Question #275245

A 700.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs




800.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied




to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting




cable and the force of the hinge on the strut.

1
Expert's answer
2021-12-03T12:43:13-0500

The tension in the supporting cable


τB=08002+70044T(3/5)=0T=1833 (N)\sum\tau_B=0\to800\cdot2+700\cdot4-4T\cdot(3/5)=0\to T=1833\ (N) . Answer


The force of the hinge on the strut.


Fh=T(4/5)=1833(4/5)=1467 (N)F_h=T\cdot(4/5)=1833\cdot(4/5)=1467\ (N)


τA=08002Fv4=0Fv=400 (N)\sum\tau_A =0\to800\cdot2-F_v\cdot4=0\to F_v=400\ (N)


F=1467+4002=1521 (N)F=\sqrt{1467^+400^2}=1521\ (N) . Answer




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