Question

Wires of length 3m and 5m are fastened to a holiday decoration that is suspended over the town square. The decoration has a mass of 20 kg. the wires, fastened as different heights, make angles of 52 degrees and 40 degrees with the horizontal. Find the tensions in each wire if the decoration is in static equilibrium.
I missed this class and have no idea how I am to do this...not only an answer but also an explanation is requested thank you!

Answer 1

Call the tension, i.e., the force in the 3m wire T1 and in the other wire T2.

As the system is static, when we resolve forces in the x and y directions, those forces should be zero.

So the force in direction x, Fx,
Fx = T1cos(52) + T2cos(40) = 0.
And
Fy = T1sin(52) + T2sin(40) - weight of the decoration = 0

This gives you two equations in two unknowns: T1 and T2, which you can solve.

Got it?

Answer 2

yeah, that works great thank you!

Answer 3

so i suposse i made an algebra error, as i am getting a negative tension, i got:
T1=-T2cos(40)/cos(52)
so
(-T2 cos40/cos52)sin52+T2sin40-20=0
then T2(sin40-(cos40*tan52))-20=0
T2=20/(sin40-(cos40tan52)) which is a negative number....

Answer 4

Sorry, there should be a negative sign the first equation as the tensions have different x directions. It doesn't matter which.

Answer 5

oh haha i shoulda caught that thank you!

Answer 6

ok, so this gives me the total x and y tension, how do i find the tension in each wire? i feel like i should know how to do this but i cannot recall

Answer 7

that will just be basic trig correct?

Answer 8

No, the T1 and T2 is the tension. So you have these two equations: solve for T1 and T2. Calculate out all of the trig functions and substitute.

Answer 9

so the first wire is T1 and the second is T2?

Answer 10

Yes

Answer 11

thank you