Question

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.

Answer 1

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

$T_A+T_B=W+350\\ T_A=500N,T_B=400N\\ 500+400=W+350\\ W=900-350=550N\\$

Taking moment about the end cable A is attached

$W\times{x}+350\times{0.75}=T_B\times{1.5}\\ 550\times{x}+262.5=600\\ 550\times{x}=600-262.5\\ 550\times{x}=337.5\\ x=\dfrac{337.5}{550}\\ x=0.614m$

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