Question #275250

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0m long and weighs 600.0N. The strut is supported by a hinge at the wall and by a cable a whoose other end is tied to the wall at point 3.0m above the left and of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

1
Expert's answer
2021-12-06T09:42:36-0500

The tension in the supporting cable


τB=06002+40044T(3/5)=0T=1167 (N)\sum\tau_B=0\to600\cdot2+400\cdot4-4T\cdot(3/5)=0\to T=1167\ (N) . Answer


The force of the hinge on the strut.


Fh=T(4/5)=1167(4/5)=933 (N)F_h=T\cdot(4/5)=1167\cdot(4/5)=933\ (N)


τA=06002Fv4=0Fv=300 (N)\sum\tau_A =0\to600\cdot2-F_v\cdot4=0\to F_v=300\ (N)


F=933+3002=980 (N)F=\sqrt{933^+300^2}=980\ (N) . Answer

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