Question #279351

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and

weighs 600.0 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
2021-12-20T19:58:29-0500

The tension in the supporting cable


τB=0600×2+400×44T×(3/5)=0T=1167 (N)\sum\tau_B=0\to600×2+400 ×4-4T ×(3/5)=0\to T=1167\ (N)


The force of the hinge on the strut.


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


τA=0600×2Fv×4=0Fv=300 (N)\sum\tau_A =0\to600 ×2-F_v ×4=0\to F_v=300\ (N)


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



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