Question

A uniform bar,12 ft long weighs 10 kg. A 5 kg is at one end and an 8 kg weigh is 2 ft from the other end. At what point will the bar be supported so that the system will remain horizontal?

Answer 1

Answer on question #54199, Physics / Solid State Physics

Question A uniform bar,12 ft long weighs 10 kg. A 5 kg is at one end and an 8 kg weigh is 2 ft from the other end. At what point will the bar be supported so that the system will remain horizontal?

Solution Let us suppose, that support is $x$ ft from the center of bar closer to 5 kg. Then, equation of balance is

$(6-x)\cdot 10+5\cdot\frac{6-x}{12}\cdot\frac{6-x}{2}=(6+x-2)\cdot 8+5\cdot\frac{6+x}{12}\cdot\frac{6+x}{2}$

$60-10x+\frac{5}{24}(6-x)^{2}=32+8x+\frac{5}{24}(6+x)^{2}$

$60-10x=32+8x+5x$

$28=23x$

$x=\frac{28}{23}\,ft=1\frac{5}{23}\,ft$

Hence, bar should be supported at $1\frac{5}{23}$ ft from the center towards the 10 kg.