A uniform rod BC of mass 4 kg is connected to a collar A by a 250-mm cord AB. Neglecting the mass of the collar and cord, determine (a) the smallest constant acceleration aA for which the cord and the rod lie in a straight line, (b) the corresponding tension in the cord.

Question

A uniform rod BC of mass 4 kg is connected to a collar A by a 250-mm cord AB. Neglecting the mass of the collar and cord, determine (a) the smallest constant acceleration [latex]{ a }_{ A }[/latex] for which the cord and the rod lie in a straight line, (b) the corresponding tension in the cord.

Accepted Answer

Geometry and kinematics:

Distance between collar and floor = AD = 250 mm + 350 mm = 600 mm

When cord and rod lie in a straight line:

AC = AB + BC = 250 mm + 400 mm = 650 mm

[latex]\cos heta=\frac{AD}{AC}=\frac{600\mathrm{\ mm}}{650\mathrm{\ mm}}[/latex]

[latex] heta=22.62^{\circ}[/latex]

Kinetics

(a) Acceleration at A.

[latex]+\circlearrowright\Sigma M_C=\Sigma(M_C)_{eff}:W(CG) heta=m\overline{{{a}}}(C G)\mathrm{cos}\, heta[/latex]

[latex]m\overline{{{a}}}=m g\ \mathrm{tan}\, heta[/latex]

[latex]\bar{a}=g an heta=(9.81\mathrm{\ m}/s^2) an22.62^{\circ}[/latex]

[latex]\mathbf{a}_A=\bar{\mathbf{a}}=4.09\mathrm{\ m}/\mathrm{s}^2\longrightarrow\blacktriangleleft[/latex]

(b) Tension in the cord.

[latex]\overset{+}{\longrightarrow } \Sigma F_x=\Sigma(F_x)_{eff} :T\operatorname{sin} heta=m\overline{a}=m g\ \mathrm{tan}\, heta[/latex]

[latex]T=\frac{mg}{\mathrm{cos}\, heta}=\frac{(4\,\mathrm{kg})(9.81)}{\cos22.62^{\circ}}[/latex]

[latex]T=42.5\mathrm{\ N}\blacktriangleleft[/latex]

Question 16.5: A uniform rod BC of mass 4 kg is connected to a collar A by ...

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