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!
2007-03-21 07:26:28 · 5 answers · asked by Holden F 2 in Science & Mathematics ➔ Physics
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.
2007-03-21 07:30:43 · answer #1 · answered by Garrett J 3 · 2⤊ 0⤋
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
2007-03-21 08:49:08 · answer #2 · answered by Anonymous · 1⤊ 0⤋
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!
2007-03-21 07:34:31 · answer #3 · answered by Cirric 7 · 0⤊ 0⤋
To artwork it out, you translate the frictional tension exuded on the container making use of the equation F = ?R F - Friction tension ? - Coefficient of friction R - generic reaction tension of the container (it fairly is comparable to the load of the container for that reason) because of the fact the rope breaks while 800N is utilized to it, this suggests the utmost friction tension might desire to be 800N as a result 800 = 0.3R rearrange 800/0.3 = R R = W = 8000/3 = 2667N <------ this might desire to be the respond
2016-10-19 06:41:46 · answer #4 · answered by ? 4 · 0⤊ 0⤋
F=ma= W= 900g
frictional force is 0.61
Min. Force required : 900*0.61*g: g s the grav force
2007-03-21 07:44:29 · answer #5 · answered by raqandre 3 · 0⤊ 1⤋