A box weighing 87 lb is being pulled with a rope that exerts a force of 60. lb at an angle of 15 degrees with the horizontal. if the coefficient of kinetic friction is .35, what is the kinetic friction?

Question

bobpursley · Answer

force vertical =60*sin15 upwards friction=.35(87-60*sin15) lbs

Henry · Answer

Fn = 87-60*sin15 = Normal force. Fk = u*Fn = 0.35 * Fn. = Force of kinetic friction.