Calculate Forces on Mr. Porter's Sign Outside Home

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To calculate the forces on Mr. Porter's sign, the weight of the sign is 495N, and the tension in the chain is 863N, while the thrust force exerted by the rod is 706.9N. The angle of the rod is 35 degrees, which affects the calculations. The tension is not simply the weight of the sign because the sign is not hanging straight down; the forces must be resolved using trigonometric functions. A diagram illustrating the forces can clarify the situation and help in understanding the calculations better.
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I can't figure out how to do this one:

"Mr. Porter has attached a sign that has a weight of 495N to a wall outside his home. Determine: a) The magnitude of the tension in the chain; b) The thrust force exerted by the rod, if the angle is 35 degrees"

I'm not seeing how the answers are a) 863N and b)706.9N (my teacher gave us the answers)
 
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Draw a diagram.
Weight is the y-component.
 
Can you show us what you've done so far on getting the answers?

From the answers, the rod must be at 35 degrees to the wall, rather than the horizontal. It's really only two simple calculations, just draw a diagram of forces, all there is to it.
 
What I'm not understanding is how the tension is 863N? If the signing is hanging straight down, shouldn't is be 495N [495sin(90)]?
 
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It isn't hanging straight down.
 
Where do you see that?
 
There's the sign, a chain, a rod and 35 degrees. I'd say the rod's placed horizontally.
 
This is all I can get.

Physics is not exactly my best subject :redface:
 

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That makes a lot more sense
Thanks
 
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