Work, energy and momentum questions
Calculate work and power, conserve signed momentum and compare collision energies.
Forces, components and equilibrium
Resolve forces, balance moments and distinguish force from torque.
Practise
Resolve horizontallyA force of 48 N acts at 20° above the positive horizontal direction. Find the magnitude of its horizontal component. Give your answer to 3 significant figures.
Larger diagram
Treat the given values as exact unless an uncertainty is stated.
Hint
The angle is measured from horizontal: horizontal component = F cos θ.
Worked solution
- Fx = 48 × cos(20°) N.
- Answer: 45.1 N (3 significant figures).
Question type: Mixed practice
Report a problem
Resolve from the stated angle
If θ is measured from horizontal, Fx = F cos θ and Fy = F sin θ. On a slope at θ to horizontal, weight has a down-slope component mg sin θ and a perpendicular component mg cos θ. The contact force equals the perpendicular weight component only when no other forces or acceleration contribute in that direction.
Moments use perpendicular distance
Moment magnitude = force × perpendicular distance from pivot to line of action. If the lever length is L and the force makes angle θ to it, the moment is FL sin θ. Balance clockwise and anticlockwise moments about the same point. A couple has zero resultant force and torque Fd, where d is the separation between the force lines.
Two equilibrium conditions
For a rigid body in equilibrium, resultant force and resultant moment are both zero. Constant velocity is compatible with equilibrium. This practice covers selected components, right-angle resultants and simple moment balances, not all free-body diagrams or centre-of-mass problems.
A-level relevance
AQA 7408: 3.4.1.1–2 (selected); OCR A H556: 2.1.1; 3.2.1, 3.2.3 (selected). Tasks cover the selected skills listed above.
AQA mechanics and materials specification · OCR A specification