00:01
In this problem, we have been given that there is a block of mass 36 kilograms and this is kept on the incline.
00:08
So we have to determine the maximum value of the angle of incline so that the block does not slide.
00:15
So in this case, if we observe the free body diagram, which shows all the forces, there will be weight acting in the vertically downward direction.
00:23
There will be normal force and there will be a static friction which will be acting along the incline.
00:30
In the upward direction.
00:32
So the work of the static friction is to just oppose the component of weight along the incline.
00:39
So we just project the weight.
00:41
So we get mg cos alpha here because this angle is alpha by just geometrical observation we can get.
00:47
And this component will be mg sine alpha.
00:51
So to figure out the angle, we just equate the forces along the horizontal directions and the vertical direction because the block is stationary.
01:00
So the friction force which is mu times the normal reaction and the normal reaction is equal to mg cos alpha from the diagram.
01:09
So this is equal to mg sine alpha.
01:12
So from here we can just cut the values of mg from both sides and we can say that mu will be equal to tan alpha.
01:21
So mu here we are going to take the static friction coefficient because the block is not moving...