Question

What is the maximum weight that can be suspended on a 25kg poll, 3/5 from the end? The maximum tension of the 30 degree guy-wire is 1500N. Find the horizontal and vertical forces exerted on the poll by the hinge.

Answer 1

Weight of the poll is W=mg=25*9.8=245,N , acting on the poll vertically downwards at L/2 from the hinge.

Weigh of JoeW'=Mg is acting on poll vertically downwards at 3L/5 from the hinge.

Components of tension in the wire, T cos acting horizontally towards the hinge, and Tsin acting vertically upwards at a distance of L from the hinge. ( L is the length of the poll)

Hinge forces fx towards right , and Fy upwards at the hinge point.

Net horizontal force on the poll is zero,Fx=Tcos ----(1)

Net up ward force on the poll = Net downward force on the poll

Fy+Tsin=mg+Mg ----------(2)

Net torque about the hinge on the system is zero,

=0

Tsin *L=mg*L/2+Mg*3L/5

Tsin =mg*L/2+Mg*3/5

M=(10Tsin-5mg) /6g=(10*1500*sin30-3*25*9.8) /6*9.8=115.05kg

Maximum weight that can be suspended on the poll is M=115.0,kg

(1) Fx=Tcos30=1500*cos30=1299N

Fy=mg+Mg-Tsin=25*9.8+115.05*9.8-1500*sin30=622.5N

Horizontal hinge force is Fx=1299N

Vertical hinge force is fy=622.5N