16.5 Resolving forces and coplanar equilibrium — Practise

Work through these on your own. Choose the two axes first, then resolve every force and set the sum in each direction to zero. Take $g = 9.8$ m s $^{-2}$ . Where a question asks for a decimal, give the number of decimal places it names.

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 $50$ N acts at $30^\circ$ above the horizontal. Find its horizontal and vertical components.
A force of $80$ N acts at $55^\circ$ above the horizontal. Find its horizontal and vertical components, each to $2$ d.p.
A weight of $441$ N acts on a body on a slope of $18^\circ$ . Find the component of the weight along the slope and the component perpendicular to the slope, each to $1$ d.p.
A weight of $98$ N acts on a body on a slope of $30^\circ$ . Find the component along the slope and the component perpendicular to the slope.
A particle of weight $50$ N hangs in equilibrium from two light strings. One string makes $60^\circ$ with the horizontal and the other makes $30^\circ$ with the horizontal. Find the tension in each string.
A block of mass $60$ kg rests on a smooth slope of $25^\circ$ . A light rope parallel to the slope holds it in equilibrium. Find the tension and the normal reaction, each to $1$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A body of weight $392$ N is held in equilibrium on a smooth slope of $30^\circ$ by a horizontal force $P$ . Find $P$ and the normal reaction, each to $1$ d.p.
A force of $120$ N acts at $35^\circ$ below the horizontal. Find the horizontal component and the downward vertical component, each to $2$ d.p.
A particle is in equilibrium under four forces. Three of them are $(5, 0)$ N, $(-3, 4)$ N and $(0, -6)$ N. Find the fourth force as a column vector.
A particle is in equilibrium under three forces. Two of them are $12$ N due east and $5$ N due north. Find the third force as a column vector and its size.
A particle of weight $30$ N hangs in equilibrium from two light strings. One string makes $40^\circ$ with the horizontal and the other makes $50^\circ$ with the horizontal. Find the two tensions, each to $2$ d.p.
A block of mass $10$ kg rests on a smooth slope of $15^\circ$ . A light rope parallel to the slope holds it in equilibrium. Find the tension and the normal reaction, each to $2$ d.p. Take $g = 9.8$ m s $^{-2}$ .

Answers and concise solutions

1. The two components

Answer: The horizontal component is $50\cos 30^\circ = 43.30$ N. The vertical component is $50\sin 30^\circ = 25.00$ N. The angle is measured from the horizontal, so the cosine goes with the horizontal component.

2. The two components

Answer: Horizontal component $= 80\cos 55^\circ = 45.89$ N. Vertical component $= 80\sin 55^\circ = 65.53$ N. The vertical component is the larger one because the angle is closer to the vertical.

3. The two components

Answer: The component down the slope is $441\sin 18^\circ = 136.3$ N. The component perpendicular to the slope is $441\cos 18^\circ = 419.4$ N. The two components add as vectors back to the $441$ N weight.

4. The two components

Answer: Along the slope: $98\sin 30^\circ = 49$ N. Perpendicular to the slope: $98\cos 30^\circ = 49\sqrt{3} \approx 84.87$ N.

5. The two tensions

Answer: Resolve vertically: $T_1\sin 60^\circ + T_2\sin 30^\circ = 50$ . Resolve horizontally: $T_1\cos 60^\circ = T_2\cos 30^\circ$ , so $T_1 = \sqrt{3}\,T_2$ . Substituting gives $2T_2 = 50$ , so $T_2 = 25$ N and $T_1 = 25\sqrt{3} \approx 43.30$ N.

6. The tension and the normal reaction

Answer: The weight is $60 \times 9.8 = 588$ N. Along the slope, the tension balances the weight component: $T = 588\sin 25^\circ = 248.5$ N. Perpendicular to the slope, the normal reaction balances the other component: $R = 588\cos 25^\circ = 533.0$ N.

7. The force and the normal reaction

Answer: Resolve along the slope: $P\cos 30^\circ = 392\sin 30^\circ$ , so $P = 392\tan 30^\circ = 226.3$ N. Resolve perpendicular to the slope: $R = 392\cos 30^\circ + P\sin 30^\circ = 452.6$ N. The normal reaction is larger than the weight, because the horizontal force pushes into the slope.

8. The two components

Answer: Horizontal component $= 120\cos 35^\circ = 98.30$ N. Downward vertical component $= 120\sin 35^\circ = 68.83$ N. The angle is still measured from the horizontal, so the cosine gives the horizontal part.

9. The fourth force

Answer: The three known forces add to $(2, -2)$ N. For equilibrium the four forces sum to zero, so the fourth force is $-(2, -2) = (-2, 2)$ N. It has size $\sqrt{2^2 + 2^2} = 2\sqrt{2} \approx 2.83$ N.

10. The third force and its size

Answer: The third force cancels the other two, so it is $(-12, -5)$ N. Its size is $\sqrt{12^2 + 5^2} = 13$ N. The three force arrows form a $5$ - $12$ - $13$ triangle.

11. The two tensions

Answer: Resolve horizontally: $T_1\cos 40^\circ = T_2\cos 50^\circ$ , and since $\cos 50^\circ = \sin 40^\circ$ , this gives $T_1 = T_2\tan 40^\circ$ . Resolving vertically and substituting gives $T_2 = 30\cos 40^\circ = 22.98$ N and $T_1 = 30\sin 40^\circ = 19.28$ N.

12. The tension and the normal reaction

Answer: The weight is $10 \times 9.8 = 98$ N. Along the slope: $T = 98\sin 15^\circ = 25.36$ N. Perpendicular to the slope: $R = 98\cos 15^\circ = 94.66$ N. The normal reaction is slightly less than the weight, because the slope carries only the perpendicular component.