Finding Cable Tension Components with Given Angles

[formulajoe]

ive been wrestling with this for an hour and a half, and all I've got it Theta y and Fy.
a tree is being supported by cables. the tension in cable AB is 4.2 KN.
all it gives me to go on is 2 angles. it 40 deg toward z axis from x axis. and the cable makes an angle of 40 deg with the ground. i found Fy to be -2.7 KN and theta y to be 130 deg.
i can even get close to Fx or Fz. if i can get those, finding their respective theta's will be a snap.

 

[Doc Al]

Can you provide a picture?

 

[formulajoe]

heres a very rough drawing i made with paint. the one 40 deg angle is made with the ground.

 

[himanshu121]

Write the components in vector form if you now the resultant and the angle

 

[HallsofIvy]

1) Where are the other cables?

2) What are you trying to find?

 

[formulajoe]

I'm just trying to find the component forces of AB, and the corresponding theta's. all that's given is those two angles, and the magnitude of the force.

 

[Doc Al]

Please define your x, y, z directions. What is the plane of the ground? What is the vertical direction?

 

[formulajoe]

does this help?
heres the exact question.
knowing the tension in cable AB is 5.2 KN, determine (a) the components of the force exerted by this cable on the tree. (b) the angles theta x, theta y, and theta z that the force forms with axes at A which are parallel to the coordinate axis.

 

[Doc Al]

Finally I see what you are doing.

You need the components of the force that the cable exerts on the tree. That force (F) makes an angle of 50 with the y-axis, so:
F_y= - F cos(50)
F_x-z= F sin(50) (magnitude only)

Then:
F_x= F sin(50)cos(40)
F_z= F sin(50)sin(40)

Make sense?

Edit: Corrected the angle from 60 to 50 degrees. (D'oh!)

 

[formulajoe]

those answers arent coming out right.
Fy = F X cos 130.
theta y makes an angle of 130 deg. i found this knowing that it made an angle of 40 deg with the ground, so it had to make a 50 deg angle with the tree. 180 - 50 = 130. this answer checks out.

but the other two dont. I've even tried working backwards from the answer, but I am not getting remotely close.

 

[formulajoe]

but when the numbers from the answers you gave me are squared, added, and then the square root is taken they equal out to 4.2. are the answers in the book wrong?

 

[Doc Al]

Oops! Where I have 60 degrees, I should have 50. Sorry. Now try it..

 

[formulajoe]

ahhh, that works perfect now. i don't completely understand how you did it though. i understand the y one completely.
so for the x, you complement the 40 deg angle and take the sin correct? and for the 40 deg angle the stake makes with the x axis, just take the cos right. just kinda treat the x and z axes as the x and y axes? same rules work?

 

[Doc Al]

Right, it's easy. Look at my original answer.

You know how to find the y-component. The projection onto the x-z axis is the component perpendicular to y. (If one is cosine, then the other is sine.) Then you just find components (of that projection) in the x-z plane (just like you would with x & y).