00:01
In our question we are given, a 2kg block is held in equilibrium on an incline of angle 60 degree, that is given as theta.
00:08
By a horizontal force f that is applied in the direction as shown in a free body diagram.
00:14
Now if the coefficient of static friction between the block and the incline is given to us as point 3, we need to determine the minimum value of the force applied and the normal force exerted by the incline on the block.
00:28
Now for equilibrium condition, the normal force is given equal to f -sign theta plus m -g -cos -theta, which we will designate as equation 1.
00:53
Also, f -coss -theta plus friction force is equal to m -g -sign -theta.
01:03
Now we have taken friction acting up the inclination as we are supposed to.
01:08
To find the minimum force f that is required to keep it in equilibrium.
01:12
Therefore we have f cos theta plus mu, substituting the value f sine theta plus mg cos theta being equal to mg sine theta in above equation...