\text{vertical} = 70\sin 25^\circ \approx 29.58 \text{ N}.

2. A weight of $200$ N acts on a body on a slope of $20^\circ$ . Find the component down the slope and the component perpendicular to the slope, each to $2$ d.p.

Hint

The angle at the slope is measured from the slope surface, so the sine goes with the component down the slope.

Next step

Down the slope: $200\sin 20^\circ$ . Perpendicular to the slope: $200\cos 20^\circ$ .

Full working

\text{down the slope} = 200\sin 20^\circ \approx 68.40 \text{ N},

\text{perpendicular} = 200\cos 20^\circ \approx 187.94 \text{ N}.

3. A block of mass $10$ kg rests on a smooth slope of $30^\circ$ . A light rope parallel to the slope holds it in equilibrium. Find the tension and the normal reaction.

Hint

First find the weight, then resolve it along and perpendicular to the slope.

Next step

The weight is $10 \times 9.8 = 98$ N. The tension balances the component down the slope.

Full working

T = 98\sin 30^\circ = 49 \text{ N},

R = 98\cos 30^\circ = 49\sqrt{3} \approx 84.87 \text{ N}.

4. A particle of weight $60$ N hangs in equilibrium from two light strings that each make $45^\circ$ with the horizontal. Find the tension in each string.

Hint

The two strings are symmetrical, so the two tensions are equal.

Next step

Resolve vertically: $T\sin 45^\circ + T\sin 45^\circ = 60$ .

Full working

2T\sin 45^\circ = 60, \qquad T = \frac{60}{2 \times 0.7071} \approx 42.43 \text{ N}.

Each tension is about $42.43$ N.

5. A particle of weight $100$ N hangs in equilibrium from two light strings. One makes $30^\circ$ with the horizontal and the other makes $60^\circ$ with the horizontal. Find the two tensions.

Hint

There are two unknown tensions, so two equations are needed: one horizontal and one vertical.

Next step

The horizontal equation is $T_1\cos 30^\circ = T_2\cos 60^\circ$ . The vertical equation is $T_1\sin 30^\circ + T_2\sin 60^\circ = 100$ .

Full working

From the horizontal equation, $T_1 = \dfrac{\cos 60^\circ}{\cos 30^\circ}T_2 = \dfrac{T_2}{\sqrt{3}}$ . Substitute into the vertical equation:

\frac{T_2}{\sqrt{3}} \times \frac{1}{2} + T_2 \times \frac{\sqrt{3}}{2} = 100.

Multiply every term by $2\sqrt{3}$ :

T_2 + 3T_2 = 200\sqrt{3}, \qquad 4T_2 = 200\sqrt{3}, \qquad T_2 = 50\sqrt{3} \approx 86.60 \text{ N}.

Then $T_1 = 50$ N. The string closer to the vertical carries the larger tension.

6. A body of weight $196$ N is held in equilibrium on a smooth slope of $30^\circ$ by a horizontal force $P$ . Find $P$ and the normal reaction.

Hint

Resolve along the slope first. The weight has a component down the slope and the horizontal force has a component up the slope.

Next step

Along the slope: $P\cos 30^\circ = 196\sin 30^\circ$ , so $P = 196\tan 30^\circ$ . Then resolve perpendicular to the slope for the normal reaction.

Full working

P = 196\tan 30^\circ \approx 113.16 \text{ N}.

R = 196\cos 30^\circ + P\sin 30^\circ \approx 169.74 + 56.58 = 226.32 \text{ N}.

7. A particle is in equilibrium under four forces. Three of them are $(4, 0)$ N, $(-1, 5)$ N and $(0, -3)$ N. Find the fourth force.

Hint

Add the three known forces as vectors first.

Next step

The three forces add to $(3, 2)$ N. The fourth force must cancel that.

Full working

The fourth force is $-(3, 2) = (-3, -2)$ N. Its size is $\sqrt{3^2 + 2^2} = \sqrt{13} \approx 3.61$ N.

8. A sign of weight $49$ N hangs from the end of a horizontal rod. A wire from the wall to the end of the rod makes $40^\circ$ with the horizontal and holds the sign in equilibrium. Find the tension in the wire and the thrust in the rod.

Hint

At the end of the rod, three forces meet: the weight, the wire tension and the rod thrust. Resolve vertically first.

Next step

The vertical component of the tension balances the weight: $T\sin 40^\circ = 49$ . Then the horizontal component balances the rod thrust.

Full working

T = \frac{49}{\sin 40^\circ} \approx 76.23 \text{ N}.

\text{thrust} = T\cos 40^\circ \approx 58.40 \text{ N}.