Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (0.0, 3.2) m, and 4.0 kg at (2.5, 0.0) m. where should a fourth object of 8.3 kg be placed so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m

The location of the fourth object such the center of gravity of the four-object arrangement will be at (0.0, 0.0) m is;

(-1.20, -1.12) m

We are given the mass distribution as;

5kg at (0, 0)

2.9 kg at (0, 3.2)

4 kg at (2.5, 0)

8.3 kg at say (x, y)

Now, to get the position of where the fourth object will be placed so that the centre of gravity will be at (0, 0), we will use the center of gravity formula. Thus;

x' = [(m1*x1) + (m2*x2) + (m3*x3) + (m4*x4)]/(m1 + m2 + m3 + m4)

Plugging in the relevant values gives;

x' = [(5*0) + (2.9*0) + (4*2.5) + (8*x)]/(5 + 2.9 + 4 + 8.3)

x' = (8.3x + 10)/20.2

Similarly;

y' = [(m1*y1) + (m2*y2) + (m3*y3) + (m4*y4)]/(m1 + m2 + m3 + m4)

Plugging in the relevant values gives;

y' = [(5*0) + (2.9*3.2) + (4*0) + (8.3*y)]/(5 + 2.9 + 4 + 8.3)

y' = (9.28 + 8.3y)/20.2

For the center of gravity to be at (0,0), then;

x' = 0 and y' = 0. Thus;

(8.3x + 10)/20.2 = 0

8.3x = -10

x = -10/8.3

x = -1.2 m

Likewise;

(9.28 + 8.3y)/20.2 = 0

8.3y = -9.28

y = -9.28/8.3

y = -1.12 m

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x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

Similarly for y direction we have

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

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