The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point

Question

The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point $(p)$ of normal bisector, as shown in the figure is ________. [2026]

Smart Achievers · Accepted Answer

(1)

$X_{cm} = \frac{2 \times (- 5 \sqrt{3}) + 3 (5 \sqrt{3}) + 15 \times 0}{20} = \frac{\sqrt{3}}{4} cm$

$Y_{cm} = \frac{2 \times (- 5 / 2) + 3 (5 / 2) + 15 \times (5 / 2)}{20} = 1.25 cm$

$Y_{cm} = 1.25 cm$

Solution

Q.
The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point $(p)$ of normal bisector, as shown in the figure is ________. [2026]

1 $(\frac{\sqrt{3}}{4}, 1.25)$

2 $(\frac{\sqrt{3}}{4}, 1.0)$

3 $(0, 0)$

4 $(1.25, 0)$

Ans.
(1)

$X_{cm} = \frac{2 \times (- 5 \sqrt{3}) + 3 (5 \sqrt{3}) + 15 \times 0}{20} = \frac{\sqrt{3}}{4} cm$

$Y_{cm} = \frac{2 \times (- 5 / 2) + 3 (5 / 2) + 15 \times (5 / 2)}{20} = 1.25 cm$

$Y_{cm} = 1.25 cm$

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Solution

Q. The position of center of mass of three masses 2 kg, 3 kg and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is ________. [2026]

1 (34,1.25)

2 (34,1.0)

3 (0,0)

4 (1.25,0)

Ans. (1) Xcm=2×(-53)+3(53)+15×020=34cm Ycm=2×(-5/2)+3(5/2)+15×(5/2)20=1.25 cm Ycm=1.25 cm