PhysicsGrade 8–11

Inclined plane

A block sits on a slope. At what angle does it start to slide? The weight is split into two parts to find out.

Loading simulation …

What is it about?

On an inclined plane with the angle of inclination α\alpha, the weight FG=m⋅gF_G = m \cdot g of a body acts vertically downwards. It is resolved into two components:

  • the downhill force FHF_H parallel to the slope (pointing down the slope) and
  • the normal force FNF_N perpendicular to the slope, which presses the body onto the surface.

The triangle of forces FGF_G, FHF_H, FNF_N is similar to the triangle of length ll, height hh and base of the plane. Therefore

FHFG=hlFH=FG⋅sin⁡αFN=FG⋅cos⁡α\frac{F_H}{F_G} = \frac{h}{l} \qquad F_H = F_G \cdot \sin\alpha \qquad F_N = F_G \cdot \cos\alpha

The steeper the plane, the larger FHF_H and the smaller FNF_N. At α=30∘\alpha = 30^\circ the downhill force is exactly half the weight.

Gradient in per cent: Road signs give the gradient as the change in height per horizontal distance, i.e. tan⁡α\tan\alpha. “12 %” means tan⁡α=0.12\tan\alpha = 0.12, so α≈6.8∘\alpha \approx 6.8^\circ.

Friction: A block stays at rest as long as static friction can balance the downhill force. Static friction is at most

FR,max⁡=μs⋅FNF_{R,\max} = \mu_s \cdot F_N

(μs\mu_s: coefficient of static friction, written μH\mu_H in the simulation). If FHF_H is larger, the block slides. While sliding, the smaller kinetic friction FR=μk⋅FNF_R = \mu_k \cdot F_N (in the simulation μG\mu_G) slows it down. The block then slides down with constant acceleration

a=FH−FRm=g⋅(sin⁡α−μk⋅cos⁡α)a = \frac{F_H - F_R}{m} = g \cdot (\sin\alpha - \mu_k \cdot \cos\alpha)

Limiting angle: Without any other force, the block starts to slide when FH=FR,max⁡F_H = F_{R,\max}, i.e. at tan⁡αG=μs\tan\alpha_G = \mu_s. The mass does not matter. Measuring the limiting angle gives the coefficient of static friction.

Force converter: A cart rolls almost without friction. To hold it on the slope or to pull it up steadily, you only need the force FHF_H instead of FGF_G – but the distance ll is longer than the height hh. This is the golden rule of mechanics: FH⋅l=FG⋅hF_H \cdot l = F_G \cdot h.

Try it

  1. Increase the angle by dragging the top end of the plane. Watch FHF_H and FNF_N in the scene and in the graph.
  2. Press Release at 20∘20^\circ and at 35∘35^\circ. Wooden block at 35°
  3. Find the limiting angle for each pair of materials, from which the block slides. Compare with the intersection of FHF_H and FR,max⁡F_{R,\max} in the graph. Rubber on asphalt · Steel on ice
  4. Double the mass. Do the limiting angle and the acceleration change? Mass 2 kg
  5. Hook the spring balance to the cart and set the pulling force so that the cart stays put. Record measurements for several angles. Cart with spring balance
  6. Pull the wooden block up. From which pulling force does it start to move? Pulling up

Tasks

Task 1: Resolving the weight

Load task

A block with m=2 kgm = 2\,\text{kg} lies on a plane with α=30∘\alpha = 30^\circ. Calculate FGF_G, FHF_H and FNF_N.

Show solution

FG=2 kg⋅9.81 Nkg=19.62 NF_G = 2\,\text{kg} \cdot 9.81\,\tfrac{\text{N}}{\text{kg}} = 19.62\,\text{N}, FH=FG⋅sin⁡30∘=9.81 NF_H = F_G \cdot \sin 30^\circ = 9.81\,\text{N}, FN=FG⋅cos⁡30∘≈16.99 NF_N = F_G \cdot \cos 30^\circ \approx 16.99\,\text{N}.

Task 2: Height and length

Load task

A ramp is l=1 ml = 1\,\text{m} long and h=34 cmh = 34\,\text{cm} high. What force do you need at least to hold a 1 kg1\,\text{kg} cart on it (ignore friction)?

Show solution

FH=FG⋅hl=9.81 N⋅0.34≈3.3 NF_H = F_G \cdot \frac{h}{l} = 9.81\,\text{N} \cdot 0.34 \approx 3.3\,\text{N} – about a third of the weight. (With α=20∘\alpha = 20^\circ the simulation shows more precisely 3.36 N3.36\,\text{N}.) Check

Task 3: Does the block slide?

Load task

A wooden block (μs=0.5\mu_s = 0.5) lies on a plane with α=25∘\alpha = 25^\circ. Does it start to slide by itself?

Show solution

tan⁡25∘≈0.47<0.5\tan 25^\circ \approx 0.47 < 0.5 – the block stays put. It only slides from αG=arctan⁡0.5≈26.6∘\alpha_G = \arctan 0.5 \approx 26.6^\circ. Check at 27°

Task 4: Acceleration

Load task

How strongly does the wooden block (μk=0.3\mu_k = 0.3) accelerate at α=35∘\alpha = 35^\circ? How long does it take for 40 cm40\,\text{cm}?

Show solution

a=9.81 ms2⋅(sin⁡35∘−0.3⋅cos⁡35∘)≈3.22 ms2a = 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot (\sin 35^\circ - 0.3 \cdot \cos 35^\circ) \approx 3.22\,\tfrac{\text{m}}{\text{s}^2}. From s=12at2s = \frac12 a t^2 we get t=2⋅0.4 m3.22 m/s2≈0.50 st = \sqrt{\frac{2 \cdot 0.4\,\text{m}}{3.22\,\text{m/s}^2}} \approx 0.50\,\text{s}.

Task 5: Measuring the coefficient of friction

A block of an unknown material starts to slide at α=31∘\alpha = 31^\circ. How large is the coefficient of static friction? Check with your own values. Check

Show solution

μs=tan⁡31∘≈0.60\mu_s = \tan 31^\circ \approx 0.60. At 30∘30^\circ the block still stays put, at 31∘31^\circ it slides.

Task 6: Pulling up

Load task

With what force must you pull the wooden block (m=1 kgm = 1\,\text{kg}, μk=0.3\mu_k = 0.3) at α=20∘\alpha = 20^\circ so that it slides up steadily? Why do you need more force to get it moving?

Show solution

FZ=FH+μk⋅FN=3.36 N+0.3⋅9.22 N≈6.1 NF_Z = F_H + \mu_k \cdot F_N = 3.36\,\text{N} + 0.3 \cdot 9.22\,\text{N} \approx 6.1\,\text{N}. To get it moving you first have to overcome the larger static friction: FH+μs⋅FN≈8.0 NF_H + \mu_s \cdot F_N \approx 8.0\,\text{N}. Try it

Task 7: Gradient in per cent

A road sign warns of a 12 %12\,\% gradient. How large is the angle of inclination? What fraction of a car’s weight acts as the downhill force?

Show solution

tan⁡α=0.12\tan\alpha = 0.12, so α≈6.8∘\alpha \approx 6.8^\circ. The fraction is FHFG=sin⁡6.8∘≈0.12\frac{F_H}{F_G} = \sin 6.8^\circ \approx 0.12, i.e. about 12 %12\,\%. For small angles sin⁡α\sin\alpha and tan⁡α\tan\alpha are almost equal. In the simulation (cart, 1 kg1\,\text{kg}, α=7∘\alpha = 7^\circ): FH≈1.20 NF_H \approx 1.20\,\text{N} out of 9.81 N9.81\,\text{N}. Check

Ideas for the classroom

  • Use: resolving forces on the inclined plane (graphically with the triangle of forces, with h/lh/l or with trigonometry), friction and limiting angle, the inclined plane as a force converter (grades 8–10); the equation of motion a=g(sin⁡α−μkcos⁡α)a = g(\sin\alpha - \mu_k\cos\alpha) rather from grade 10.
  • Terms: Here the component of the weight perpendicular to the plane is called the normal force FNF_N. Some textbooks distinguish it from the equally large force that the surface exerts on the body.
  • Model: plane 1 m1\,\text{m} long, g=9.81 Nkgg = 9.81\,\tfrac{\text{N}}{\text{kg}}. The coefficients of friction are typical values; real values depend strongly on surface, moisture and cleanliness. The cart has a very small rolling friction (0.0050.005). Before release, a hand holds the body; that is why the static friction shown is at most FR,max⁡F_{R,\max}. At the bottom end a stop holds the body.
  • Series of measurements: With the spring balance attached, a measurement can only be recorded if the body stays at rest with this pulling force. For the cart, the points lie on the curve FH(α)F_H(\alpha); for the block there is a whole range of holding forces because of static friction – a good reason for discussion.
  • Common stumbling blocks:
    • “The normal force always equals the weight.” – On the inclined plane it is smaller: FN=FG⋅cos⁡αF_N = F_G \cdot \cos\alpha.
    • “Heavy bodies start sliding earlier.” – The limiting angle depends only on μs\mu_s, not on the mass.
    • “Friction is always μ⋅FN\mu \cdot F_N.” – Static friction is only as large as needed, at most μs⋅FN\mu_s \cdot F_N.
    • “The inclined plane saves work.” – It saves force, but you have to cover a longer distance.