# A block weighing 10 kg is on a plane with an incline of pi/6 and friction coefficient of 1/2. How much force, if any, is necessary to keep the block from sliding down?

Mar 1, 2018

$6.65 N$

#### Explanation:

Here downward component of weight of the block acting to pull the block down along the plane is $m g \sin 30 = 10 \left(\frac{1}{2}\right) g = 5 g$

And,frictional force acting upwards to balance that force is $\mu m g \cos 30 = 4.32 g$(where, $\mu$ is the coefficient of friction)

So,net force acting on the object is $5 g - 4.32 g = 0.68 g = 6.65 N$ downwards along the plane,

So,this much force must be supplied to the block upwards along the plane to balance this extra force.