16.1 Force and Newton's first law — Practise with help

Work each one first, closed book. Decide whether the question wants a list of forces, a resultant, a size or a direction. Draw a quick sketch before you calculate.

Warm-up: remember

Add the vectors $(2, 3)$ and $(5, -1)$ .
Find the size of the vector $6\mathbf{i} + 8\mathbf{j}$ .
Work out $\sqrt{3^2 + 4^2}$ .

Answers

$(2 + 5, 3 - 1) = (7, 2)$ .
$\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ .
$\sqrt{9 + 16} = \sqrt{25} = 5$ .

If question 2 was hard, revise the size of a vector in Section 10.2 before you continue.

Practise with help

1. A crate is at rest on a horizontal floor. Name the forces that act on the crate, and say what Newton's first law tells you about them.

Hint

One force comes from the Earth without touching the crate. One force comes from the floor and acts where they touch.

Next step

The crate stays at rest, so the resultant force is zero. Write down what that says about the sizes and the directions of the two forces.

Full working

Two forces act on the crate:

its weight, acting downwards;
the normal reaction from the floor, acting upwards.

The crate is at rest, so by Newton's first law the resultant force is zero. The two forces have equal sizes and opposite directions, so the normal reaction equals the weight.

2. A force of $14$ N acts due east and a force of $6$ N acts due west on a particle. Find the resultant force.

Hint

The two forces act along the same straight line, so the sizes combine by addition or subtraction.

Next step

The larger force points east, so take $14$ subtract $6$ and keep the east direction.

Full working

The forces are opposite, so subtract:

14 - 6 = 8.

The larger force points east, so the resultant is $8$ N due east.

3. Forces of $9$ N and $12$ N act at right angles at a point. Find the size of the resultant and the angle it makes with the $12$ N force.

Hint

Two perpendicular forces make a right-angled triangle. The resultant is the hypotenuse.

Next step

Work out $\sqrt{9^2 + 12^2}$ first. Then use $\tan^{-1}$ of the $9$ N force over the $12$ N force.

Full working

\text{size} = \sqrt{9^2 + 12^2} = \sqrt{225} = 15 \text{ N}.

\text{angle} = \tan^{-1}\left(\frac{9}{12}\right) = 36.9^\circ

to $1$ d.p., measured from the $12$ N force towards the $9$ N force.

4. A particle is in equilibrium under three forces. Two of them are $10$ N due east and $24$ N due north. Find the third force as a column vector, and find its size.

Hint

Equilibrium means the resultant is zero, so the three force arrows form a closed triangle. The third force cancels the other two.

Next step

Write the two known forces as $(10, 24)$ N. The third force is the negative of that vector.

Full working

The two known forces are $(10, 24)$ N. The third force is

-(10, 24) = (-10, -24) \text{ N}.

Its size is

\sqrt{10^2 + 24^2} = \sqrt{676} = 26 \text{ N}.

5. A car travels at a constant velocity of $20$ m/s along a straight road. The driving force is $600$ N. Find the total resistance.

Hint

A constant velocity is the second case in Newton's first law.

Next step

The resultant force is zero, so the resistance balances the driving force.

Full working

The velocity is constant, so the resultant force is zero. The driving force and the resistance are equal and opposite, so the total resistance is $600$ N backwards.

6. Forces $(3, 4)$ N and $(5, -2)$ N act on a particle. Find the resultant force and its size.

Hint

Add the matching components first.

Next step

The resultant is $(3 + 5, 4 - 2)$ N. Then take the square root of the sum of the squares of the two components.

Full working

(3, 4) + (5, -2) = (8, 2) \text{ N}.

\text{size} = \sqrt{8^2 + 2^2} = \sqrt{68} = 2\sqrt{17} \approx 8.25 \text{ N}.

7. Forces $(2, 5)$ N, $(-4, 1)$ N and $(1, -3)$ N act at a point. Find the resultant force, and state whether the particle is in equilibrium.

Hint

Add all three vectors component by component.

Next step

The $x$ -component is $2 - 4 + 1$ and the $y$ -component is $5 + 1 - 3$ .

Full working

(2 - 4 + 1)\mathbf{i} + (5 + 1 - 3)\mathbf{j} = -\mathbf{i} + 3\mathbf{j} \text{ N}.

The resultant is not the zero vector, so the particle is not in equilibrium. It accelerates in the direction of $-\mathbf{i} + 3\mathbf{j}$ .

8. A box is at rest on a horizontal table. A student says the only force on the box is its weight, so the box should accelerate downwards. Explain what is missing from this reasoning.

Hint

The box touches the table. Think about the force the table exerts where they touch.

Next step

If the box is at rest, the resultant force on it must be zero. Compare the size of the weight with the size of the missing force.

Full working

The student has left out the normal reaction from the table. The table pushes up on the box at the place where they touch. The box is at rest, so the resultant force is zero, which means the normal reaction equals the weight. The two forces cancel, so the box does not accelerate.