SOLUTION: A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is ti

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Question 1187414: A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut. Show your solution.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The forces exerted by cable T and hinge H can be decomposed into their vertical components, T%5Bv%5D and H%5Bv%5D , and their horizontal components, T%5Bh%5D and H%5Bh%5D .
As the strut is uniform, we can work as if all of its weight is applied at its midpoint, at %224.0+m%22%2F2=%222.0+m%22 from the hinge.
The strut (horizontal), wall (vertical), and cable form a 3-4-5 right triangle, so the tension on the cable and its components (forming a similar triangle) are in the same T%5Bv%5D%3AT%5Bh%5DT=3%3A4%3A5 ratio.
We have 4 significant figures for the weight and only 2 significant figures for the distances, so I would not know how many significant figures to use in the results. Use however many you think your teacher wants to see.

BALANCING TORQUES:
T%5Bv%5D%284.0m%29=%28400.0N%29%284.0m%29%2B%28600.0N%29%282.0m%29
T%5Bv%5D%284.0m%29=2800Nm --> T%5Bv%5D=700.0N
T=%285%2F3%29T%5Bv%5D=%285%2F3%29%28700.0N%29=1166.7N and
T%5Bh%5D=%284%2F3%29T%5Bv%5D=%284%2F3%29%28700.0N%29=933.3N

BALANCING HORIZONTAL (COMPONENT) FORCES:
H%5Bh%5D=T%5Bh%5D=933.3N (opposing forces with the same magnitude)

BALANCING VERTICAL FORCES/COMPONENTS:
Upward forces H%5Bv%5D and T%5Bv%5D=700.0N balance the downward weights 400.0N and 600.0N , so
H%5Bv%5D%2B700.0N=400.0N%2B600.0N --> H%5Bv%5D=300.0N
The magnitude of the force exerted by the hinge is
H=sqrt%28%28933.3N%29%5E2%2B%28300.0N%29%5E2%29=980.3N
The angle it makes with the horizontal strut, alpha is such that
tan%28alpha%29=300.0N%2F%22933.3+N%22=0.32144 --> alpha=17.8%5Eo

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