A block weighing #9 kg# is on a plane with an incline of #(7pi)/12# and friction coefficient of #1/3#. How much force, if any, is necessary to keep the block from sliding down?

To find the force necessary to keep the block from sliding down the incline, we need to consider the forces acting on the block. These forces include the gravitational force pulling the block downward, the normal force perpendicular to the incline, and the frictional force opposing the motion of the block.

The gravitational force can be calculated as ( F_{\text{gravity}} = mg ), where ( m ) is the mass of the block (9 kg) and ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )).

The normal force can be determined using trigonometry as ( F_{\text{normal}} = mg \cos(\theta) ), where ( \theta ) is the angle of the incline (( \frac{7\pi}{12} )).

The maximum frictional force that can be exerted is given by ( F_{\text{friction}} = \mu \cdot F_{\text{normal}} ), where ( \mu ) is the coefficient of friction (1/3).

If the block is on the verge of sliding down, the force applied parallel to the incline (( F_{\text{applied}} )) will be equal to the maximum frictional force, as the applied force must counteract the frictional force to prevent sliding.

Therefore, ( F_{\text{applied}} = F_{\text{friction}} = \mu \cdot F_{\text{normal}} ).

Substituting the given values:

[ F_{\text{applied}} = \frac{1}{3} \cdot (9 , \text{kg} \cdot 9.8 , \text{m/s}^2 \cdot \cos(\frac{7\pi}{12})) ]

[ F_{\text{applied}} = \frac{1}{3} \cdot (9 \cdot 9.8 \cdot \cos(\frac{7\pi}{12})) ]

[ F_{\text{applied}} \approx \frac{1}{3} \cdot (88.2 \cdot 0.2588) ]

[ F_{\text{applied}} \approx \frac{1}{3} \cdot 22.81016 ]

[ F_{\text{applied}} \approx 7.60339 , \text{N} ]

Therefore, approximately ( 7.60339 , \text{N} ) of force is necessary to keep the block from sliding down the incline.