A stone block of mass m = 200 kg lies on an incline of angle…

A stone block of mass $m = 200\ \mathrm{kg}$ lies on an incline of angle $\alpha = 15^\circ$. To move it down the incline at constant speed, a force $F = 490\ \mathrm{N}$ along the incline is needed. (a) What is the coefficient of friction? (b) What force along the incline moves the block up at constant speed?

Explanation of Method

Pushing a stone block down and up an incline.

Step 1: Components of the weight

\[ mg\sin\alpha = 1962 \cdot 0.2588 \approx 507.8\ \mathrm{N}, \]

\[ mg\cos\alpha = 1962 \cdot 0.9659 \approx 1895.1\ \mathrm{N}. \]

Step 2: (a) Moving down

The push $F$ and the slope component both act down the incline; friction, opposing the motion, acts up it: $F + mg\sin\alpha = \mu\,mg\cos\alpha \;\Longrightarrow\; \mu = \frac{490 + 507.8}{1895.1}$.

\[\mu \approx 0.53\]

Step 3: (b) Moving up

Now both the slope component and friction act down the incline: $F_{\text{up}} = mg\sin\alpha + \mu mg\cos\alpha = 507.8 + 997.8$.

\[F_{\text{up}} \approx 1506\ \mathrm{N}\]

Check

Friction $ \mu mg\cos\alpha = 997.8\ \mathrm{N} $ is the same both ways; $ F_{\text{up}} - F_{\text{down}} = 2mg\sin\alpha = 1015.6\ \mathrm{N}\ \checkmark $

$\mu = 0.53 \gt \tan 15^\circ = 0.27$: the block stays put on its own, which is why it has to be pushed down $\checkmark$

$15^\circ$ $\text{(a) pushed down}$ $\text{(b) pulled up}$ $490\ \mathrm{N}$ $f$ $1506\ \mathrm{N}$ $f$ $f = \mu mg\cos\alpha \approx 998\ \mathrm{N}, \quad mg\sin\alpha \approx 508\ \mathrm{N}$

Friction always opposes the motion: up the slope when the block is pushed down, down the slope when it is pulled up. The slope component $mg\sin\alpha$ always points down.

Result

\[\boxed{\,\mu \approx 0.53\,}\]

\[\boxed{\,F_{\text{up}} \approx 1506\ \mathrm{N}\,}\]

Remark

About the original solution. It wrote $F + f = mg\sin\alpha$, i.e. friction pointing down the slope together with the push. Friction opposes the motion, so it points up. The resulting $\mu = 0.01$ would mean the block slides down by itself ($\tan 15^\circ = 0.27$), and no push would be needed. The correct values are $\mu \approx 0.53$ and $F_{\text{up}} \approx 1506\ \mathrm{N}$ (not 527.75 N).