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Pavlova-9 [17]
3 years ago
6

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.5 kg at (

0.0, 3.3) m, and 4.0 kg at (2.8, 0.0) m. Where should a fourth object of 7.1 kg be placed so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m?
Physics
1 answer:
nekit [7.7K]3 years ago
8 0

Answer:  x = -1.6  y= -1.2

Explanation:

By definition the x- and –y coordinates of the center of mass for a discrete set of bodies, are given by the following expressions:

Xi = ∑ mi. xi  / ∑ mi

Yi = ∑ mi. yi  / ∑ mi

In order to the x- and –y coordinates of the center of mass of all the system, be (0, 0),

both num.erators in the Xi, Yi expressions must be equal to 0.

So, replacing by the values, and taking into account that the 5.0 Kg is located in the origin, we can write the following:

Xi = 2.8 . 4 Kg + X4 . 7.1 Kg = 0   → X4 = -1.6

Yi = 3.3 . 2.5 Kg + Y4 . 7.1 Kg = 0  → Y4 = -1.2

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| Impedance | = √ [R² +(ωL)²]

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Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











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