Determining Reactions in Three Dimensions The bar AB in Fig. 5.32 is supported by the cables BC and BD and a ball and socket support at A. Cable BC is parallel to the z axis, and cable BD is parallel to the x axis. The 200-N weight of the bar acts at its midpoint. What are the tensions in the

Question

Determining Reactions in Three Dimensions

The bar AB in Fig. 5.32 is supported by the cables BC and BD and a ball and socket support at A. Cable BC is parallel to the z axis, and cable BD is parallel to the x axis. The 200-N weight of the bar acts at its midpoint. What are the tensions in the cables and the reactions at A?

Strategy
We must obtain the free-body diagram of the bar AB by isolating it from the support at A and the two cables. Then we can use the equilibrium equations to determine the reactions at A and the tensions in the cables.

Accepted Answer

Draw the Free-Body Diagram In Fig. a we isolate the bar and show the reactions that may be exerted on it. The ball and socket support can exert three components of force, [latex] A_{x}, A_{y}, and A_{z} [/latex]. The terms [latex] T_{BC} and T_{BD}[/latex] represent the tensions in the cables.

Apply the Equilibrium Equations The sums of the forces in each coordinate direction equal zero:

[latex] \sum{F_{x} } = A_{x} - T_{BD} = 0 [/latex],

[latex] \sum{F_{y} } = A_{y} - 200 = 0 [/latex] ,

[latex] \sum{F_{z} } = A_{z} - T_{BD} = 0 [/latex]. (5.22)

Let [latex] r_{AB}[/latex] be the position vector from A to B. The sum of the moments about A is

[latex]\sum{M_{(point A)} } = [r_{AB} imes (- T_{BC}K )] + [r_{AB} imes (- T_{BD}i )]+ [\frac{1}{2} r_{AB} imes (- 200j)] = \left| \begin{matrix} i & j & k \ 1 & 0.6 & 0.4 \ 0 & 0 & - T_{BC} \end{matrix} ight| + \left|\begin{matrix} i & j & k \ 1 & 0.6 & 0.4 \ - T_{BD} & 0 &0 \end{matrix} ight| + \left|\begin{matrix} i & j & k \ 0.5 & 0.3 & 0.2 \ 0 & - 200 &0 \end{matrix} ight| = (- 0.6 T_{BC} + 40)i + (T_{BC} - 0.4 T_{BD})j + (0.6T_{BD} - 100)k.[/latex]

The components of this vector (the sums of the moments about the three coordinate
axes) each equal zero:

[latex]\sum{M_{x} } = - 0.6T_{BC} + 40 = 0, [/latex]

[latex]\sum{M_{y} } = T_{BC} - 0.4T_{BD}= 0, [/latex]

[latex]\sum{M_{z} } = - 0.6T_{BD} - 100 = 0. [/latex]

Solving these equations, we obtain the tensions in the cables:

[latex]T_{BC} = 66.7 N , T_{BD} = 166.7 N[/latex].

(Notice that we needed only two of the three equations to obtain the two tensions. The third equation is redundant.)

Then from Eqs. (5.22) we obtain the reactions at the ball and socket support:

[latex]A_{x} = 166.7 N, A_{y} = 200 N , A_{z} = 166.7 N [/latex] .

Discussion

Notice that by summing moments about A we obtained equations in which the unknown reactions [latex]A_{x}, A_{y}, and A_{z}[/latex] did not appear. You can often simplify your solutions in this way.

Question 5.8: Determining Reactions in Three Dimensions The bar AB in Fig....

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