A crate of mass 100 kg rests on the floor. The coefficient of static friction is 0.4. If a force of 250 N(parallel to the floor) is applied to the create, what's the magnitude of the force of static friction on the crate ?

Explanation

The normal force on the object balances its weight, so `F_(N)=mg=(100 kg)(10 m//s^(2))=1,000 N`. Therefore, `F_("static friction, max")=F_(f("static"),"max")=mu_(f)F_(N)=(0.4)(1,000 N)=400 N` This is the maximum force that static friction can exert, but in this case it isn't the actual value of the static friction force. Since the applied force on the crate is only 250 N, which is less than the `F_(f("static"),"max")`, the force of static friction will be less also : `F_(f("static"))=250 N`, and the crate will not slide.