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AS & A Level · Practice questions · Forces, components and resultants

Resolve, add, rebuild.

Six original Cambridge-style questions on force diagrams, components and resultants, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Two directions, signs, then Pythagoras.

01
Recall
[2 marks]

A force of 40 N acts at \(25^\circ\) above the horizontal. Find its horizontal and vertical components.

  • Horizontal: \(40\cos 25^\circ = 36.3\) N. ✓
  • Vertical: \(40\sin 25^\circ = 16.9\) N. ✓
02
Calculation
[3 marks]

Forces of 9 N horizontally to the right and 12 N vertically downwards act on a particle. Find the magnitude and direction of the resultant.

  • \(R = \sqrt{9^2 + 12^2} = 15\) N. ✓
  • \(\tan\alpha = \dfrac{12}{9}\), so \(\alpha = 53.1^\circ\). ✓
  • The resultant is 15 N at \(53.1^\circ\) below the horizontal, to the right. ✓
03
Calculation
[5 marks]

Three coplanar forces act at a point: 25 N along the positive \(x\)-axis, 18 N along the positive \(y\)-axis and 30 N at \(30^\circ\) below the negative \(x\)-axis. Find the magnitude of the resultant and the angle it makes with the positive \(x\)-axis.

  • \(X = 25 - 30\cos 30^\circ = 25 - 25.98 = -0.981\). ✓
  • \(Y = 18 - 30\sin 30^\circ = 18 - 15 = 3\). ✓
  • \(R = \sqrt{0.981^2 + 3^2} = 3.16\) N. ✓
  • \(X < 0,\ Y > 0\): the resultant points up and to the left, at \(\tan^{-1}\dfrac{3}{0.981} = 71.9^\circ\) above the negative \(x\)-axis. ✓
  • Angle with the positive \(x\)-axis \(= 180^\circ - 71.9^\circ = 108.1^\circ\). ✓
04
Analysis
[4 marks]

Forces of \(P\) N and 20 N act at a point, and the angle between their directions is \(120^\circ\). The resultant is perpendicular to the \(P\) N force. Find \(P\) and the magnitude of the resultant, giving the resultant in exact form.

  • Resolve along the \(P\) N force: \(P + 20\cos 120^\circ = 0\), so \(P - 10 = 0\), \(P = 10\). ✓
  • Resolve perpendicular: \(R = 20\sin 120^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\) N (17.3 N). ✓
05
Analysis
[4 marks]

A particle of weight 60 N is on a plane inclined at \(35^\circ\) to the horizontal. A force of 25 N acts on it at \(20^\circ\) above a line of greatest slope, pointing up the plane. Find the total component of these two forces (a) parallel to the plane, (b) perpendicular to the plane. State the direction of each.

  • (a) Up the plane: \(25\cos 20^\circ - 60\sin 35^\circ = 23.49 - 34.41 = -10.9\). ✓
  • So 10.9 N down the plane. ✓
  • (b) Away from the plane: \(25\sin 20^\circ - 60\cos 35^\circ = 8.55 - 49.15 = -40.6\). ✓
  • So 40.6 N into the plane (this is what the normal reaction must balance). ✓
06
Analysis
[5 marks]

A barge is pulled along a straight canal by two ropes. One rope has tension 300 N and makes \(25^\circ\) with the direction of the canal; the other has tension \(T\) N and makes \(40^\circ\) with the canal, on the other side. The resultant of the two tensions acts along the canal. Find \(T\) and the magnitude of the resultant.

  • Across the canal the components cancel: \(300\sin 25^\circ = T\sin 40^\circ\). ✓
  • \(T = \dfrac{300 \times 0.4226}{0.6428} = 197\) N (3 s.f.). ✓
  • Along the canal: \(R = 300\cos 25^\circ + 197.2\cos 40^\circ = 271.9 + 151.1\). ✓
  • \(R = 423\) N (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.