Question 2 10 O Priv References Mc Graw Han WQHD Required information The system shown can undergo vertical motion only and the springs are unstretched when 5A 5B=0 NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part k₁ ky Determine 54, 5, and the force supported by each spring if mA = 40 kg. mg= 65 kg. k=100 2 10 O Priv References Mc Graw Han WQHD Required information The system shown can undergo vertical motion only and the springs are unstretched when 5A 5B=0 NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part k₁ ky Determine 54, 5, and the force supported by each spring if mA = 40 kg. mg= 65 kg. k=100 N/mm, ky 120 N/mm, and kg - 140 N/mm (Round the final answers to four decimal places) 54, 5, and the forces supported by each spring are as follows: 8A- mm mim < Prev 2 of 6 Next > ög= 165H2 to MA k₂ ma

GL9A7T The Asker · Mechanical Engineering

Transcribed Image Text: 2 10 O Priv References Mc Graw Han WQHD Required information The system shown can undergo vertical motion only and the springs are unstretched when 5A 5B=0 NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part k₁ ky Determine 54, 5, and the force supported by each spring if mA = 40 kg. mg= 65 kg. k=100 N/mm, ky 120 N/mm, and kg - 140 N/mm (Round the final answers to four decimal places) 54, 5, and the forces supported by each spring are as follows: 8A- mm mim < Prev 2 of 6 Next > ög= 165H2 to MA k₂ ma

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from the condition of equillibrium,\begin{aligned}& m_{A} g=K_{1} \delta_{A}+k_{2}\left(\delta_{A}-\delta_{B}\right) \\\Rightarrow & (40 \times 9.81)=\delta_{A}\left(10^{5}+1.2 \times 10^{5}\right)-K_{2} \delta_{B}\end{aligned}\begin{array}{l}\Rightarrow 40 \times 9.81=\delta_{A}\left(2.2 \times 10^{5}\right)-1.2 \times 10^{5} \times \delta_{B} \\ \Rightarrow 0.00392=2.2 \delta_{A}-1.2 \delta_{B}-(i)\end{array}from F \cdot B \cdot D of " m_{B} "from equillibuium,\begin{aligned}K_{2}\left(\delta_{B}-\delta_{A}\right) & =m_{B} g+K_{3} \delta_{B} \\\Rightarrow 65 \times 9.81 & =\delta_{B}\left(K_{2}-K_{3}\right)-K_{2} \delta_{A} \\& =\delta_{B}\left(1.2 \times 10^{5}-1.4 \times 10^{5}\right)-1.2 \times 10^{5} \times \delta_{A} \\\Rightarrow 65 \times 9.81 & =-0.2 \times 10^{5} \delta_{B}-1.2 \times 10^{5} \delta_{A} \\\Rightarrow-0.00637 & =0.2 \delta_{B}+1.2 \delta_{A}-\text { (ii) }\end{aligned}By solving equation - (iv and (ii)\begin{array}{l}\delta_{A}=4 \mathrm{~mm} \\\delta_{B}=10 \mathrm{~mm}\end{array}Now force supported by(i) Spring 1F_{1}=K_{1} \delta_{A}=100 \times 4=400 \mathrm{~N}(ii) Spring 2F_{2}=\left(\delta_{B}-\delta_{A}\right) K_{2}=(10-4) \times 120=720 \mathrm{~N}(iii) spring 3F_{3}=K_{3}\left(\delta_{B}\right)=140 \times 10=1400 \mathrm{~N} ...