16.6 Resultant forces and dynamics in a plane — Practise

Work through these on your own. Choose one axis along the acceleration and one perpendicular to it. Use the perpendicular direction for the normal reaction and the other direction for the acceleration. Take $g = 9.8$ m s $^{-2}$ . Where a decimal is wanted, give the number of decimal places the question 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 $60$ N acts at $30^\circ$ above the horizontal and a force of $100$ N acts at $120^\circ$ . Find the size of the resultant and its direction from the positive $x$ -axis.
A crate of mass $45$ kg is pulled along a horizontal floor by a rope that makes $25^\circ$ with the horizontal. The tension is $200$ N and a resistance of $80$ N opposes the motion. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
For the crate of mass $45$ kg pulled by a rope at $25^\circ$ with a tension of $200$ N, find the normal reaction from the floor. Take $g = 9.8$ m s $^{-2}$ .
A body of mass $20$ kg is pulled along a smooth horizontal floor by a rope at $30^\circ$ to the horizontal with a tension of $100$ N. Find the acceleration.
A block of mass $10$ kg slides down a slope of $30^\circ$ . A friction force of $17$ N acts up the slope. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $5$ kg slides down a slope of $40^\circ$ . A friction force of $18.8$ N acts up the slope. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
A block of mass $6$ 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 $4$ kg. A friction force of $14.7$ N acts on the block. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
For the $6$ kg block on the table with $14.7$ N of friction and the $4$ kg hanging mass, find the tension in the string. Take $g = 9.8$ m s $^{-2}$ .
Two forces of $30$ N each act at a point with an angle of $60^\circ$ between them. The forces act on a mass of $4$ kg. Find the resultant force and the acceleration.
Forces $(10, 0)$ N, $(0, 15)$ N and $(-5, -5)$ N act on a mass of $2$ kg. Find the resultant force and the acceleration.
A block of mass $8$ kg is pulled up a smooth slope of $20^\circ$ by a rope parallel to the slope with a tension of $60$ N. Find the normal reaction and the acceleration, each to $2$ d.p. Take $g = 9.8$ m s $^{-2}$ .
A body of mass $3$ kg rests on a smooth horizontal floor. A rope at $30^\circ$ to the horizontal pulls it with a tension of $20$ N. Find the normal reaction and the acceleration.

Answers and concise solutions

1. The resultant and its direction

Answer: The components are $x = 51.96 - 50.00 = 1.96$ N and $y = 30.00 + 86.60 = 116.60$ N. The size is $\sqrt{1.96^2 + 116.60^2} \approx 116.62$ N, and the direction is $\tan^{-1}\left(\dfrac{116.60}{1.96}\right) \approx 89.04^\circ$ . The resultant is almost vertical, because the two vertical parts add while the horizontal parts nearly cancel.

2. The acceleration

Answer: The horizontal component of the tension is $200\cos 25^\circ = 181.26$ N. The resultant along the floor is $181.26 - 80 = 101.26$ N, so $a = \dfrac{101.26}{45} \approx 2.25$ m s $^{-2}$ .

3. The normal reaction

Answer: There is no vertical acceleration, so the vertical forces balance. Resolving upwards: $R + 200\sin 25^\circ = 45 \times 9.8$ , so $R = 441 - 84.52 = 356.48$ N. The rope lifts the crate, so the normal reaction is less than the weight.

4. The acceleration

Answer: Only the horizontal component of the tension moves the body: $100\cos 30^\circ = 86.60$ N. So $a = \dfrac{86.60}{20} \approx 4.33$ m s $^{-2}$ . The vertical component does not change the horizontal motion.

5. The acceleration

Answer: The weight is $98$ N. Its component down the slope is $98\sin 30^\circ = 49$ N. The friction of $17$ N opposes the sliding, so the resultant down the slope is $49 - 17 = 32$ N. Then $a = \dfrac{32}{10} = 3.2$ m s $^{-2}$ .

6. The acceleration

Answer: The weight is $49$ N. Its component down the slope is $49\sin 40^\circ = 31.50$ N. The resultant is $31.50 - 18.8 = 12.70$ N, so $a = \dfrac{12.70}{5} \approx 2.54$ m s $^{-2}$ .

7. The acceleration

Answer: The hanging weight $4 \times 9.8 = 39.2$ N drives the system against the friction of $14.7$ N. The resultant is $39.2 - 14.7 = 24.5$ N on a total mass of $10$ kg, so $a = 2.45$ m s $^{-2}$ .

8. The tension

Answer: For the hanging mass, moving down: $4g - T = 4a$ . With $a = 2.45$ m s $^{-2}$ , $T = 4(9.8 - 2.45) = 29.4$ N. Check with the block: $T - 14.7 = 6 \times 2.45 = 14.7$ , and $14.7 + 14.7 = 29.4$ N, which agrees.

9. The resultant and the acceleration

Answer: The resultant is $\sqrt{30^2 + 30^2 + 2(30)(30)\cos 60^\circ} = 30\sqrt{3} \approx 51.96$ N. Then $a = \dfrac{51.96}{4} \approx 12.99$ m s $^{-2}$ in the direction of the resultant.

10. The resultant and the acceleration

Answer: The resultant is $(10 + 0 - 5)\mathbf{i} + (0 + 15 - 5)\mathbf{j} = (5, 10)$ N. Its size is $\sqrt{5^2 + 10^2} = 5\sqrt{5} \approx 11.18$ N, so $a = \dfrac{11.18}{2} \approx 5.59$ m s $^{-2}$ .

11. The normal reaction and the acceleration

Answer: Perpendicular to the slope, $R = 8 \times 9.8\cos 20^\circ = 73.66$ N. The weight component down the slope is $8 \times 9.8\sin 20^\circ = 26.82$ N. The resultant up the slope is $60 - 26.82 = 33.18$ N, so $a = \dfrac{33.18}{8} \approx 4.15$ m s $^{-2}$ .

12. The normal reaction and the acceleration

Answer: The vertical component of the pull is $20\sin 30^\circ = 10$ N upwards, so $R = 3 \times 9.8 - 10 = 19.4$ N. The horizontal component is $20\cos 30^\circ = 17.32$ N, so $a = \dfrac{17.32}{3} \approx 5.77$ m s $^{-2}$ .