The uniform smooth rod shown in Fig. 5–17a is subjected to a force and couple moment. If the rod is supported at A by a smooth wall and at B and C either at the top or bottom by rollers, determine the reactions at these supports. Neglect the weight of the rod.

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Free-Body Diagram. As shown in Fig. 5–17 b , all the support reactions act normal to the surfaces of contact since these surfaces are smooth. The reactions at B and C are shown acting in the positive y′ direction. This assumes that only the rollers located on the bottom of the rod are used for support.
Equations of Equilibrium. Using the x, y coordinate system in Fig. 5–17 b , we have

[latex] \stackrel{+}{ ightarrow} \Sigma F_x=0 ; \quad C_{y^{\prime}} \sin 30^{\circ}+B_{y^{\prime}} \sin 30^{\circ}-A_x=0 [/latex] (1)

[latex] +\uparrow \Sigma F_y=0 ; \quad-300 N +C_{y^{\prime}} \cos 30^{\circ}+B_{y^{\prime}} \cos 30^{\circ}=0 [/latex] (2)

[latex] ⤿ +\Sigma M_A=0 ; \quad-B_{y^{\prime}}(2 m )+4000 N· m -C_{y^{\prime}}(6 m ) \ +\left(300 \cos 30^{\circ} N ight)(8 m )=0 [/latex] (3)

When writing the moment equation, it should be noted that the line of action of the force component 300 sin 30° N passes through point A , and therefore this force is not included in the moment equation.
Solving Eqs. 2 and 3 simultaneously, we obtain

[latex]B_{y^{\prime}} = -1000.0 N = -1 kN [/latex]
[latex]C_{y^{\prime}} = 1346.4 N = 1.35 kN[/latex]

Since [latex]B_{y^{\prime}}[/latex] is a negative scalar, the sense of [latex]\mathbf{B} _{y^{\prime}}[/latex] is opposite to that shown on the free-body diagram in Fig. 5–17 b . Therefore, the top roller at B serves as the support rather than the bottom one. Retaining the negative sign for [latex]B_{y^{\prime}}[/latex] (Why?) and substituting the results into Eq. 1 , we obtain

[latex] \begin{array}{c} 1346.4 \sin 30^{\circ} N +\left(-1000.0 \sin 30^{\circ} N ight)-A_x=0 \ A_x=173 N \end{array} [/latex]

Question 5.10: The uniform smooth rod shown in Fig. 5–17a is subjected to a...

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