Asked by sara

A homogeneous rod AB=100 cm of mass m=4kg is suspended from a fixed roof by means of two very light strings OA and O'C having the same length point C is at 20cm from B , rod AB is in a horizontal position of equilibrium
Make the inventory of the forces acting on the rod AB.
Answered by Damon
define + is up

Draw a picture

x = 0 at A , Fa up
x = 80 at C , Fc up

x = 50 at CG, mg =4*9.81=39.24 down so Fcg = -39.24 N

sum of forces = 0
Fa + Fc + Fcg = 0
so
Fa+Fc = 39.24 N
now do moments about 0 for example
Fc * 80 = 39.24 * 50
Fc = 24.5 N
so
Fa = 39.24 - 24.5 = 14.7 N
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