A 12 kg block is pushed up a rough plane inclined at 37° by a horizontal force of 200 N. If the coefficient of friction is 0.20 and $\sin 37^\circ = 0.6$, $\cos 37^\circ = 0.8$, and $g = 10\,\text{m/s}^2$, what is the acceleration of the block?

What you need to do here

Because the applied force is horizontal, it has one component up the slope and another component pushing the block into the slope. That second component increases the normal reaction, which increases friction, so we must compute $N$ before using $f=\mu N$.

Resolve the 200 N horizontal force relative to the plane

The plane makes angle $\theta = 37^\circ$ with the horizontal, so the horizontal force makes angle $\theta$ with the plane.

• Component up the plane: $$F_{\parallel} = 200\cos\theta = 200(0.8)=160\,\text{N}$$
• Component into the plane: $$F_{\perp} = 200\sin\theta = 200(0.6)=120\,\text{N}$$

Weight components

With $m=12\,\text{kg}$ and $g=10\,\text{m/s}^2$, $mg=120\,\text{N}$.

• Down the plane: $$mg\sin\theta = 120(0.6)=72\,\text{N}$$
• Into the plane: $$mg\cos\theta = 120(0.8)=96\,\text{N}$$

Normal reaction and friction

The normal reaction balances the total force into the plane: $$N = mg\cos\theta + F_{\perp} = 96 + 120 = 216\,\text{N}$$

Friction opposes the motion (block is pushed up, so friction acts down): $$f = \mu N = 0.20(216)=43.2\,\text{N}$$

Net force up the plane and acceleration

Take up the plane as positive: $$F_{\text{net}} = 160 - 72 - 43.2 = 44.8\,\text{N}$$

Then $$a = \frac{F_{\text{net}}}{m} = \frac{44.8}{12} = 3.73\,\text{m/s}^2$$

So the block accelerates at about $3.7\,\text{m/s}^2$ up the plane.