Trig: Vector?

Question

Look at this picture. What are the tensions in the two cables?

http://i2.photobucket.com/albums/y6/VietIronChef/math.jpg

Answer 1

Draw a vertical line passing thro` eye bolt on weight.
Line meets girder at D
A is vertex at 25° angle
B is vertex at 40° angle
C is vertex at 115° angle

Vertical Forces
F1 cos 65° + F2 cos 50° = 1000

Horizontal Forces
F1 cos 25° - F2 cos 40° = 0

Equations give solutions:-
F1 = 845 lbs
F2 = 1000 lbs

Answer 2

Lets call the one on the right T1 and the left T2

First draw your free-body diagram, and you'll see that each respective cable has the same angle between their respective tension vector and the x-axis. We also know that there is a 1000 lb. force acting downward on the y-axis.

Set up your force equations as x-y components:

Fnet (x)= ma(x) = T2[cos(angle 2)] - T1[cos(angle 1)]

Set ma(x) equal to 0 since it is at equilibrium and solve for one of the Tensions.

* It is Tcos(...) because if you try to draw the sides of a triangle, each side will represent a vector for each tension; the above is because it is along the adjacent side of the angle and that the vector is pointing along the x-axis.

Fnet (y) = ma(y) = T1[sin(angle 1)] + T2[sin(angle 2)] - 1000

Set ma(y) equal to 0 again due to equilibrium. And just substitute from there.

* Tsin(...) for similar reasons but its just along the opposite side of the angle and that its along the y-axis.

You can do it!

Answer 3

It must be that the vertical tension totals 1000 lbs. Then:
a sin40 + b sin25 = 1000
It also must be that the horizontal tensions cancel, or else the system would be moving. Then:
a cos40 + b cos25 = 0
We have two equations in two unknowns. Lets solve:
a = -b*(cos25/cos40) From second eq.
-b*(cos25/cos40)*sin40 + b sin25 = 1000 From first eq.
b = 1000/(sin25 -(cos25/cos40)*sin40)
b = -2 959.76845 lbs
a = -3501.7044 lbs
The negative signs are meaningless, they only indicate direction.