A.2 · IB Physics
Inclined plane and friction
A block on a rough slope. Change the angle, mass and both friction coefficients, push it with a force along the slope or flick it up the slope, then release it. The free-body diagram is drawn to scale, so you can see which force wins.
Predict first
Start from the default settings (θ = 30°, μs = 0.40, μd = 0.30, m = 2.0 kg, P = 0). If you double the mass to 4.0 kg and release, what happens to the block's acceleration?
Weight along the slope and friction are both proportional to m, so m cancels: a = g(sinθ − μd cosθ), about 2.3 m s⁻² down the slope for any mass.
Check it in the model: Press Release and note Acceleration a, then move the Mass m slider from 2.0 to 4.0 kg, release again and compare.
Up the slope is positive. The free-body diagram is drawn to scale.
at rest: slips if |P − mg sinθ| > μs N
sliding: f = μd N, opposing the motion
a = (P − mg sinθ ∓ μd mg cosθ) ÷ m
with P = 0, slips when tanθ > μs
- Will it slip from rest?
- Acceleration a
- Friction force
- Normal force N
- Velocity v
- Time t
- Slip angle arctan μs (P = 0)
Try this
- Keep P = 0 and u = 0 with μs = 0.40 and release the block at several angles. At what whole-degree angle does it first slip, and how does that compare with arctan μs in the readout?
- Use the “Slides (35°)” preset, then change the mass from 2 kg to 8 kg and release again. Compare the accelerations: what does that say about mg and friction?
- Use “Flick up and slide back” and compare the gradient of the v–t graph on the way up with the gradient on the way down. Which is steeper, and which way does friction act in each part?
Common error: writing static friction as μs N every time, when μs N is only its largest value; at rest the friction is exactly what balances the other forces (here mg sinθ − P), and N on a slope is mg cosθ, not mg.