00:01
Let us begin with subpart a.
00:03
It is given a box of mass n equals to 10 kilogram rest on a 35 degree inclined plane with the horizontal.
00:14
So let us draw the free body diagram.
00:20
We have the inclined plane which is a 35 degree inclined to be horizontal.
00:30
We have the mass box of mass and mass.
00:37
M equals to 10 hologram resting on the surface.
00:42
A string is used to keep the box in equilibrium.
00:46
So the string is given in this direction.
00:49
So there is a tension in the string p.
00:53
So let this be mg, which is the weight of the mass.
01:01
The normal of the mass acting of the mass is n.
01:06
There is a downward force acting against the normal which is the g cost theta your theta is not 5 degree string we can resolve it into two components so we can draw the two components which is p post alpha or alpha is the angle which will be the exponent of the string and this is t sine alpha so the string makes an ankle alpha which is given as 25 degrees alpha equals 25 degrees with the inclined plane now the coefficient of friction between the box and the inclined place is given as 0 .3 so here we have mu this is the coefficient of friction so here the no for the normal force it will become mu mg plus theta so we have t sine alpha p force alpha p p and then there is the frictional force which is acting which is acting in the opposite side so here we have f s which is the static frictional force here we have m .g.
02:38
Sign peta ng sine theta where peta is 35 degrees so this is the free body free body diadem of the rocks of mass so this is the answer for sub part a.
03:04
Now let us move on to subart b...