Question

A uniform boom is supported by a perpendicular cable attached at its length from the boom's hinge on the floor and makes a angle of with horizontal floor. At the top end of the boom is attached a weight. What is the tension in the supporting cable?

Answer 1

Answer:

1200 N)×(L/2)×cos(65) + (2000 N)×(L)×cos(65) = T×(3L/4)×sin(65+25)

T = 1465.08 N

T = 1470 N (to 3 sf) < - - - - - - - - - - - - - - - - - - - - - - tension in the cable

Rx = (1465.08 N)×cos(25)

Rx = 1327.81 N (to the right)

Rx = 1330 N (to 3 sf) < - - - - - - - - - - - - - - - - - - - - - horizontal reaction

Ry = (1200 N) + (2000 N) - (1465.08 N)×sin(25)

Ry = 2580.83 N (upwards)

Ry = 2580 N (to 3 sf) < - - - - - - - - - - - - - - - - - - - - - vertical reaction