The cord shown in Fig. 5–13a supports a force of 100 lb and wraps over the frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical components of reaction at pin A.

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Accepted Answer

Free-Body Diagrams. The free-body diagrams of the cord and pulley are shown in Fig. 5–13b. Note that the principle of action, equal but opposite reaction must be carefully observed when drawing each of these diagrams: the cord exerts an unknown load distribution p on the pulley at the contact surface, whereas the pulley exerts an equal but opposite effect on the cord. For the solution, however, it is simpler to combine the free-body diagrams of the pulley and this portion of the cord, so that the distributed load becomes internal to this “system” and is therefore eliminated from the analysis, Fig. 5–13c.

Equations of Equilibrium. Summing moments about point A to eliminate [latex] \mathbf{A}_{x}[/latex] and [latex] \mathbf{A}_{y}[/latex], Fig. 5–13 c , we have

[latex] \begin{array}{c} ⤿ +\Sigma M_A=0 ; \quad 100 lb (0.5 ft ) - T(0.5 ft )=0 \T=100 lb \end{array} [/latex]

Using this result,

[latex] \begin{array}{c} \stackrel{+}{ ightarrow} \Sigma F_x=0 ; \quad-A_x+100 \sin 30^{\circ} lb =0 \ A_x=50.0 lb \end{array} [/latex]

[latex] \begin{array}{c} +\uparrow \Sigma F_y=0 ; \quad A_y-100 lb -100 \cos 30^{\circ} lb =0 \ A_y=187 lb \end{array} [/latex]

NOTE: From the moment equation, it is seen that the tension remains constant as the cord passes over the pulley. (This of course is true for any angle u at which the cord is directed and for any radius r of the pulley.)

Question 5.6: The cord shown in Fig. 5–13a supports a force of 100 lb and ...

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