Question 4.5: A 400-lb weight is attached at A to the lever shown. The con......
A 400-lb weight is attached at A to the lever shown. The constant of the spring BC is k = 250 lb/in., and the spring is unstretched when u = 0. Determine the position of equilibrium.
Free-Body Diagram. We draw a free-body diagram of the lever and cylinder. Denoting by s the deflection of the spring from its undeformed position, and noting that s = ru, we have F = ks = kru.
Equilibrium Equation. Summing the moments of W and F about O, we write
+\mathrm l\Sigma M_{O}=0:\qquad W l\sin \mathrm u-\,\mathrm{r(k r u)}=0\qquad\sin~\mathrm u={\frac{k r^{2}}{W l}}~\mathrm uSubstituting the given data, we obtain
\sin\mathrm{u}={\frac{(250\ \mathrm {lb} /i n.)(3\mathrm{~in.})^{2}}{(400\ \mathrm {lb})(8\mathrm{~in.})}}\mathrm{u}\quad\sin\mathrm{u}=0.703\mathrm{~}\mathrm{u}Solving by trial and error, we find u = 0 u = 80.3˚
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