Find Forces at Hooks on 42N Flexible Chain

[Jacob87411]

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.

(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?

 

[Doc Al]

Your approach is not quite correct. Realize that the force exerted on the chain is always tangent to the chain (in other words, parallel to the line of the chain); it is equal to the tension force exerted by the chain at the ends. Since the chain is in equilibrium, the sum of the vertical forces must equal zero. (So find the vertical components of the forces on the chain.)

 

[Jacob87411]

Yes the force from each chain minus the chain should be zero because it is in equilibrium..so do you find the vertical force from the chain then that combined with the forces from the hooks is zero?

 

[Doc Al]

There are three forces acting on the chain: force from the left hook, force from the right hook, and the weight. Find the vertical component of each of these forces.