- #1
physicsma1391
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[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
[tex]\Sigma[/tex] F[tex]_{}y[/tex]=ma[tex]_{}y[/tex]
+Fa[tex]_{}1[/tex]sin60 + Fa[tex]_{}2[/tex]sin60 = 20 kg (9.81 m/s[tex]^{}2[/tex])
I divided by sin60
Fa[tex]_{}1[/tex] + Fa[tex]_{}2[/tex] = 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.
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
[tex]\Sigma[/tex] F[tex]_{}y[/tex]=ma[tex]_{}y[/tex]
+Fa[tex]_{}1[/tex]sin60 + Fa[tex]_{}2[/tex]sin60 = 20 kg (9.81 m/s[tex]^{}2[/tex])
I divided by sin60
Fa[tex]_{}1[/tex] + Fa[tex]_{}2[/tex] = 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.