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lgorman20101 lgorman20101
wrote...
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11 years ago
I cannot figure out the answer. The question is "What force should a winch exert on a 900 kg box to keep it moving at a constant rate if kinetic coefficient of friction is .61?" If you could also explain how you got the answer it would be greatly appreciated. Thanks!
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wrote...
11 years ago
Easy-peasy: 5380.2 N

The 900 kg box weighs:

900 * g Newtons, where g is the gravitational force of the earth, 9.8 Newtons per kilogram.  (Remember weight and mass are not exactly the same.)

That means it's weight is 8820 N.

The coefficient of friction is 0.61, meaning that given a gravitation force (weight) of 8820 N the force of friction is:

8820 N * 0.61 = 5380.0 N.

That's the force that would have to be exerted by a winch to keep up with friction.  Any less and the box would slow down and eventually stop.  And more and the box would speed up.
wrote...
11 years ago
Hi. It's been a while, but I think it is just 900 times .61.

Edit: First answer sounds correct. Newtons is the force unit. The cable would exert force, not mass. I love to learn!
wrote...
11 years ago
F=ma= W= 900g

frictional force is 0.61

Min. Force required : 900*0.61*g: g s the grav force
wrote...
11 years ago
If you sum the forces in the y to equal, mass times the acceleration in the y, you get:

Fn - mg = 0
Fn =mg

Now the equation for friction is:
f = muFn
We are told that mu = .61, and we found that Fn = mg:
f = .61 * 900* 9.8
f = 5380.2N
And as we know friction always opposes the direction of motion.

Now you would have to provide a force equal to that to keep it moving with a constant velocity.  So the answer is 5380.2N
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