Question

A 12.0 m uniform beam is hinged to a vertical wall and held horizontally by a 5.00 m cable attached to the wall 4.00 m above the hinge, as shown in the figure below (Figure 1).The metal of this cable has a test strength of 0.900 kN which means that it will break if the tension in it exceeds that amount. You may want to review For related problem solving tips and strategies, you may want to view a Video Tutor Solution of Equilibrium and pumping iron What is the heaviest beam that the cable can support with the given configuration? m = _______ Find the horizontal component of the force the hinge exerts on the beam. F_b = ______

Answer 1

Given: - length of beam = 12.0 m cable length = 5 m Tensile strength of cable = 0.9 kN Now, Part A Heaviest beam that the cable can support with given configuration. weight of beam is act at centre of beam (bam is uniform) for equilibrium Moment about hinge = 0 \r\n Mg times 6 - T sin 6 = 0 Mr times 6 = T sin 6 sin 5 = 4/s and g = 10 m/s^2 M = T sin 6/6 g = 0.9 times 1000 times 4/s/ 6 times 10 = 12 kg Maximum weight of beam = 12 kg Now part B) horizontal force of hinge. for horizontal Equilibrium sigma fx = 0 T cps 6 - hinge force (F_n) = 0 cos 6 = 3/s Fh = T cos 6 = 0.9 times 1000 times 3/s F-n = 540 N