Why is using Newton's First Law failing me?

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Homework Help Overview

The problem involves a factory worker pushing a crate along a level floor at constant velocity while considering the effects of friction. The subject area pertains to Newton's First Law and the forces acting on an object in motion.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Newton's First Law, questioning the use of the sum of forces being equal to zero in the context of constant velocity. There are attempts to calculate the force required to overcome friction, with some participants suggesting the need for a free body diagram to clarify the forces involved.

Discussion Status

The discussion is ongoing, with participants providing guidance on the importance of correctly identifying the friction force and clarifying the equations used. There is recognition of flawed execution in the original poster's reasoning, but no consensus has been reached on the correct approach.

Contextual Notes

Participants note the importance of accurately defining the friction force and the potential confusion arising from the angle of the applied force. The original poster expresses uncertainty about using the sum of forces in their calculations.

Murtuza Tipu
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A factory worker pushes a 29.7kg crate a distance of 5.0m along a level floor at constant velocity by pushing downward at an angle of 32∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

What magnitude of force must the worker apply to move the crate at constant velocity?

So because constant velocity
∑ f=0
Fx *cos(360-32) -fk = 0
Fx *cos(360-32) -(mg)(24) = 0
Fx *cos(328) - 69.8544N = 0
Fx = 69.8544N/cos(328∘)
Fx = 82.3708N

That turns out to be wrong, the expected answer is fk the force of kinetic friction. Why? Shouldn't I be able to use ∑F=0 in this problem to find the answer?

I also tried to find the square root of Fx2 + Fy2 but that didn't help either.
 
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You didn't use the right friction force. Start with the definition of friction and maybe draw yourself a free body diagram.
 
Murtuza Tipu said:
A factory worker pushes a 29.7kg crate a distance of 5.0m along a level floor at constant velocity by pushing downward at an angle of 32∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

What magnitude of force must the worker apply to move the crate at constant velocity?

So because constant velocity
∑ f=0
Fx *cos(360-32) -fk = 0
I'm going to just use ##32^\circ## below; it doesn't make a difference. Your equation isn't correct. You should have
$$F_x - f_k = 0$$ or
$$F\cos 32^\circ - f_k = 0.$$ The x-component of F is ##F_x = F\cos 32^\circ##. Writing ##F_x \cos 32^\circ## doesn't make logical sense.

Why? Shouldn't I be able to use ∑F=0 in this problem to find the answer?
Yes, you can. Your thinking here is correct. It's just your execution is flawed for the reason paisello2 has hinted at.
 
You will save a lot of time and grief if you draw a free body diagram. have you done that?
 

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