How to Calculate Tension in a Hanging Sign Supported by Two Ropes

[physicsma1391]

[SOLVED] Tension- Hanging Sign

A 20.0 kg sign is being held up by 2 ropes. Each rope makes a 60 degree angle with the sign. So there is an upright triangle, each angle 60 degrees. I have to calculate the tenstion in the two ropes.



So far, I have
$$\Sigma$$ F$$_{}y$$=ma$$_{}y$$
+Fa$$_{}1$$sin60 + Fa$$_{}2$$sin60 = 20 kg (9.81 m/s$$^{}2$$)
I divided by sin60
Fa$$_{}1$$ + Fa$$_{}2$$ = 226.55 N
Each rope has 112.78 N


I'm not sure if i have to do the sum of the forces in the x direction or if the work i already did is correct.

 

[stewartcs]

Draw a FBD then break it down into components and use Newton's second law...

x: -T1cos(theta) + T2cos(theta) = 0

y: T1sin(theta) + T2sin(theta) - Mg = 0

Now solve the equations simultaneously.

 

[physicsma1391]

when you say solve them simultaneously do you mean set them equal to each other since they both = 0?

 

[stewartcs]

when you say solve them simultaneously do you mean set them equal to each other since they both = 0?

The easiest way is to substitute.

Hint: Solve the x component equation for T1 and substitute into the y component equation.

 

[physicsma1391]

OK! i did this and got each rope had 113.28 N of tension.

 

[stewartcs]

OK! i did this and got each rope had 113.28 N of tension.

Looks about right.

You should have come up with T1 = T2 = (Mg)/(2sin(theta)) = (20*9.8)/(2*sin(60)) = 113.16 N.

 

[physicsma1391]

yep! thanks so much!