Answer to Question #86759 in Mechanics | Relativity for Edidiong Michael inyang

Question

A uniform metre rule of weight 1.0N, is pivoted at the 0.40m mark. A 2.0N weight is hung from the 0.15m mark. Where must a 2.0N weight be placed to balance the rule

Answer 1

$P=1~\text{N};~L=1\text{m};~x_0=0.4~\text{m};\\ P_1=P_2=2~\text{N};~x_1=0.15~\text{m};~x_2~-~? \\ P_1(x_0-x_1) + P_l(x_0-x_l) = P_2(x_2-x_0) + P_r(x_r-x_0); \\ x_l=\frac{x_0}{2};~x_r=x_0+\frac{L-x_0}{2};~P_l=P\frac{x_0}{L};~P_r=P\frac{L-x_0}{L}; \\ x_2 = x_0 + \frac{1}{P_2}(P_1(x_0-x_1)+P\frac{x_0}{L}\frac{x_0}{2}-P\frac{L-x_0}{L}\frac{L-x_0}{2}) = x_0 + \frac{1}{2P_2L}(2P_1L(x_0-x_1)+2PLx_0-PL^2) = 0.4~\text{m}~+~\frac{1}{2*2~\text{N}*1~\text{m}}(2*2~\text{N}*1~\text{m}*(0.4~\text{m}-0.15~\text{m})~+~1~\text{N}*1~\text{m}*(2*0.4~\text{m}-1~\text{m})) = 0.60~\text{m}.$