# Homework Help: Solving force systems

1. Jan 29, 2012

### SUCRALOSE

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.

2. Jan 29, 2012

### SUCRALOSE

Please see this picture for above question.

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3. Jan 29, 2012

### PeterO

What is the sum of the horizontal components of the two, fully defined forces?

4. Jan 29, 2012

### SUCRALOSE

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: Jan 29, 2012
5. Jan 29, 2012

### PeterO

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

70(cos140o) + 80(cos165o) + 150(cosθ) = 0

6. Jan 29, 2012

### SUCRALOSE

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.