Tension and Angles of Frame on Wall

[Heat]

Homework Statement

A frame hung against a wall is suspended by two wires attached to its upper corners.

If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Neglect any friction between the wall and the picture frame.)

Homework Equations

F=ma
w=mg

The Attempt at a Solution

I think it looks something like this:

http://img405.imageshack.us/img405/6182/46664005mf5.jpg

It says .75 weight of frame, so I did the following to convert to N.

w=.75(9.8) = 7.35N

That is 7.35 on each side, and this is the part that get's me "if the tension in each wire is equal to 0.75 of the weight of the frame?"

I am lost, here. but for angle I think it should be something like this arc cos (14.7/7.35)

well at least i tried :D

 

[Doc Al]

You are not given the weight of the frame so don't assume any particular value. Instead, just call that weight W.

You are given the tension in the wires in terms of that weight. The tension equals 0.75*weight, so T = 0.75*W.

Now apply the condition for equilibrium, namely that vertical forces acting on the frame must add to zero. (What are the vertical components of the wire tension?)

 

[Heat]

so the tension on each wire is .75w, where w is unknown.

"condition for equilibrium, namely that vertical forces acting on the frame must add to zero."

vertical must add up to w, for them to cancel out, right?

if so where do I go from there.

the angle would not be:::: arc cosine (w/.75w)?

 

[Doc Al]

Almost! Realize that there are two wires pulling up on the frame.

 

[Heat]

so that means they are sharing the weight, so it should be half of w. ?

 

[Doc Al]

so that means they are sharing the weight, so it should be half of w. ?

You can certainly think of it that way.

 

[Heat]

so it would be arc cosine of .5w/.75w = arc cosine .66 = 48.19 ?

 

[Doc Al]

Yep. Looks good.

 

[Heat]

thank you :)