00:01
Hi, in this question, the weight of the chest w is given as 1 ,400 newtons.
00:12
And this chest is being pushed in an upward direction in a ramp of slope 1 by 4.
00:20
That means the height is given as from the horizontal position, the height is given as 1 meters and the length of the ramp is given as 4 meters.
00:29
So we need to determine the angle.
00:31
This angle alpha.
00:34
Now, the value of mu k, that is the coefficient of kinetic friction, mu k is given as 0 .2, that is the coefficient of kinetic friction between the chest and the ramp, and the angle of inclination of this force, f, is given as 30 degree.
00:58
Let us say, theta is given as 30 degree.
01:01
With respect to horizontal this is with respect to horizontal now let us find out this angle alpha the sign alpha can be written as sine alpha is equal to 1 by 4 that is 1 meter height divided by 4 meters the length of the ramp on calculation we obtained alpha is equal to 14 point 48 degrees.
01:35
Now coming to the beta, angle beta, this angle beta we need to determine.
01:43
This force is making an inclination of 30 degree with respect to horizontal.
01:48
Therefore the angle beta will be equal to 30 degree with respect to horizontal minus the inclination of the ramp is 14 .48 degrees.
02:02
Therefore, peta will be 15 .52 degrees.
02:08
15 .52 degrees.
02:11
Now for euclibrium, let us resolve the forces perpendicular to the plane.
02:18
When we resolve the forces perpendicular to the plane, we can write the equation n is acting in the upward direction of the perpendicular, upward direction on the rapid direction on the rapid.
02:32
Rapp is equal to w.
02:38
Kos alpha minus f sine beta.
02:47
This w...