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

AI Thread Summary
The discussion revolves around calculating the net force exerted by two crewmen pulling a raft at different angles. One crewman pulls with a force of 130N at 34 degrees below the horizontal, while the second crewman pulls at 45 degrees above the horizontal. Participants debate the use of sine and cosine in their calculations, with emphasis on the need to cancel the Y components of the forces for the net force to be directed forward. Clarification is provided that sine is used for the vertical components, while cosine is applicable for horizontal components. A diagram is suggested to better visualize the problem and confirm it is a 2D scenario.
agentlee
Messages
18
Reaction score
0

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 ?
 
Physics news on Phys.org
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(∠)
 

Attachments

  • boat.jpg
    boat.jpg
    54.1 KB · Views: 1,194
Last edited:
Back
Top