(360Deg/6.28rad) * 0.87rad = 49.9 Deg.
Wb = mg = 2.75kg * 9.8N/kg = 26.95 N. =
Wt. of block P.
Fb = 26.95N @ 49.9 Deg.
Fp = 26.95*sin49.9 = 20.61 N. = Force
parallel to incline.
Fv = 26.95*cos49.9 = 17.36 N. = Force
perpendicular to incline.
Fn = Fb2 - Fp - Fs = 0.
Fb2 - 20.61 - 0.39*17.36 = 0.
Fb2 - 27.38 = 0.
Fb2 = 27.38 N. = Force of block Q.
mg = Fb2 = 27.38.
m = 27.38/g = 27.38/9.8 = 2.79 kg = Mass 0f block Q.
A block P of mass 2.75 kg is lying on a rough inclined plane of angle è=0.87 radians. Block P is attached via a model string that passes over a model pulley to a second block Q of unknown mass M kg hanging vertically. The coefficient of static friction between the block P and the inclined plane is ì=0.39.
Calculate the value of the mass of block Q, in kg, that is just sufficient to initiate the sliding motion of block P up the inclined plane.
Thanks!
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