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What is the coefficient of friction between the box and the floor A 10kg box is pulled at a uniform velocity with a rope that forms a angle of 60 degrees A tension of 80N is constantly maintained?
HowieDiddot
Before tackling this one, I must clean it up a bit.
-- I'll assume that the floor is horizontal under the box.
-- "Angle" means the difference between two directions, but the question specifies
only one of them ... the direction of the rope. I'll assume that the 60° is the angle
between the rope and the horizontal travel of the box, and that the rope and the
tension in it are both directed above the horizontal, i.e., sloped toward the ceiling,
not toward the floor.
Now we have something we can work with.
-- The horizontal component of the tension in the rope is 80 cos(60) = 40 N.
-- The box is sliding along at constant speed, so the horizontal forces on it are balanced.
That means that the friction force is also 40 N but backwards.
-- The weight of the box is (m g) = (10 x 9.8) = 98 N.
-- The coefficient of friction is friction force/weight = 40/98 = 40.8 %
=======================
Why this solution is bogus, at least in part:
The other component of the tension in the rope ... the vertical one ... is 80 sin(60) = about 69.3 N.
That force is applied to the box at the point where the rope connects, and pulls
straight up at that point. Its effect must be to reduce the box's apparent weight
at that end, and by some complicated amount everywhere along the length of the
box. So the force of friction is also distributed along the length of the box in some
non-uniform and complicated way, and the aggregate apparent coefficient of friction
is some ugly integral of the contributions due to an element of weight at every
element of length/area from one end of the box to the other.
Am I over-thinking this ? ? Perhaps it would be best if I take a nap.
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∙ 11y ago
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