2.2 Resolving and adding vectors — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Say which direction each angle is measured from before you use a sine or a cosine.

Warm-up: remember

A right-angled triangle has shorter sides 5 and 12. Find the hypotenuse.
Work out $50\cos 60^\circ$ .
Work out $50\sin 60^\circ$ to three significant figures.

Answers

$\sqrt{5^2 + 12^2} = \sqrt{169} = 13$ .
$50 \times 0.5 = 25$ .
$50 \times 0.866 = 43.3$ .

If question 1 goes wrong, add the squares of the two shorter sides and take the square root. If question 2 or 3 goes wrong, check which ratio goes with the angle. The side opposite the angle uses the sine. The side next to it uses the cosine.

Practise with help

1. Resolve a force of 60 N at 40° to the horizontal into horizontal and vertical components.

Hint

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

Next step

Write $F_x = 60\cos 40^\circ$ and $F_y = 60\sin 40^\circ$ , then evaluate each.

Full working

\begin{aligned} F_x &= 60\cos 40^\circ = 46.0\,\text{N} \\ F_y &= 60\sin 40^\circ = 38.6\,\text{N} \end{aligned}

Both components are smaller than the 60 N force, which is a useful check.

2. A vector has components $-3.0$ and $+4.0$ . Find its magnitude and its direction measured anticlockwise from the positive $x$ -axis.

Hint

Find the magnitude with Pythagoras first. Then use the tangent to find the angle inside the triangle, and place the vector in the correct quadrant.

Next step

The magnitude is $\sqrt{(-3.0)^2 + 4.0^2} = 5.0$ . The negative $x$ -component and positive $y$ -component put the vector in the second quadrant.

Full working

\begin{aligned} \text{magnitude} &= \sqrt{(-3.0)^2 + 4.0^2} = 5.0 \\ \text{angle in the triangle} &= \tan^{-1}\!\left(\frac{4.0}{3.0}\right) = 53.1^\circ \end{aligned}

In the second quadrant the direction from the positive $x$ -axis is $180^\circ - 53.1^\circ = 126.9^\circ$ .

3. Two forces of 7.0 N and 24 N act at right angles. Find the magnitude and direction of the resultant.

Hint

The two forces are already perpendicular components of their resultant, so this is a right-angled triangle to solve.

Next step

Use $R = \sqrt{7.0^2 + 24^2}$ , then $\tan\theta = \dfrac{24}{7.0}$ .

Full working

\begin{aligned} R &= \sqrt{7.0^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\,\text{N} \\ \theta &= \tan^{-1}\!\left(\frac{24}{7.0}\right) = 73.7^\circ \end{aligned}

The resultant is 25 N at 73.7° to the 7.0 N force.

4. A weight of 800 N acts on a slope at 20° to the horizontal. Find the component along the slope and the component perpendicular to the slope.

Hint

The weight acts vertically down. The slope makes 20° with the horizontal, so the weight makes 20° with the perpendicular to the slope.

Next step

The component along the slope uses the sine of the slope angle, and the perpendicular component uses the cosine.

Full working

\begin{aligned} W_{\parallel} &= 800\sin 20^\circ = 273.6\,\text{N} \\ W_{\perp} &= 800\cos 20^\circ = 751.8\,\text{N} \end{aligned}

The perpendicular component is the larger one, because the slope is shallow.

5. Resolve a velocity of 15 m s⁻¹ at 35° above the horizontal into horizontal and vertical components.

Hint

The angle is measured from the horizontal. Decide which component uses the cosine before you reach for the calculator.

Next step

Write $v_x = 15\cos 35^\circ$ and $v_y = 15\sin 35^\circ$ .

Full working

\begin{aligned} v_x &= 15\cos 35^\circ = 12.3\,\text{m}\,\text{s}^{-1} \\ v_y &= 15\sin 35^\circ = 8.60\,\text{m}\,\text{s}^{-1} \end{aligned}

Check that $\sqrt{12.3^2 + 8.60^2} = 15.0$ , which recovers the original speed.

6. Explain why Pythagoras cannot be used on its own to add two vectors at 55°, and name a method that works.

Hint

Think about the triangle formed by the two vectors and their resultant, and about the condition Pythagoras needs.

Next step

State the condition, then say whether the 55° triangle meets it. Then name a method that does not need a right angle.

Full working

Pythagoras applies only to a right-angled triangle. Two vectors at 55° do not form a right angle, so Pythagoras cannot give the resultant on its own. A scale drawing works for any angle: draw the vectors head to tail, then measure the resultant. Resolving each vector into perpendicular components and adding them also works.