Question

7. In the figure below, for the green-colored uniform beam to be in static equilibrium: a) where should a red-colored block be suspended ? b) what should be the tension in the cable? L = 7.0m, d = 1.7m (10 pts Cable Mass of person=80 kg L/2 Mass of beam =800 kg L Mass of the block= 160 kg

Answer 1

Solution :- Mass of block (M) = 160kg Mass of person (m) = 80 kg L=7.om Mass of beam (M, ) = pooky d =1.7m For Static equilibrium :- FBD :- - ½ T in Cable) 0 ki kd Mg M, I mg (a Zero From Condition of equiltibwum :- Moment at point o EM: 160x 9-2X* = 80*9.8 (750-17) 0 mg x (1 -d) ; Mg xa 1568 Xx 1411.2 = 0.9m k=o.am 14/12 1568 From mid a point of beam. { fy=0 ; T-Mg - Mig-mg to T-(M+M+m) g = (160 + 800+80]X9.9 Tension in Cable (T) = 104on T=1040 N.