Finding Tension and Forces in a Cable-Supported Sign System

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The discussion revolves around calculating the tension in a cable supporting a sign and the forces exerted on a boom by a hinge. The initial equations presented by the user included incorrect torque calculations, leading to confusion. A correction was made regarding the torque equation, emphasizing the need to account for the moments of each weight separately. After resolving the equations, the tension in the cable was determined to be 350 N. The conversation highlights the importance of accurate torque calculations in solving physics problems related to static equilibrium.
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Homework Statement



A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.0 N. A cable, inclined at an angle of 35 with the boom, is attached at a distance of 2.38 m from the hinge in the wall (see figure below). The weight of the sign is 120 N. What is the tension in the cable and what are the horizontal and vertical forces Fx and Fy exerted on the boom by the hinge?

Homework Equations



I had 3 equations:
fx =Tcos 35
fy = 180+120-Tsin 35
Torque = Tsin 35 (2.38)-200(1.5)
solved for torque n got ...17.826...which is far from accurate! PLEASE HELP!

The Attempt at a Solution


Im sooooooo lost...pic attached!
 

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agadag said:
I had 3 equations:
fx =Tcos 35
fy = 180+120-Tsin 35
Torque = Tsin 35 (2.38)-200(1.5)
solved for torque n got ...17.826...which is far from accurate! PLEASE HELP!

Hi agadag! :smile:

(your fy should start 80, but I expect that's a typo?)

Your 200(1.5) is wrong …

you must find the moments (torques) of each weight separately. :wink:
 
yup that was a typo.
sooo what ur saying is..in order to find Torque:
Torque = Tsin 35(2.38) - [80(1.5) + 120(3)]
480 = T sin 35(2.38)
T = 350!
Thankyou!
 

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