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.


Expert's answer

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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