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# Forces and Equilibrium watch

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1. Hello. I'm having a bit of a problem with Forces and Equilibirum.

Imagine a Vertical Normal, with an angle of 45 degrees to the left of it, with a Newton Meter of reading 120N. Then, to the right of the normal is a 90 degree angle, with a newton meter of reading 47N. There is a mass suspended from this arrangement (from the point where the 30 degree angle starts).

What is the weight of the mass? Do i have to use Trig?

Thank you very much for your help.
2. (Original post by celeritas)
Firstly, draw a simple diagram.

Rewrite each force in i and j components.

120 N: - 120cos45 i + 120sin45 j

47 N: 47 i

Mass: x i + y j

Since they are in equilibrium, sum of i and j components is 0.

i: -120cos45 + 47 + x = 0

so x = 37.8528

j: 120sin45 + y = 0

so y = - 84.8528
Admittedly I was talking nonsense before (post now deleted), I misread the question, but mass is not a vector, so why do you have it as such?
3. (Original post by F1 fanatic)
Admittedly I was talking nonsense before (post now deleted), I misread the question, but mass is not a vector, so why do you have it as such?
Well mass is not a vector, true. But you find the mass from its weight, which acts vertically downwards.

Actually, I doubt my answer too (lol, deleted too).
4. Ha!!

Thanks guys! I'll have a look and see what i can do with it...

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