WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Basic Physics
Resolving forces on an inclined plane
- 3 things to know
- AS Unit 1
Overview
Recall questions on Resolving forces on an inclined plane.
What you need to know
Answer each one before you reveal it.
At what angle of slope are the down-slope and perpendicular components of a body's weight equal?
An 8.0 kg sledge slides down a frictionless slope inclined at 15° to the horizontal. Calculate its acceleration, taking g as 9.81 m s⁻².
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the component of a body's weight acting down a slope inclined at angle θ to the horizontal.
Model answer
It is W sin θ, where W is the weight of the body in newtons.
- 2
State the component of a body's weight acting perpendicular to the surface of a slope inclined at angle θ to the horizontal.
Model answer
It is W cos θ, directed into the surface.
- 3
Explain the geometry that makes the down-slope component of a weight W sin θ rather than W cos θ.
Model answer
The weight acts vertically and the normal to the slope is tilted from the vertical by the slope angle, so the angle between the weight and the normal is θ. That puts W cos θ along the normal and leaves W sin θ along the surface.
- 4
A 12 kg crate sits on a slope at 20° to the horizontal. Calculate the component of its weight acting down the slope, taking g as 9.81 N kg⁻¹.
Model answer
The weight is 12 × 9.81 = 118 N and the down-slope component is 118 sin 20° = 40.3 N.
- 5
A 12 kg crate rests without moving on a slope at 20° to the horizontal. Calculate the normal contact force from the surface, taking g as 9.81 N kg⁻¹.
Model answer
The normal contact force balances the perpendicular component of the weight, giving 12 × 9.81 × cos 20° = 111 N.
- 6
A crate rests in equilibrium on a rough slope. State the magnitude and direction of the frictional force acting on it.
Model answer
The friction is equal in magnitude to the down-slope component of the weight and acts up the slope.
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