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.

Expert's answer

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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