Forces and friction

Tilt a slope slowly. Nothing, nothing, then it slides.

A block sits still on a gentle slope and stays put as you raise it, right up to one particular angle. Friction explains both the waiting and the moment it gives way.

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Split the weight

The block has mass 2 kg, so its weight is $2g$ , straight down. On a slope at $\alpha$ , split it along and into the slope:

\begin{aligned} &2g\sin\alpha \text{ down the slope} \\ &2g\cos\alpha \text{ into it} \end{aligned}

The slope pushes back with $R = 2g\cos\alpha$ .

Just enough friction

Friction supplies exactly what is needed, $2g\sin\alpha$ up the slope, but it can never exceed $\mu R$ . With $\mu = 0.5$ , the two meet when

2g\sin\alpha = 0.5 \times 2g\cos\alpha

that is $\tan\alpha = 0.5$ , at $\alpha = 26.6^\circ$ . The block is on the point of sliding.

It slides

Tilt it to $\tan\alpha = \tfrac{3}{4}$ , so $\sin\alpha = \tfrac{3}{5}$ and $\cos\alpha = \tfrac{4}{5}$ . The slope now pulls with $1.2g = 11.76$ N, and friction is stuck at its limit, $0.8g = 7.84$ N. The difference accelerates the block.

Any mass, same angle

On the point of sliding, $mg\sin\alpha = \mu mg\cos\alpha$ . The mass cancels, leaving

\tan\alpha = \mu

A heavier block starts sliding at exactly the same angle.

Chapter 2

Friction does just enough, until it can't.

Resolving forces at angles, resultants, particles hanging from strings, sledges pulled at an angle, friction changing direction, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›