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Question 16
A straight uniform rod, AB, has length 6 m and mass 0.2 kg. A particle of weight w newtons is fixed at A. A second particle of weight 3w newtons is fixed at B. Th... show full transcript
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Answer
To solve the problem, we first take moments about the point of suspension. Let ( T ) be the tension in the string and calculate the moments about point B:
For the weight at B (3w), its moment about point B is: ( 3w \cdot 0 = 0 )
For the weight at A (w) located 6 m from B, its moment about point B is: ( w \cdot 6 )
The distance from B to the suspension point is x, hence the moment due to tension T is: ( T \cdot x )
Using these moments, we set up the equation:
[ 3w \cdot 0 + w(6) = T \cdot x ]
Next, using the equilibrium of forces in the vertical direction, we have: [ T = w + 3w = 4w ]
Substituting for T in the moment equation gives: [ 6w = 4w \cdot x ]
Rearranging the equation, we find: [ x = \frac{6w}{4w} = \frac{6}{4} = 1.5 ]
However, we account for the weights at A and the mass of the rod which is 0.2 kg (equivalent to 0.2g). Thus, considering total moments:
For the system considering the combined weights and distances: [ 3w \cdot 0 + w(6) + 0.2g \cdot (3) = 4w \cdot x + 0.2g \cdot (x) ]
We then simplify using total moments: [ 6w + 0.6g = (2w + 0.1g) x ]
Finally, rearranging gives: [ x = \frac{3w + 0.3g}{2w + 0.1g} ]
This shows that the required relation holds true.
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