00:01
In this problem, a block of mass 29 kg is sitting on a surface inclined at 44 degrees and the coefficient of static friction is 0 .56 between block and surface.
00:15
Using this we have to find out the minimum force necessary to prevent slipping.
00:21
So the forces acting on the block are the gravitational force in the downward direction and the normal force perpendicular to the slope and friction force along the direction up the slope because the block tends to slip down the slope.
00:47
So let's suppose the force acting on the block is f parallel to the incline.
01:00
So now the angle of the gravitational force which is vertical downward with a direction perpendicular to incline is theta where theta is the angle of the incline.
01:16
So the component of gravitational force perpendicular to incline is mg cos theta and the component of gravitational force parallel to incline is mg sin theta.
01:42
So these are the two components of gravitational force on the block.
01:50
So for this problem we will take the direction up the incline as positive x axis and direction perpendicular to incline as positive y axis.
02:04
So here the block is prevented from slipping which means the block is in equilibrium.
02:13
So the forces along y axis must add to 0.
02:18
Now forces along y axis are normal force and minus mg cos theta along negative y axis and this will be equal to 0...