7.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

A free body diagram needs to be drawn.

 

The length of the bar, l = 2 m

T1 and T2

At translational equilibrium, we have $\stackrel{?}{L_q}=\stackrel{?}{L_R}$ = $T_1$

(T1 / T2) = ( $T_2$ / $T_1\sin ?36.9°$ = 4/3

T1 = (4/3)T2

For rotational equilibrium, on taking the torque about the centre of gravity, we have

T1 $T_2\sin ?53.1°$ x d = T2 $\sin ?53.1°$  (2-d)

T1 x 0.8d = T2 x 0.6 (2-d)

(4/3)T2 x 0.8d = T2 x 0.6 (2-d)

(4/3) x 0.8d = 0.6 (2-d)

1.07d = 1.2 – 0.6d

d = 0.72

So the c.g. of the given bar lies at 0.72 m from its left end.

<p>A free body diagram needs to be drawn.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1729061750php45aGHR_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1729061750php45aGHR.jpeg" alt="" width="250" height="158"></picture></div></div><p>&nbsp;</p><p>The length of the bar, l = 2 m</p><p>T1 and T2</p><p>At translational equilibrium, we have&nbsp;<span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi>L</mi></mrow></mrow><mrow><mrow><mi>q</mi></mrow></mrow></msub></mrow></mrow><mo>? </mo></mover><mo>=</mo><mi></mi><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi>L</mi></mrow></mrow><mrow><mrow><mi>R</mi></mrow></mrow></msub></mrow></mrow><mo>? </mo></mover></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span></p><p> (T1 / T2) = (&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;/&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mi></mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>36.9</mn><mo>°</mo></mrow></mrow></math></span>&nbsp;= 4/3</p><p>T1 = (4/3)T2</p><p>For rotational equilibrium, on taking the torque about the centre of gravity, we have</p><p>T1&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mi></mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>53.1</mn><mo>°</mo></mrow></mrow></math></span>&nbsp;x d = T2&nbsp;<span title="Click to copy mathml"><math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mn>53.1</mn><mo>°</mo></mrow></mrow></math></span>&nbsp; (2-d)</p><p>T1 x 0.8d = T2 x 0.6 (2-d)</p><p> (4/3)T2 x 0.8d = T2 x 0.6 (2-d)</p><p> (4/3) x 0.8d = 0.6 (2-d)</p><p>1.07d = 1.2 – 0.6d</p><p>d = 0.72</p><p>So the c.g. of the given bar lies at 0.72 m from its left end.</p>

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