Question

A mass of 6 kg is placed on an incline with friction and no additional forces acting on it and it begins to slide at some angle when the angle is slowly increased. Now, the same mass is placed on the same incline, but an additional force of 17 Newtons is applied to it parallel to the ground away from the incline (it's always in that direction... also it's parallel to the ground, not the incline.). As the angle of the incline is slowly increased, it slides at a different angle with this force. If the angle which it begins to slide with the force is 24 degrees, how much further can the angle be raised with the force than without the force? Answer in degrees.

Answer 1

The angle at which block slides is called angle of repose

$\mu = tan\theta$ = tan 24 = 0.45

dN 2

Using Fy=0

R - mgcos$\theta$ - dNsin$\theta$ = 0

R - 58.8cos$\theta$ - 17sin$\theta$ = 0 ...i

Using Fx=0

dNcos$\theta$ + 0.45R - mgsin$\theta$ = 0

17cos$\theta$ + 0.45R - 58.8sin$\theta$ = 0

R = (58.8sin$\theta$ - 17cos$\theta$) / 0.45 ....ii

put ii in i

(58.8sin$\theta$ - 17cos$\theta$) / 0.45 - 58.8cos$\theta$ - 17sin$\theta$ = 0

solving for $\theta$ we get

$\theta$ = 40.35 deg