Question

a) Find the resultant couple acting on the Z-channel, express your answer in Cartesian vector format. b) Calculate the coordinate direction angles of the resultant couple. 1201b 80 lb 1751b 1751b 8 in. 80 lb 12016 8 in. 6 in. < 6 in. K 8 in. * 8 in. +

Answer 1

part a

$\boldsymbol{couple \,\,due \,\,to\,\,120\,\,lb\,\,forces}$

F1 = 120 k 1b, it's position r1 = (6 i-8j + 8k) in

$F_2=-120\,\boldsymbol{\hat{k}}\,lb,it's\,\,position\,\,r_2=(-6\,\boldsymbol{\hat{i}}+8\boldsymbol{\hat{j}}-8\boldsymbol{\hat{k}})\,in$

Couple due Fi and F2 is given by, M1 = ri x Fi +r2 x F2

M1 = (6 i-8j + 8k) x 120 k +(-6i+89 -8k) x (-120 k)

$M_1=-1920\boldsymbol{\hat{i}}-1440\boldsymbol{\hat{j}}\,lb.in$

$\boldsymbol{couple \,\,due \,\,to\,\,175\,\,lb\,\,forces}$

$F_3=-175\,\boldsymbol{\hat{j}}\,lb\,\,and\,\,it's\,\,position\,\,r_3=8\,\,\boldsymbol{\hat{k}}\,in$

$F_4=175\,\boldsymbol{\hat{j}}\,lb,it's\,\,position\,\,r_4=(-6\,\boldsymbol{\hat{i}}+8\boldsymbol{\hat{k}})\,in$

$Couple\,\,due\,\,F_3\,\,and\,\,F_4\,\,is\,\,given\,\,by,M_2=r_3\,\,\mathbf{x}\,\,F_3+r_4\,\,\mathbf{x}\,\,F_4$

$M_2=(8\boldsymbol{\hat{k}})\,\,\mathbf{x}\,\,(-175\,\boldsymbol{\hat{j}})+(-6\,\boldsymbol{\hat{i}}+8\boldsymbol{\hat{k}})\,\,\mathbf{x}\,\,(175\,\boldsymbol{\hat{j}})$

$M_2=-1050\,\boldsymbol{\hat{k}}\,\,lb.in$

$\boldsymbol{couple \,\,due \,\,to\,\,80\,\,lb\,\,forces}$

$F_5=80\,\boldsymbol{\hat{i}}\,lb\,\,and\,\,it's\,\,position\,\,r_5=(6\boldsymbol{\hat{i}}+8\,\,\boldsymbol{\hat{k}})\,in$

$F_6=-80\,\boldsymbol{\hat{i}}\,lb\,\,and\,\,it's\,\,position\,\,r_6=(-6\boldsymbol{\hat{i}}-8\boldsymbol{\hat{j}}+8\,\,\boldsymbol{\hat{k}})\,in$

$M_3=(6\boldsymbol{\hat{i}}+8\boldsymbol{\hat{k}})\,\,\mathbf{x}\,\,(80\,\boldsymbol{\hat{i}})+(-6\,\boldsymbol{\hat{i}}-8\boldsymbol{\hat{j}}+8\boldsymbol{\hat{k}})\,\,\mathbf{x}\,\,(-80\,\boldsymbol{\hat{i}})$

M3 = -640k lb.in

$M=M_1+M_2+M_3$

$\boldsymbol{M=(-1920\boldsymbol{\hat{i}}-1440\boldsymbol{\hat{j}}-1690\boldsymbol{\hat{k}}})\,lb.in$

$\bg_white \boldsymbol{part-b}$

$Magnitude\,\,of\,\,resultant\,\,couple\,\,M,\left | M \right |=\sqrt{(-1920^2)+(-1440^2)+(-1690^2)}\,lb.in$

$\left | M \right |\approx2935.32\,lb.in$

Let, a, 6 and 7 be direction angles of M with coordinate r,y and a respectively

$\cos(\alpha)=\frac{x-component\,\,of\,\,M}{\left | M \right |}$

$\cos(\alpha)=\frac{-1920}{2935.32}$

$\boldsymbol{\Rightarrow \alpha=cos^{-1}\left (\frac{-1920}{2935.32} \right )\approx130.85\degree}$

$\cos(\beta)=\frac{y-component\,\,of\,\,M}{\left | M \right |}$

$\cos(\beta)=\frac{-1440}{2935.32}$

-1440 1 B = = COS 119.38° 2935.32

$\cos(\gamma)=\frac{z-component\,\,of\,\,M}{\left | M \right |}$

$\cos(\gamma)=\frac{-1690}{2935.32}$

$\Rightarrow\boldsymbol{ \gamma=\cos^{-1}\left (\frac{-1690}{2935.32} \right )\approx125.16\degree}$