How to find friction on a slope at constant speed

La domanda

La soluzione

1. 01The key words are "at constant speed". Constant speed means no acceleration, so the forces cancel: $v = \text{constant} \Rightarrow a = 0 \Rightarrow \sum F = 0$. Write this down before anything else.
2. 02Draw the forces before resolving anything, so that you do not invent a force that is not there or miss one that is. The tension $T = 1.7\times 10^{4}\ \text{N}$ acts up the slope along the cable.
3. 03The weight $W = 2.0\times 10^{4}\ \text{N}$ acts straight down towards the centre of the Earth, not into the slope, and drawing it into the slope costs marks. The normal force acts perpendicular to the surface of the ramp.
4. 04Friction acts down the slope, because it opposes the motion and the slab is moving up. Decide which way the slab is going before you draw friction; if the slab slid back, friction would point up the slope.
5. 05Resolve the weight. The angle between the weight and the normal force is the same 15° as the ramp, so the component pressing into the slope takes the cosine and the component along the slope takes the sine: $W\cos\theta$ perpendicular to the ramp and $W\sin\theta$ along it.
6. 06Along the slope the tension balances friction plus the weight component: $T = f + W\sin\theta$. The normal force does not appear in this equation because it has no component along the slope.
7. 07Rearrange and substitute: $f = T - W\sin\theta = 1.7\times 10^{4} - 2.0\times 10^{4}\sin 15^\circ$. Since $W\sin 15^\circ = 5.2\times 10^{3}\ \text{N}$, the friction force is $f = 1.2\times 10^{4}\ \text{N}$.
8. 08To find the coefficient of friction, first find the normal force: $N = W\cos\theta = 2.0\times 10^{4}\cos 15^\circ = 1.9\times 10^{4}\ \text{N}$. Divide by $N$, not by the weight, because friction is proportional to the normal force and the two differ on a slope: $\mu = \frac{f}{N} = \frac{1.2\times 10^{4}}{1.9\times 10^{4}} = 0.61$. Using $W$ would give 0.60, which is wrong.
9. 09A value of 0.61 is close to the figure of about 0.6 for stone dragged over stone, so the result is physically sensible. If the winch pulls the slab at 0.25 m s⁻¹, the power it must deliver is $P = Tv = 1.7\times 10^{4} \times 0.25 = 4.3\ \text{kW}$.

Risposta
The friction force is 1.2 × 10⁴ N (12 kN), acting down the slope, and the coefficient of friction is 0.61.