Question

Where must a 500 N weight be hang on a uniform beam with a length of 5 m and weighing 150 N so that a woman at one end of the beam carries only one fourth as much load as a man at the other end?

Answer 1

$F_{woman}= \frac{1}{4}* F_{man}\\ \sum F_y = 0\\ 500+150= F_{woman}+F_{man}\\ 650=F_{woman}+4F_{woman}\\ 4F_{woman}=650\\ F_{woman}130 N\\ \sum M_{man}= 0\\ F_{woman} * l -150 * \frac{l}{2}-500 * x =0 \\ 130*5 -150 * \frac{5}{2}-500*x =0\\ x= 0.55 m$

So, the weight needed should be suited at 0.55 m away from the man.