Question

The end a of a uniform rod AB of weight W is hinged a fixed point. The rod is held in equilibrium in a horizontal position by means of a string attached to B and to a point c vertically above A. The string is inclined at 60 degrees to the rod. Find the tension in the string and the reaction of the hinge.

Answer 1

Question #74184, Physics / Mechanics | Relativity

The end A of a uniform rod AB of weight W is hinged a fixed point. The rod is held in equilibrium in a horizontal position by means of a string attached to B and to a point C vertically above A. The string is inclined at 60 degrees to the rod. Find the tension in the string and the reaction of the hinge.

Solution

$\sum M_{A} = 0$

$$T \sin 6 0 {}^ {\circ} \times (A B) - W \times \frac {A B}{2} = 0;$$ $$T = \frac {W}{2 \sin 6 0 {}^ {\circ}}$$ $$\sum F _ {x} = 0;$$ $$R _ {A x} - T \cos 6 0 {}^ {\circ} = 0;$$ $$R _ {A x} = T \cos 6 0 {}^ {\circ} = \frac {W \cos 6 0 {}^ {\circ}}{2 \sin 6 0 {}^ {\circ}} = \frac {W}{2 \tan 6 0 {}^ {\circ}}$$ $$\sum F _ {y} = 0;$$ $$R _ {A y} + T \sin 6 0 {}^ {\circ} - W = 0;$$ $$\begin{array}{l} R_{Ay} = W - T \sin 60{}^{\circ} = W - \frac{W \sin 60{}^{\circ}}{2} = \frac{W}{2}; \\ R_{A} = \sqrt{R_{Ax}^{2} + R_{Ay}^{2}} = \sqrt{\frac{W^{2}}{4 \tan^{2} 60{}^{\circ}} + \frac{W^{2}}{4}} = \sqrt{\frac{W^{2} \cot^{2} 60{}^{\circ}}{4} + \frac{W^{2}}{4}} = \frac{W}{2 \sin 60{}^{\circ}} \\ \end{array}$$

Answer: $T = \frac{W}{2 \sin 60{}^{\circ}}; R_{A} = \frac{W}{2 \sin 60{}^{\circ}}$

Answer provided by https://www.AssignmentExpert.com