Physics · Simulation and virtual lab Free
Forces on a slope: friction, components and the angle of sliding
Tilt the plank until the block slides, see every force on the block and measure the coefficient of friction.
- You will learn to
- Identify weight, normal reaction and friction on a slope and resolve the weight into components.
- Find the angle of sliding and calculate μ = tan θ.
- Measure and calculate the acceleration of a block sliding down a slope.
- You will measure
- Sliding angle (°), time to slide 1.2 m (s), acceleration (m/s²).
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: plank, wooden block, protractor, metre rule, stopwatch, rubber and plastic surfaces.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To identify the forces acting on a block on an inclined plane, to resolve the weight into components, to find the angle at which the block starts to slide and so the coefficient of static friction, and to measure the acceleration of the sliding block.
What to do
- Keep wood on wood and 1.0 kg. Set the angle to 10° and tap Release the block. Does it move? Record.
- Increase the angle by 2° at a time. Find the smallest angle at which the block slides. Record each try.
- Calculate μ (static) = tan θ at the sliding angle. Repeat with 2.0 kg: does the angle change?
- Make the slope steeper than the sliding angle. Use the time to slide 1.2 m to find a = 2s ÷ t².
- Change to rubber on concrete and to steel on ice. Find each sliding angle.
- Compare your measured acceleration with a = g(sin θ − μ cos θ) using μ (sliding).
Results
Questions
Name the three forces acting on a block resting on a slope.
Its weight (mg, vertically down), the normal reaction (perpendicular to the slope) and friction (along the slope, opposing the motion or the tendency to move).
Write the components of the weight along and perpendicular to the slope.
Along the slope: mg sin θ. Perpendicular to the slope: mg cos θ (balanced by the normal reaction).
Why does the block stay still at small angles?
Static friction can grow up to μN. At small angles mg sin θ is smaller than this maximum, so friction balances it.
Show that a block starts to slide when tan θ = μ.
It slides when mg sin θ = μ mg cos θ, so tan θ = μ. The mass cancels.
Why do lorries skid more easily on wet or icy roads?
The coefficient of friction between tyre and road is smaller, so the maximum friction force is smaller.
Why is the acceleration less than g sin θ?
Sliding friction μN acts up the slope, so the resultant force is mg sin θ − μ mg cos θ.
For teachers: expected results and how the model works
Model values (typical, for teaching): wood on wood μs = 0.50, μk = 0.35 (block starts to slide between 26° and 27°, since tan⁻¹ 0.50 = 26.6°); rubber on concrete μs = 0.90, μk = 0.70 (42.0°); steel on ice μs = 0.10, μk = 0.05 (5.7°). g = 9.78 m/s². The block slides 1.20 m. Wood at 30°: a = 1.93 m/s², t = 1.12 s; at 40°: a = 3.66 m/s², t = 0.81 s. The mass changes the size of every force but not the sliding angle or the acceleration. Arrows are drawn to scale; the dashed red arrows are the components of the weight.
Exam practice Exam set
National-exam style questions on this simulation (16 marks). Try each one on paper first, then open the marking scheme.
A 4 kg box rests on a slope at 30°. Take g = 9.8 m/s². Calculate (a) the component of its weight down the slope, (b) the normal reaction, (c) the friction force if the box does not move. (6 marks)
Marking scheme: (a) mg sin 30° = 4 × 9.8 × 0.5 = 19.6 N (2). (b) mg cos 30° = 4 × 9.8 × 0.866 = 33.9 N (2). (c) Friction balances the pull: 19.6 N up the slope (2).
A block just starts to slide when a plank is raised to 31°. Find the coefficient of static friction. (2 marks)
Marking scheme: μ = tan 31° (1) = 0.60 (1).
A 2 kg block slides down a 40° slope where μ (sliding) = 0.35. Calculate its acceleration (g = 9.8 m/s²). (4 marks)
Marking scheme: Down the slope: mg sin 40° = 12.6 N (1); friction = 0.35 × 2 × 9.8 × cos 40° = 5.3 N (1); resultant = 7.3 N (1); a = 7.3 ÷ 2 = 3.7 m/s² (1).
State two ways of increasing friction and two ways of reducing it, with an everyday example of each. (4 marks)
Marking scheme: Increase: rougher surfaces / tyre treads (1), larger normal force / heavier load (1). Reduce: lubrication with oil (1), ball bearings or smooth surfaces (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Forces on a slope: friction, components and the angle of sliding
Aim: To identify the forces acting on a block on an inclined plane, to resolve the weight into components, to find the angle at which the block starts to slide and so the coefficient of static friction, and to measure the acceleration of the sliding block.
Apparatus (real laboratory): plank, wooden block, protractor, metre rule, stopwatch, rubber and plastic surfaces.
What to do (with the simulation on www.donatstem.co.rw)
- Keep wood on wood and 1.0 kg. Set the angle to 10° and tap Release the block. Does it move? Record.
- Increase the angle by 2° at a time. Find the smallest angle at which the block slides. Record each try.
- Calculate μ (static) = tan θ at the sliding angle. Repeat with 2.0 kg: does the angle change?
- Make the slope steeper than the sliding angle. Use the time to slide 1.2 m to find a = 2s ÷ t².
- Change to rubber on concrete and to steel on ice. Find each sliding angle.
- Compare your measured acceleration with a = g(sin θ − μ cos θ) using μ (sliding).
Results
| # | Surfaces | Angle (°) | Moves? (yes/no) | mg sin θ (N) | Time to slide 1.2 m (s) | Acceleration (m/s²) |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
| 9 | | | | | | |
| 10 | | | | | | |
Graph
Plot acceleration (m/s²) against angle (°) for one pair of surfaces. Mark the angle at which sliding starts.
Questions
- Name the three forces acting on a block resting on a slope. [2]
- Write the components of the weight along and perpendicular to the slope. [2]
- Why does the block stay still at small angles? [2]
- Show that a block starts to slide when tan θ = μ. [2]
- Why do lorries skid more easily on wet or icy roads? [2]
- Why is the acceleration less than g sin θ? [2]
- A 4 kg box rests on a slope at 30°. Take g = 9.8 m/s². Calculate (a) the component of its weight down the slope, (b) the normal reaction, (c) the friction force if the box does not move. [6]
- A block just starts to slide when a plank is raised to 31°. Find the coefficient of static friction. [2]
- A 2 kg block slides down a 40° slope where μ (sliding) = 0.35. Calculate its acceleration (g = 9.8 m/s²). [4]
- State two ways of increasing friction and two ways of reducing it, with an everyday example of each. [4]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Forces on a slope: friction, components and the angle of sliding
Expected results
Model values (typical, for teaching): wood on wood μs = 0.50, μk = 0.35 (block starts to slide between 26° and 27°, since tan⁻¹ 0.50 = 26.6°); rubber on concrete μs = 0.90, μk = 0.70 (42.0°); steel on ice μs = 0.10, μk = 0.05 (5.7°). g = 9.78 m/s². The block slides 1.20 m. Wood at 30°: a = 1.93 m/s², t = 1.12 s; at 40°: a = 3.66 m/s², t = 0.81 s. The mass changes the size of every force but not the sliding angle or the acceleration. Arrows are drawn to scale; the dashed red arrows are the components of the weight.
Answers
- Name the three forces acting on a block resting on a slope. [2]
Its weight (mg, vertically down), the normal reaction (perpendicular to the slope) and friction (along the slope, opposing the motion or the tendency to move).
- Write the components of the weight along and perpendicular to the slope. [2]
Along the slope: mg sin θ. Perpendicular to the slope: mg cos θ (balanced by the normal reaction).
- Why does the block stay still at small angles? [2]
Static friction can grow up to μN. At small angles mg sin θ is smaller than this maximum, so friction balances it.
- Show that a block starts to slide when tan θ = μ. [2]
It slides when mg sin θ = μ mg cos θ, so tan θ = μ. The mass cancels.
- Why do lorries skid more easily on wet or icy roads? [2]
The coefficient of friction between tyre and road is smaller, so the maximum friction force is smaller.
- Why is the acceleration less than g sin θ? [2]
Sliding friction μN acts up the slope, so the resultant force is mg sin θ − μ mg cos θ.
- A 4 kg box rests on a slope at 30°. Take g = 9.8 m/s². Calculate (a) the component of its weight down the slope, (b) the normal reaction, (c) the friction force if the box does not move. [6]
(a) mg sin 30° = 4 × 9.8 × 0.5 = 19.6 N (2). (b) mg cos 30° = 4 × 9.8 × 0.866 = 33.9 N (2). (c) Friction balances the pull: 19.6 N up the slope (2).
- A block just starts to slide when a plank is raised to 31°. Find the coefficient of static friction. [2]
μ = tan 31° (1) = 0.60 (1).
- A 2 kg block slides down a 40° slope where μ (sliding) = 0.35. Calculate its acceleration (g = 9.8 m/s²). [4]
Down the slope: mg sin 40° = 12.6 N (1); friction = 0.35 × 2 × 9.8 × cos 40° = 5.3 N (1); resultant = 7.3 N (1); a = 7.3 ÷ 2 = 3.7 m/s² (1).
- State two ways of increasing friction and two ways of reducing it, with an everyday example of each. [4]
Increase: rougher surfaces / tyre treads (1), larger normal force / heavier load (1). Reduce: lubrication with oil (1), ball bearings or smooth surfaces (1).