16.1 Force and Newton's first law — Practise

Work through these on your own. For each question, decide whether the answer is a single force, a size or a direction, and finish the calculation in that form. Add forces along the same line by adding or subtracting sizes. Add forces at an angle as vectors.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Two horizontal forces of $12$ N and $5$ N act on a particle in the same direction. Find the resultant force.
A horizontal force of $20$ N acts to the right and a horizontal force of $8$ N acts to the left on a particle. Find the resultant force, giving its size and direction.
Forces of $6$ N and $8$ N act at right angles at a point. Find the size of the resultant force.
A force of $9$ N acts due east and a force of $40$ N acts due north at a point. Find the size of the resultant force and its direction from east.
Two of the three forces on a particle in equilibrium are $5$ N due east and $3$ N due north. Find the third force as a column vector.
Forces $(2, 3)$ N, $(-5, 1)$ N and $(4, -6)$ N act at a point. Find the resultant force as a column vector.
A car travels at a constant velocity. The driving force is $800$ N and the total resistance is $800$ N. Find the resultant force on the car.
A skydiver of weight $700$ N falls at terminal velocity. Find the resultant force on the skydiver and the size of the air resistance.
Two forces of $10$ N each act at a point with an angle of $60^\circ$ between them. Find the size of the resultant force.
Forces of $15$ N and $20$ N act at right angles at a point. Find the size of the resultant and the angle it makes with the $20$ N force.
A particle is held in equilibrium by three forces. Two of them are $8$ N east and $6$ N north. Find the size of the third force.
Forces $(4, -1)$ N and $(-1, 5)$ N act at a point. Find the resultant force and its size.

Answers and concise solutions

1. The resultant force

Answer: The forces act in the same direction, so add them: the resultant is $12 + 5 = 17$ N in that direction.

2. The resultant force

Answer: The forces oppose each other, so subtract: $20 - 8 = 12$ N. The larger force points to the right, so the resultant is $12$ N to the right.

3. The size of the resultant

Answer: The two forces are perpendicular, so use Pythagoras: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ N.

4. The size and the direction

Answer: The size is $\sqrt{9^2 + 40^2} = \sqrt{1681} = 41$ N. The direction is $\tan^{-1}\left(\dfrac{40}{9}\right) = 77.3^\circ$ north of east, to $1$ d.p.

5. The third force

Answer: The forces add to the zero vector, so the third force is $-(5\mathbf{i} + 3\mathbf{j}) = -5\mathbf{i} - 3\mathbf{j}$ N. It has size $\sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83$ N.

6. The resultant

Answer: Add the matching components: the resultant is $(2 - 5 + 4)\mathbf{i} + (3 + 1 - 6)\mathbf{j} = \mathbf{i} - 2\mathbf{j}$ N.

7. The resultant force

Answer: The driving force and the resistance are equal and opposite, so the resultant is $800 - 800 = 0$ N. A zero resultant is exactly what Newton's first law needs for a constant velocity.

8. The resultant and the air resistance

Answer: At terminal velocity the skydiver moves at a constant velocity, so the resultant force is $0$ N. The air resistance must balance the weight, so it is $700$ N upwards.

9. The size of the resultant

Answer: The size is $\sqrt{10^2 + 10^2 + 2(10)(10)\cos 60^\circ} = \sqrt{300} = 10\sqrt{3} \approx 17.3$ N. The two forces partly cancel, so the answer is less than $20$ N.

10. The size and the angle

Answer: The size is $\sqrt{20^2 + 15^2} = \sqrt{625} = 25$ N. The angle with the $20$ N force is $\tan^{-1}\left(\dfrac{15}{20}\right) = 36.9^\circ$ to $1$ d.p.

11. The size of the third force

Answer: The third force is the opposite of the sum of the other two, so it is $(-8, -6)$ N. Its size is $\sqrt{8^2 + 6^2} = 10$ N. The three force arrows, drawn head to tail, close up to make a triangle.

12. The resultant and its size

Answer: The resultant is $(4 - 1)\mathbf{i} + (-1 + 5)\mathbf{j} = 3\mathbf{i} + 4\mathbf{j}$ N. Its size is $\sqrt{3^2 + 4^2} = 5$ N.