Force question on pulling worker

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Homework Help Overview

The discussion revolves around a physics problem involving a worker pulling a crate on a rough surface, focusing on the forces at play, particularly friction. The crate has a mass of 10 kg, and the coefficients of static and kinetic friction are provided.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore whether the crate moves based on the applied force and the frictional forces. There are attempts to calculate the maximum static friction and compare it with the applied force to determine the motion of the crate.

Discussion Status

There is an ongoing examination of the relationship between the applied force and the frictional forces. Some participants suggest that the crate does not move based on their calculations, while others question the reasoning behind the frictional force values and the implications for motion.

Contextual Notes

Participants are working within the constraints of the problem's parameters, including the coefficients of friction and the mass of the crate. There is a focus on understanding the conditions under which the crate would move or remain stationary.

StephenDoty
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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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