A horizontal uniform beam AB of length 4m and a mass of 20 kg is supported at the end B by means of a string which passes over a fixed, smooth pulley supporting a counterbalancing weight of 8 kg on the other side. What force, F, when applied at the point A in a suitable direction, will hold the beam in static equilibrium

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F = 160 ​ N

no force F, as specified, can hold the beam in static equilibrium

F = 40 \sqrt{19} N

F = (200 - 80 \sqrt{3}) N

answer is B.

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Detailed Solution

${\bar{F}}_{1} = 80 \cos 30 \hat{i} - 80 \sin 30 \hat{j}$

$= 80. \frac{\sqrt{3}}{2} \hat{i} - 80. \frac{1}{2} \hat{j}$

$= 40 \sqrt{3} \hat{i} - 40 \hat{j}$

${\bar{F}}_{2} = - 200 \hat{j}$

${\bar{F}}_{1} + {\bar{F}}_{2} + {\bar{F}}_{3} = 0$

${\bar{F}}_{3} = - (+ 40 \sqrt{3} \hat{i} + 40 \hat{j} - 200 \hat{j})$

$= - 40 \sqrt{3} \hat{i} + 160 \hat{j}$

$| F_{3} | = \sqrt{4800 + 25600}$

$= 40 \sqrt{19}$

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