Question

centre of mass of three particles 10 kg 20 kg and 30 kg is at (0,0,0) where should be particleof mass 40 kg be placed so that the center of mass of combination will be at (3,3,3)​

Answer 1

(30/4;30/4;30/4)

Explanation:

3=10(0)+20(0)+30(0)+40(x)/100

so; x=30/4

Answer 2

The particle of 40 kg placed at (7.5, 7.5, 7.5).

Explanation:

Given that,

Mass of first particle = 10 kg

Mass of second particle = 20 kg

Mass of third particle = 30 kg

We need to calculate the total mass of the system

$M=m_{1}+m_{2}+m_{3}$

$M= 10+20+30$

$M=60\ kg$

The center of mass of the total mass 60 kg of the system placed at (0, 0, 0).

40 kg mass placed at (x, y, z).

The center of mass of combination will be at (3,3,3)​

We need to calculate the value of x, y and z

Using formula of center of mass

For x,

$x_{cm}=\dfrac{Mx_{1}+M'x_{2}}{M+m'}$

Put the value into the formula

$3=\dfrac{60\times0+40\times x}{60+40}$

$x=\dfrac{30}{4}$

$x=7.5$

For y,

Similarly

$y=7.5$

For z,

$z=7.5$

Hence, The particle of 40 kg placed at (7.5, 7.5, 7.5).

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Topic : center of mass

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