Question

the drawing shows a block at rest on an incline. the mass of the block is 8.0 kg. what is the static frictional force that acts on the block?
73 n
32 n
78 n
29 n
there is not enough information to calculate the frictional force.

Explanation:

Step1: Analyze forces along incline

Since the block is at rest, the static - frictional force $f_s$ balances the component of the gravitational force along the incline. The gravitational force is $F_g = mg$, where $m = 8.0\ kg$ and $g=9.8\ m/s^2$. The component of the gravitational force along the incline is $F_{g\parallel}=mg\sin\theta$, with $\theta = 22^{\circ}$.

Step2: Calculate the component of gravitational force

$F_{g\parallel}=mg\sin\theta=8.0\times9.8\times\sin(22^{\circ})$.
We know that $\sin(22^{\circ})\approx0.375$.
So $F_{g\parallel}=8.0\times9.8\times0.375 = 8\times9.8\times\frac{3}{8}=29.4\ N\approx29\ N$. And $f_s = F_{g\parallel}$.

Answer:

29 N