Question

For the system shown below The ranges of force P such that the block does not move up or down the slope (i.e. stationary) 4. 75° Lo 15 μ=0.7

Answer 1

Angle of the slope is:

$\theta = tan^{-1}\frac{8}{15} = 28.07\degree$

Angle of force P with respect to horizontal is given to be: 75^o

so, angle of P with respect to the slope is: 75 - 28.07 = 46.93^o

now, the block does not move and so the net force on the block is zero.

$Pcos46.93\degree - \mu R - mgsin28.07\degree = F_{net} = 0$

where, R = normal reaction for the block = mgcos28.07^o - Psin46.93^o

=> $Pcos46.93\degree - \mu mgcos28.07\degree +\mu Psin46.93\degree - mgsin28.07\degree = 0$

=> P = 486.06 N

this is the required force to keep the block at rest.