Equilibrium Force: Solving Tension in Cable B

[matadorqk]

Homework Statement

34. Joe wishes to hang a soign weighing 750N so that cable A attached to the store makes a 30 degrees angle, as shown in Figure 7-16. Cable B is horizontal and attached to an adjoining building. What is the tension in Cable B.

Figure 7-16 was roughly sketched in Paint from the textbook. All pertinent information is presented.

The Attempt at a Solution

Well, my only attempt to a solution was, first: Do The square root of 750cos30^2 +750sin30^2. This simply lead me to getting 750N as my answer. Well, I did that for 750cos0 etc, following same procedure, and answer is 750N. I did it for 750cos120 and I still got the same answer( 120 because that's the angle between A-B). I really don't think the answer is 750 because wouldn't the force split between both wires.. Unless A would be the only thing holding the sign up, but then the sign would.. crash into the left building. Umm.. I really don't know what else I can do to solve it, so if u could give me a hand? Thanks

** Sorry for the misleading title, I was going to present another problem but then i figured out how to do it and moved on, and yet sturggled with another one so I just had the page open and presented the problem, yet forgot to change its title, which is still a little pertinent to what its about.

 

[robb_]

Sorry, can't see your sketch yet.
Draw a free body diagram for the sign. Is it accelerating?

 

[matadorqk]

Sorry, can't see your sketch yet.
Draw a free body diagram for the sign. Is it accelerating?

Well the sketch is basically just two buildings

||||\ .
||||-\
||||30\ Wire A
||||---\
||||----\ ______________________
!--------- Wire B (above)
!| Rob's | (just imagine exclamation marks are not there)
!---------

Its not accelerating

- The sketch is attached to the first post

 

[matadorqk]

Free body diagram?

After I draw the free body diagram what should I do?

 

[robb_]

Choose a coordinate system and then break up the force vectors into components. Write down F_net = ma for each direction, and you just said a is zero, so what does that imply for F_net in either direction?

 

[matadorqk]

Hmm, from what I understand, Aceleration is zero, so all i have left is fnet equals the sum of the two forces - gravity which equals zero so what I get is the sum of the force on both = gravity which is nothing... :S this is really confusing

The thing is I don't know the force vectors, how am I suppose to find those?

 

[robb_]

OKay, to apply F_net = ma, you need to choose an object to look at. (I told you to pick the sign, as it will solve the problem, ) Then look at anything that is pushing or pulling on that object. Draw a diagram showing all of these pushes, i.e. forces, and remember to consider direction carefully. If the object is not accelerating, all the pushes to the left must be balanced by pushes to the right, same up and down. This is found using F_net = ma. F_g is not zero here, the net force is zero.