Explain to me why the net force is found this way?

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Homework Help Overview

The problem involves two crewmen pulling a raft at specified angles and forces, with the goal of determining the necessary force for one crewman to ensure the net force is directed forward. The context is within the subject area of vector resolution and forces in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of sine and cosine functions in resolving the forces acting on the raft. There are differing opinions on which trigonometric function is appropriate for calculating the components of the forces based on their angles.

Discussion Status

The discussion is ongoing, with participants questioning the reasoning behind the choice of sine versus cosine in their calculations. Some suggest that a diagram could help clarify the situation, while others express confusion about the definitions of the angles and the direction of the net force.

Contextual Notes

There is uncertainty regarding the definitions of "forward direction" and the angles used in the problem. Participants are also considering the implications of the forces being resolved into their respective components.

agentlee
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Homework Statement


Two crewmen pull a raft through a lock. One crewman pulls with a force of 130N Southwest at an angle of 34 degrees below the horizontal. The second crewman pulls at an angle of 45 degrees above the horizontal Northwest. With what force should the second crewman pull so that the net force of the two crewmen is in the forward direction?


Homework Equations





The Attempt at a Solution



I am pretty sure I am missing the obvious here but why would you do:
Fsin45 = 130sin34 ?
 
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Answer looks correct. Do you have any problem in the method used?
 
Yeah, I don't quite understand why you would do it that way.

I was thinking of Fcos45 = 130cos214 since it's on the x-axis.
 
agentlee said:

Homework Statement


Two crewmen pull a raft through a lock. One crewman pulls with a force of 130N Southwest at an angle of 34 degrees below the horizontal. The second crewman pulls at an angle of 45 degrees above the horizontal Northwest. With what force should the second crewman pull so that the net force of the two crewmen is in the forward direction?
What is the 'forward direction'? West?
When you write below/above horizontal, do you mean, respectively, South and North of West?
 
agentlee said:
Yeah, I don't quite understand why you would do it that way.

I was thinking of Fcos45 = 130cos214 since it's on the x-axis.


"Cos" gives you the force in the dirction the boat is moving.

"Sin" gives you the side to side force on the boat. If you want the boat to go straight ahead what must that sum to?

[STRIKE]The equation..

Fsin45 = 130sin34

..is actually missing two terms (one on each side) that have been canceled because they are equal.[/STRIKE]

Edit: Scratch that last bit.
 
Last edited:
I think a diagram would save time.
 
agentlee said:
http://www.csun.edu/~phys100a/solutions_cadavid/hw4-ch5.pdf
I found the problem here #28. I just don't understand why you would use sin and not cos. My professor said that whichever the angle is being made with, you would use cos with that axis.
If you wanted to calculate the net force in the forward (X) direction you would use cos, but that's not what you're after here. What you need is for the Y components of the two forces to cancel, so the angle of interest in each case is to the Y axis.
 
The diagram clarifies the problem. It confirms it's just a 2D problem not a 3D.

I have split the vectors into two components which might help. Remember that..

Sin(∠) = Opposite / Hypotenuse

so

Opposite = Hypotenuse * Sin(∠)
 

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