A 200 pound box of books on the floor requires a 140 pound horizontal force to make it start sliding on the floor, but only a 50 pound force to keep it sliding at a constant speed. What is the coefficient of static friction between the box and the floor?
Explain answer please
2006-10-25 15:40:02 · 2 answers · asked by ??? 2 in Science & Mathematics ➔ Physics
f=uN, where f is the force of friction in N, u the coefficient of static friction, and N is the force normal to the floor which in this problem is equal to the weight of the box of 200pounds.
If the box starts to move with a 140-pound horizontal force, it means that the force of friction is also equal to 140 pounds.
Substitute the known values:
f=uN
140=u*200 (No need to convert to Newton because
anyway the conversion factors will
cancel out.)
u=140/200
=0.70
Once the box moves at constant speed the coefficient of friction decreases (it's now called coefficient of kinetic
friction). This is a known fact: coefficient of kinetic friction is always smaller than the coefficient of static friction. If you look at the formula: f=uN, if N remains the same as in this problem, and u becomes smaller, then
f becomes smaller as well. In this problem f is given as 50 pounds (it's equal to the applied force of 50 pounds). Now substitute in the formula: 50=u*200; u=50/200; u=0.25. This is the coefficient of kinetic friction. Compare it to the coefficient of static friction of 0.70. The former is much smaller. That's why the corresponding applied force is much smaller too.
2006-10-27 06:02:48 · answer #1 · answered by tul b 3 · 0⤊ 0⤋
Refer here for the (eventual) answer:
http://answers.yahoo.com/question/index;_ylt=AhxSpYa7XgdL9vMQyUbWw0bzy6IX?qid=20061025194237AAvFOX8
2006-10-25 16:04:51 · answer #2 · answered by jacinablackbox 4 · 0⤊ 0⤋