Question 4.9: A uniform pipe cover of radius r = 240 mm and mass 30 kg is ......
A uniform pipe cover of radius r = 240 mm and mass 30 kg is held in a horizontal position by the cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reactions at A and B.
Free-Body Diagram. A free-body diagram is drawn with the coordinate axes shown. The forces acting on the free body are the weight of the cover,
{W}\,=\,-m~g{~\mathrm j}\,=\,-(30\mathrm{~k}g)(9.81\;{\mathrm{m/s}}^{2}){\mathrm{j}}\;=\;-(294\;{\mathrm{N}}){\mathrm{j}}and reactions involving six unknowns, namely, the magnitude of the force T exerted by the cable, three force components at hinge A, and two at hinge B. The components of T are expressed in terms of the unknown magnitude T by resolving the vector \overrightarrow{DC} into rectangular components and writing
\overrightarrow{DC} ~=~-(480\mathrm{~mm}) \mathrm i\,+\,(240\mathrm{~mm}){\mathrm j}\,-\,(160\mathrm{~mm}) {\mathrm k}\qquad{ D}C\,=\,{5}60\mathrm{~mm}{T}=T{\frac{{\overrightarrow{{D C}} }}{D C}}=-{\frac{6}{7}}T{{\mathrm{i}}}+{\frac{3}{7}}T{ {\mathrm{j}}}-{\frac{{2}}{{7}}}T{\mathrm{k}}
Equilibrium Equations. We express that the forces acting on the pipe cover form a system equivalent to zero:
\Sigma{F}\,=\,0\colon\qquad\qquad A_{x}\mathrm{i}\ \,+\,A_{y}\mathrm{j}\,+\,A_{z}\mathrm{k}\,+\,{ {B}_{x}}\mathrm{i}\,+\,B_{y}\mathbf{j}\,+\, {T}\,-\,(294~{N})\mathrm{j}\,=\,0 \\ \\ \qquad\qquad (A_{x}\,+\,B_{x}\,-\,{\textstyle\frac{6}{7}}T)\mathrm{i}\;+\;({{A}}_{y}\,+\,B_{y}\,+\,{\textstyle\frac{3}{7}}T\,-\, {29}{ {4}}\,\mathrm{N})\mathrm{j}\,+\,(A_{z}\,-\,{\textstyle\frac{2}{7}}T)\mathrm{k}\,=\,0 (1)
\Sigma{ M}_{B}\,=\,{\Sigma}({r}\,\times\,{F})\,=\,0: \\ \\ 2r\mathrm{k}\ \times\ (A_{x}\mathrm{i}\,+\,A_{y}\mathrm{j}\ +\,A_{z}\mathrm{k}) \\ \\ \qquad\qquad\qquad\qquad{}+\;(2\mathrm{i}+\,r\mathrm{k})\times(-{\textstyle\frac{6}{7}}{ T}\mathrm{i}\,+\,{\textstyle\frac{3}{7}}{T}\mathrm j-\,{\textstyle\frac{2}{7}}{ {T}}\mathrm{k}) \\ \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad {+\,(r\mathrm i+r\mathrm k)}\times(-294\,\mathrm{N)\mathrm j}=0 \\ \\ \qquad (-2{ A}_{y}-\textstyle{\frac{3}{7}}T+294\mathrm{~N})r\large{\mathrm i}\ +(2{ A}_{x}-\textstyle{\frac{2}{7}}T)r\large{\mathrm j} +(\textstyle{\frac{6}{7}}T-\ 294\mathrm{ ~N})r \mathrm{ k}=0 (2)
Setting the coefficients of the unit vectors equal to zero in Eq. (2), we write three scalar equations, which yield
A_{x}\,=\,+49.0\ \mathrm{N}\qquad A_{y}\,=\,+73.5\ \mathrm{N}\qquad T\,=\,343\ \mathrm{N}Setting the coefficients of the unit vectors equal to zero in Eq. (1), we obtain three more scalar equations. After substituting the values of T,A_{x},{\mathrm{~and~}}A_{y} into these equations, we obtain
A_{z}\,=\,+98.0\,\,{ N}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,B_{x}\,=\,\,+245\,\,{ N}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,B_{y}\,=\,\,+73.5\,\,{ N}\,The reactions at A and B are therefore
A = +(49.0 N)i + (73.5 N)j + (98.0 N)k
B = +(245 N)i + (73.5 N)j
