Force question on pulling worker

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A worker pulls a 10-kg crate with a force of 40 N on a rough floor, where the coefficients of static and kinetic friction are 0.5 and 0.3, respectively. The maximum static friction force is calculated to be 49 N, which exceeds the applied force, indicating that the crate does not move. Since the crate remains stationary, the kinetic friction force is not applicable. The net force acting on the crate is zero, meaning the frictional force must equal the applied force of 40 N. Thus, the frictional force exerted by the surface is 40 N.
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A worker pulls horizontally on a rope that is attached to a 10-kg crate resting on a rough
floor. The coefficients of static and kinetic friction are 0.5 and 0.3, respectively. The
worker pulls with a force of 40 N. The frictional force exerted by the surface is...??

the box is moving because fk<fs

If the Fnet is not 50N, the applied force then
Fnet= T - fk
Fnet= 50N - fk

and of cource fk= .3 * 10 * 9.8 = 29.4N

Is this right??
 
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The first thing to figure out is: Does it move?
 
Yes it does move because fk= umg and fs=umg and since us>uk fk<fs which means that the object moves and accelerates, right??
 
I don't understand your reasoning. Instead, figure out the maximum force that static friction can deliver given the nature of the surfaces. Compare that maximum value with the applied force.
 
ok...the fs=49N which is < the 40N push. So the box is not moving. So is the fk= 40N so fnet= 0 = fp - fk?
 
StephenDoty said:
ok...the fs=49N which is < the 40N push. So the box is not moving.
Good. The maximum possible static friction force is 49 N, which is more than enough to handle the applied force of 40 N. So the box doesn't move.

So is the fk= 40N so fnet= 0 = fp - fk?
Since it doesn't move, fk (kinetic friction) is irrelevant.

But the net force is zero, so what must be the friction?
 
40N


thank you very much
 
Perfecto!
 
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