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

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To determine the angle that makes the resultant of three forces vertical, the horizontal components of the forces must sum to zero. The calculations show that the sum of the horizontal components is -131 lb, while the vertical components total 66 lb, leading to a resultant of 147 lbs. By using the angles of 140° and 165°, the equation 70(cos140°) + 80(cos165°) + 150(cosθ) = 0 can be established to find the unknown angle. The discussion highlights the importance of balancing horizontal and vertical forces to achieve a vertical resultant. This problem-solving approach has clarified the concept for participants.
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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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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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