Question
The three A-36 steel wires each have a diameter of $2 \mathrm{mm}$ and unloaded lengths of $L_{A C}=1.60 \mathrm{m}$ and $L_{A B}=L_{A D}=2.00 \mathrm{m} .$ Determine the force in each wire after the 150 -kg mass is suspended from the ring at $A$
Step 1
00 mm = 0.00200 m, A = πd^2/4 = π(0.002)^2/4 = π·1.0e-6 = 3.1416×10^-6 m^2. - E (A-36 steel) ≈ 200 GPa = 2.00×10^11 N/m^2. - W = mg = 150(9.81) = 1471.5 N. - Given lengths: L_AC = 1.60 m (vertical), L_AB = L_AD = 2.00 m. - Horizontal distance from A to B (or D): h Show more…
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The A-36 steel wires $A B$ and $A D$ each have a diameter of $2 \mathrm{mm}$ and the unloaded lengths of each wire are $L_{A C}=1.60 \mathrm{m}$ and $L_{A B}=L_{A D}=2.00 \mathrm{m} .$ Determine the required diameter of wire $A C$ so that each wire is subjected to the same force when the 150 -kg mass is suspended from the ring at $A$
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Two steel wires, each having a cross-sectional area of $2 \mathrm{mm}^{2}$ are tied to a ring at $C,$ and then stretched and tied between the two pins $A$ and $B$. The initial tension in the wires is $50 \mathrm{N}$. If a horizontal force $\mathbf{P}$ is applied to the ring, determine the force in each wire if $P=20$ N. What is the smallest force $P$ that must be applied to the ring to reduce the force in wire $C B$ to zero? Take $\sigma_{Y}=300$ MPa. $E_{\mathrm{st}}=200 \mathrm{GPa}$.
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