A crate of 46.8kg - tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 347N . After that you must reduce your push to 209N to keep it moving at a steady 28.8 cm/s .
a) What is the coefficient of static friction between the crate and the floor?
b) What is the coefficient of kinetic friction between the crate and the floor?
a) What is the coefficient of static friction between the crate and the floor?
b) What is the coefficient of kinetic friction between the crate and the floor?
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a) So we're trying to solve for coefficient of static friction which we will call (us) in this case. We know that
us = force of static friction/normal force
To find normal force we look at the vertical forces which in this case is normal force + force of gravity. Since the object isn't accelerating up and down the verical force is 0. Therefore,
force of gravity = normal force
mg = Fn
(46.8 kg)*(9.8 m/s^2) = 458.64 N = normal force
The question gives us the force of static friction which is 347 N so now we can solve.
coefficient of static friction = 347 N/ 458.64 N = 0.76
b) using the same method in a, this time you use the kinetic friction which is 209 N
coefficient of kinetic friction = 209 N/ 458.64 N = 0.46
us = force of static friction/normal force
To find normal force we look at the vertical forces which in this case is normal force + force of gravity. Since the object isn't accelerating up and down the verical force is 0. Therefore,
force of gravity = normal force
mg = Fn
(46.8 kg)*(9.8 m/s^2) = 458.64 N = normal force
The question gives us the force of static friction which is 347 N so now we can solve.
coefficient of static friction = 347 N/ 458.64 N = 0.76
b) using the same method in a, this time you use the kinetic friction which is 209 N
coefficient of kinetic friction = 209 N/ 458.64 N = 0.46