Two people, A and B carry a load of 1000 kg by hanging it on a pole, supported on their shoulders. If A can exert a force 4 times as much as B and the pole is 2.5 m long, where are the loads suspended?

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Solution

With respect to A,
$1000 \times (2.5 - x) = 4 F \times (2.5) 1000$
with repeat to B,
$1000 \times x = F \times 2.5$
$\frac{1000 \times (2.5 - x)}{1000 \times (x)} = \frac{(4 F \times 2.5)}{F \times 2.5}$
$\frac{(2.5 - x)}{x} = 4$
$4 x = 2.5 - x$
$5 x = 2.5$
$x = 0.5 m$

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Two people, A and B carry a load of 1000 kg by hanging it on a pole, supported on their shoulders. If A can exert a force 4 times as much as B and the pole is 2.5 m long, where are the loads suspended?

With respect to A,1000×(2.5−x)=4F×(2.5)1000with repeat to B,1000×x=F×2.51000×(2.5−x)1000×(x)=(4F×2.5)F×2.5(2.5−x)x=44x=2.5−x5x=2.5x=0.5m

With respect to A,
$1000 \times (2.5 - x) = 4 F \times (2.5) 1000$
with repeat to B,
$1000 \times x = F \times 2.5$
$\frac{1000 \times (2.5 - x)}{1000 \times (x)} = \frac{(4 F \times 2.5)}{F \times 2.5}$
$\frac{(2.5 - x)}{x} = 4$
$4 x = 2.5 - x$
$5 x = 2.5$
$x = 0.5 m$