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The deflection and the forces supported by springs are calculated and the results are shown below in detail. Given:(1)\begin{array}{ll}m_{A}=56 \mathrm{~kg} & k_{1}=100 \mathrm{~N} / \mathrm{mm} \\m_{B}=86 \mathrm{~kg} & k_{2}=120 \mathrm{~N} / \mathrm{mm} \\k_{3}=140 \mathrm{~N} / \mathrm{mm}\end{array}Solution:Drow the free body diagram of mass A.\begin{array}{l}\text { Draw the free body diagram of mas } \\100 \delta_{A}-(56 \times 9.81)-120\left(\delta_{B}-\delta_{A}\right)=0 \\100 \delta_{A}-549.36-120 \delta_{B}+120 \delta_{A}=0 . \\220 \delta_{A}-120 \delta_{B}=549.36 \\\left.220 \delta_{A}=549.36+120 \delta_{B}-\delta_{A}\right)=0 \\\delta_{A}=2.497+0.545 \delta_{B} .\end{array}m_{B} g \quad k_{2}\left(\delta_{B}-\delta_{A}\right)Draw the free body diagram of mass B. m\begin{array}{l}\Sigma F_{y}=0 \\k_{2}\left(\delta_{B}-\delta_{A}\right)+k_{3} \delta_{B}-m_{B} g=0 . \\120\left(\delta_{B}-\delta_{A}\right)+140 \delta_{B}-(86 \times 9.81)=0 . \\120 \delta_{B}-120\left(2.497+0.545 \delta_{B}\right)+140 \delta_{B} \\-843.66=0 . \\120 \delta_{B}-299067-65.4 \delta_{B}+140 \delta_{B}-843.66=0 . \\194.6 \delta_{B}=1143.33 . \\\left.* \delta_{B}-\delta_{A}\right) \\* \delta_{B}=5.875 \mathrm{~mm} .\end{array}k_{B} \delta_{B} \quad M_{B g}sub \delta_{B} in eqn (1).(2)\begin{aligned}\delta_{A} & =2.497+0.545 \delta_{B} \\& =2.497+0.545(5.875) . \\* \delta_{A} & =5.699 \mathrm{~mm} *\end{aligned}Force supported by spring 1:\begin{aligned}F_{1}=k_{1} \delta_{A} & =100 \times 5.699 . \\F_{1} & =569.9 \mathrm{~N} .\end{aligned}Force supported by spring 2:\begin{array}{c}F_{2}=k_{2}\left(\delta_{B}-\delta_{A}\right)=120(5.875-5.699) \\F_{2}=21.12 \mathrm{~N}\end{array}Force supported by spring 3:\begin{array}{c}F_{3}=K_{3} \delta_{B}=140(5.875) \\F_{3}=822.5 \mathrm{~N}\end{array} ...