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Mathematical Requirements (Appendix C)

Resolving vectors, Pythagoras and trigonometry

  • 3 things to know
  • All three components

Overview

Recall questions on resolving a vector into perpendicular components, recombining components with Pythagoras and the inverse tangent, when to use the sine and when the cosine, resolving along a slope, and worked calculations with directions stated.

What you need to know

Answer each one before you reveal it.

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Key term

How is a vector resolved into two perpendicular components?

Key term

Why must a direction always be stated when a vector quantity is quoted?

Key term

A block rests on a slope inclined at 20° to the horizontal. Give the components of its weight parallel and perpendicular to the slope.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    How is a vector resolved into two perpendicular components?

    Model answer

    The component along a chosen direction is the magnitude multiplied by the cosine of the angle to that direction, and the perpendicular component uses the sine of the same angle. The two components together are fully equivalent to the original vector.

  2. 2

    A force of 250 N acts at 35° above the horizontal. Calculate its horizontal and vertical components.

    Model answer

    The horizontal component is 250 cos 35° = 205 N and the vertical component is 250 sin 35° = 143 N. Working back, √(205² + 143²) = 250 N and tan⁻¹(143/205) = 35°.

  3. 3

    A body is displaced 3.0 m east and then 4.0 m north. Calculate the magnitude and direction of the resultant displacement.

    Model answer

    The magnitude is √(3.0² + 4.0²) = 5.0 m, and the direction is tan⁻¹(3.0/4.0) = 37° east of north. Working back, 5.0 sin 36.9° = 3.0 m east and 5.0 cos 36.9° = 4.0 m north.

  4. 4

    Why must a direction always be stated when a vector quantity is quoted?

    Model answer

    A vector is not fully specified by its magnitude alone, and two vectors of equal size in different directions have entirely different effects. A momentum of 5 kg m s⁻¹ north is not the same as 5 kg m s⁻¹ south.

  5. 5

    A block rests on a slope inclined at 20° to the horizontal. Give the components of its weight parallel and perpendicular to the slope.

    Model answer

    The component down the slope is mg sin 20° and the component pressing into the slope is mg cos 20°. Here m is the mass in kg and g is 9.81 N kg⁻¹.

  6. 6

    A 40 kg block sits on a slope of 20°. Calculate the component of its weight acting down the slope.

    Model answer

    The weight is 40 × 9.81 = 392 N, so the component is 392 sin 20° = 134 N. Working back, 134/sin 20° = 392 N, the full weight.

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