2.2 Resolving and adding vectors — Practise

Work through these on your own. For each one, write the direction the angle is measured from before you choose a sine or a cosine. For a resultant, state the angle from one of the two vectors.

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

A force of $50\,\text{N}$ acts at $30^\circ$ to the horizontal. Find its horizontal and vertical components.
A projectile is launched at $100\,\text{m}\,\text{s}^{-1}$ at $75^\circ$ above the horizontal. Find the horizontal and vertical components of its velocity.
A vector has an $x$ -component of $-5.0\,\text{m}$ and a $y$ -component of $-12.0\,\text{m}$ . Find its magnitude and its direction, measured anticlockwise from the positive $x$ -axis.
Two forces act at right angles. One force is $9.0\,\text{N}$ and the other is $40\,\text{N}$ . Find the magnitude and direction of the resultant force.
A block of weight $600\,\text{N}$ rests on a slope that makes $15^\circ$ with the horizontal. Resolve the weight into a component along the slope and a component perpendicular to the slope.
A force of $200\,\text{N}$ acts at $135^\circ$ from the positive $x$ -axis. Find its $x$ -component and its $y$ -component, including the sign of each.
A ferry points straight across a river at $4.0\,\text{m}\,\text{s}^{-1}$ . The current flows at $3.0\,\text{m}\,\text{s}^{-1}$ at right angles to the crossing. Find the ferry velocity over the ground, as a magnitude and a direction.
A force of $12\,\text{N}$ has a horizontal component of $8.0\,\text{N}$ and an upward vertical component. Find the angle the force makes with the horizontal, and the vertical component.
A plane flies due east through the air at $60\,\text{m}\,\text{s}^{-1}$ . A wind blows due north at $20\,\text{m}\,\text{s}^{-1}$ . Find the velocity of the plane over the ground, as a magnitude and a direction.
A displacement is $(4.0, -2.0)\,\text{m}$ . Find the displacement $3.0$ times larger in the same direction.
Two forces act on a point at $50^\circ$ to each other. Explain which method you would use to find the resultant, and why a calculation with Pythagoras on its own is not enough.
A displacement has components $5.0\,\text{m}$ east and $12.0\,\text{m}$ north. Find its magnitude and the angle it makes with the eastward direction.

Answers and concise solutions

1. The two perpendicular components

Answer: The horizontal component is $50\cos 30^\circ = 43.3\,\text{N}$ . The vertical component is $50\sin 30^\circ = 25.0\,\text{N}$ .

2. The two components of the velocity

Answer: Horizontal component $= 100\cos 75^\circ = 25.9\,\text{m}\,\text{s}^{-1}$ . Vertical component $= 100\sin 75^\circ = 96.6\,\text{m}\,\text{s}^{-1}$ .

3. The magnitude and direction

Answer: Magnitude $= \sqrt{(-5.0)^2 + (-12.0)^2} = 13.0\,\text{m}$ . Both components are negative, so the vector points into the third quadrant: $\theta = 180^\circ + \tan^{-1}\!\left(\dfrac{12.0}{5.0}\right) = 247.4^\circ$ .

4. The resultant force

Answer: Magnitude $= \sqrt{9.0^2 + 40^2} = 41\,\text{N}$ . Direction $= \tan^{-1}\!\left(\dfrac{40}{9.0}\right) = 77.3^\circ$ from the $9.0\,\text{N}$ force.

5. The two components of the weight

Answer: The weight makes $75^\circ$ with the along-slope direction, so the component along the slope is $600\cos 75^\circ = 155.3\,\text{N}$ , which equals $600\sin 15^\circ$ . The perpendicular component is $600\sin 75^\circ = 579.6\,\text{N}$ , which equals $600\cos 15^\circ$ .

6. The two components, with signs

Answer: The $x$ -component is $200\cos 135^\circ = -141.4\,\text{N}$ . The $y$ -component is $200\sin 135^\circ = +141.4\,\text{N}$ . The force points into the second quadrant, so the $x$ -component is negative.

7. The velocity over the ground

Answer: Magnitude $= \sqrt{4.0^2 + 3.0^2} = 5.0\,\text{m}\,\text{s}^{-1}$ . Direction $= \tan^{-1}\!\left(\dfrac{3.0}{4.0}\right) = 36.9^\circ$ from the line straight across, on the downstream side.

8. The angle and the vertical component

Answer: From $\cos\theta = \dfrac{8.0}{12} = 0.667$ , the angle is $\theta = 48.2^\circ$ . The vertical component is $12\sin 48.2^\circ = 8.9\,\text{N}$ .

9. The velocity over the ground

Answer: The two velocities are at right angles. Magnitude $= \sqrt{60^2 + 20^2} = 63.2\,\text{m}\,\text{s}^{-1}$ . Direction $= \tan^{-1}\!\left(\dfrac{20}{60}\right) = 18.4^\circ$ north of east.

10. The scaled displacement

Answer: Multiplying a vector by a positive scalar multiplies each component by that scalar: $3.0 \times (4.0, -2.0) = (12.0, -6.0)\,\text{m}$ . The direction is unchanged, and the magnitude is three times larger.

11. The method and its reason

The two forces are not at right angles, so the resultant cannot be found from Pythagoras alone. The triangle formed by the two forces has no right angle.
Draw the two forces head to tail as a scale drawing, choose a scale such as 1 cm to 10 N, and measure the resultant with a ruler and a protractor.

Answer: The forces are not at right angles, so Pythagoras does not apply directly. Draw them head to tail to scale and measure the resultant. A calculation by resolving each force into components also works, but it is longer than a careful scale drawing for two forces.

12. The magnitude and the angle

Answer: Magnitude $= \sqrt{5.0^2 + 12.0^2} = 13.0\,\text{m}$ . The angle from the eastward direction is $\tan^{-1}\!\left(\dfrac{12}{5.0}\right) = 67.4^\circ$ north of east.