Question
A force of 580 N is needed to be applied on a rope at an angle of 30 to the horizontal in pulling a crate with a constant velocity across a horizontal plank. What is the mass of the crate if the coefficient of kinetic friction is o.25?
Answers
Henry
Fapcosa - u*Fv = 0,
580cos30 - 0.25Fv = 0,
502.3 - 0.25Fv = 0,
0.25Fv = 502.3,
Fv = 2009N.
mg = 2009,
m = 2009 / g = 2009 / 9.8 = 205kg.
580cos30 - 0.25Fv = 0,
502.3 - 0.25Fv = 0,
0.25Fv = 502.3,
Fv = 2009N.
mg = 2009,
m = 2009 / g = 2009 / 9.8 = 205kg.