Please check my solutions -- Mass being pulled with an angled rope

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Homework Help Overview

The discussion revolves around a physics problem involving a block of mass being pulled by a force at an angle, with considerations for work, friction, and normal force. The subject area includes dynamics and forces acting on objects.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the definition of displacement in the context of work and question the units involved. There is also a focus on the correct formulation of the friction force and normal force, with some participants suggesting corrections to the original equations presented.

Discussion Status

The discussion is active, with participants providing guidance on the formulation of equations and questioning assumptions about the forces involved. There is an acknowledgment of potential errors in the original setup, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are navigating through the implications of the angle of the applied force and its effect on the normal force, as well as the friction coefficient in the context of the problem. There is an emphasis on ensuring the correct interpretation of forces and their directions.

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A block of mass M = 4 kg is pulled by a force F = 100 N forming an angle of 60 degrees with themhorizontal plane with friction coefficient 0.3.

Determine the work of force F, friction force and normal force.

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Welcome to PF. :smile:

On your first solution, what are units of "S" for work? What units should Work be in?
 
I assume S to be the displacement?
 
rsk said:
I assume S to be the displacement?
Oh, thanks. I guess the displacement is indeed a variable -- I missed that.
 
The friction force is $$f=\mu N$$ and NOT $$f=\mu mg$$. So to correctly determine the friction force you need to find the normal force N first.
Your equation for the normal force N doesn't seem correct either (though the final result might be correct), it should be $$N-100\sin60=4\cdot 9.81$$ i.e. the force of 100N is actually forming an angle of -60 degrees with the horizontal, that is it is pointing down not up.
 

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