Answer to Question #223934 in Civil and Environmental Engineering for syra

Question #223934
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?
1
Expert's answer
2021-08-09T08:03:49-0400

"F_{woman}= \\frac{1}{4}* F_{man}\\\\\n\\sum F_y = 0\\\\\n500+150= F_{woman}+F_{man}\\\\\n650=F_{woman}+4F_{woman}\\\\\n4F_{woman}=650\\\\\nF_{woman}130 N\\\\\n\\sum M_{man}= 0\\\\\nF_{woman} * l -150 * \\frac{l}{2}-500 * x =0 \\\\\n130*5 -150 * \\frac{5}{2}-500*x =0\\\\\nx= 0.55 m"

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


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