Answer to Question #214693 in Mechanics | Relativity for rodomani

Question #214693

A uniform board that is 1.50m long and weighs 3.50x10^2 N is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 5.00x10^2 N without breaking, and cable B can support up to 4.00x10^2 N. You want to place a small weight on this bar. Find the heaviest weight you can put on without breaking either cable and its location.


1
Expert's answer
2021-07-09T08:36:00-0400

One of the condition of equilibrium is that sum of upward force must be equal to sum of downward force

TA+TB=W+350TA=500N,TB=400N500+400=W+350W=900350=550NT_A+T_B=W+350\\ T_A=500N,T_B=400N\\ 500+400=W+350\\ W=900-350=550N\\

Taking moment about the end cable A is attached

W×x+350×0.75=TB×1.5550×x+262.5=600550×x=600262.5550×x=337.5x=337.5550x=0.614mW\times{x}+350\times{0.75}=T_B\times{1.5}\\ 550\times{x}+262.5=600\\ 550\times{x}=600-262.5\\ 550\times{x}=337.5\\ x=\dfrac{337.5}{550}\\ x=0.614m

The heaviest weight is 550N placed at 0.614m from cable A


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