Total force up
=100(9.81) = T1 sin37 + T2 sin 53
Total horizontal force
= 0 = T1 cos 37 - T2 cos 53
use that to get T1 and T2
now for example moments about left end
T2 sin 53 *5 = 100*9.81* X
A non-homogeneous trunk, 5 meters long and 100 kilograms mass is in equilibrium and is hung at the ends by two strings; One form an angle of 37 degrees and the other an angle of 53 degrees, both with the horizontal.
Determine the location of the center of mass of the trunk and the tension of both strings.
I need helo, I already did the free boda diagrama but I don't know how TO calcula te the CM as it is not a regular object. Thanks.
2 answers
You could have done that, patience :)
1 answer