A flexible chain weighing 42.0 N hangs between two hooks located at the same height (Fig. P12.19). At each hook, the tangent to the chain makes an angle = 41.5° with the horizontal.(adsbygoogle = window.adsbygoogle || []).push({});

(a) Find the magnitude of the force each hook exerts on the chain.

(b) Find the tension in the chain at its midpoint.

So the hook exerts both a X and Y force.

Fx=Cos(41.5)(42)=31.5

Fy=Sin(41.5)(42)27.4

using pythagorean theorem you get total force=42N, or you could use cos^2+sin^2=1 to get the same answer..so 21N on each hook?

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# Homework Help: Chain Problem

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