What angle will make the resultant of three forces vertical? [with picture]

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Homework Help Overview

The discussion revolves around determining the angle that will make the resultant of three forces vertical, with participants referencing a provided picture for context. The subject area is primarily focused on vector resolution and force equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the horizontal components of the forces and their sums, with one participant expressing confusion about how to start the problem. Others attempt to calculate the resultant forces and question the setup of the equations.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing calculations and insights into the horizontal components. While one participant expresses a sense of understanding, there is no explicit consensus on the approach to finding the angle.

Contextual Notes

Participants mention the challenge of rearranging equations for the unknown angle and express frustration with the problem's complexity. The discussion reflects a mix of attempts and clarifications regarding the forces involved.

SUCRALOSE
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I have tried everything on this one that I can think of.

I understand that the forces in the horizontal are 0, but I don't know where to start.

says: determine the angle for which the resultant of the three forces is vertical. See Picture.
 
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Please see this picture for above question.
 

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SUCRALOSE said:
I have tried everything on this one that I can think of.

I understand that the forces in the horizontal are 0, but I don't know where to start.

says: determine the angle for which the resultant of the three forces is vertical. See Picture.

What is the sum of the horizontal components of the two, fully defined forces?
 
The sum of the horizontal components, in the x direction, is:

70(cos140) + 80(cos165) = -131lb

to take it further;

70(sin140) + 80(sin165) = 66lb

r= (-131)2 + (66)2 = 147 lbs

inverse tan = 66/-131 = -27 degrees(I think I get it now) the resultant is vertical up.. so the sum of the left side is the same as right side...
 
Last edited:
SUCRALOSE said:
The sum of the horizontal components, in the x direction, is:

70(cos140) + 80(cos165) = -131lb

to take it further;

70(sin140) + 80(sin165) = 66lb

r= (-131)2 + (66)2 = 147 lbs

inverse tan = 66/-131 = -27 degrees


(I think I get it now) the resultant is vertical up.. so the sum of the left side is the same as right side...

Exactly, and with your creative use of 140o and 165o angles, you could simply have said

70(cos140o) + 80(cos165o) + 150(cosθ) = 0
 
Thanks lots Peter,

At first I was trying to rearrange " 70(cos140o) + 80(cos165o) + 150(cosθ) = 0" for the unknown angle and was frustrated as to why that wouldn't work.

tricky trick question.

Thanks, now I can sleep, lol.
 

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