00:02
In this problem a block of steel is at rest on the horizontal table and the coefficient of a static friction between the block and the table is given and we have to find out the magnitude of the horizontal force that will put the block on the verge of moving.
00:24
So the forces acting on the block are normal force in the upward direction, weight mg in the downward direction, applied force and friction force opposite to the applied force.
00:46
In the vertical direction normal force balances the weight so friction force will be maximum because it is on the verge of moving.
00:57
So the maximum static friction force is coefficient of static friction times normal force.
01:07
Now since it is on the verge of moving the applied force must be equal to the friction force.
01:20
Coefficient is mass g so we get force 56 newton.
01:39
For part b the applied force is upward 60 degree from the horizontal.
02:03
Forces are applied force, normal force, gravitational force and the friction force.
02:11
Now we break the applied force into two components.
02:18
The horizontal component will be f cos theta where theta is the angle with the horizontal and vertical component f sin theta.
02:28
Now in the vertical direction there is no acceleration so normal force plus f sin theta in the upward direction will balance the weight in the downward direction.
02:40
So normal force is friction force coefficient of friction times normal force and in the horizontal direction the horizontal component of applied force will be equal to friction force when it is on the verge of moving.
03:26
So we get f cos theta plus so applied force is putting the values coefficient mass g angle and we get force 59 newton...