What is the formula for finding the tension of the boom in this problem?

[physics10189]

Homework Statement

A 1820N uniform boom is suppoted by a cable. The boom is pivoted at the bottom, and a 1100N object hangs from its end.
The boom has a length L=13 m and is at an angle of 48 degree above the horizontal. At support cable is attached to the boom at a distance of .77L from the foot of the boom and its tension is perpendicular to the boom.
Find the tension in the cable holding up the boom.

Homework Equations

The Attempt at a Solution

1100N * L * cos(48)
+ 1820N* 0.5L * cos(48)
- T * 0.77L
=0

Since L is not relevant in this equation you can just cancel it out.
I got answer to be
1746.691583N
which of course is wrong.
Could you please them me what I am doing wrong.

 

[Doc Al]

Looks right to me.

 

[physics10189]

ok well...similar to his question

Find the vertical components of the reaction
force on the boom by the floor.

i found that the horizontal component of the reaction
force on the boom by the floor is just Tsin(theta).

so how do i find the vertical components?

 

[Doc Al]

Hint: What does the net vertical force on the boom equal?

 

[physics10189]

is it the
Normal of y= Fg(boom)+Fg(block)-Tcos(theata)?

 

[nvn]

Yes, that's right.