Three objects are located at the following positions in atwo-dimensional (x, y)

Question

Three objects are located at the following positions in atwo-dimensional (x, y) coordinate system: m1 at (2.3 m, 4.4 m), m2at (12.2 m, 89.4 m), and m3 at    (74.1 m, -0.8m).
If the masses are 1 kg, 1.5 kg, and 2.1 kg, respectively, what isthe position of the center of mass of this system? Please give step by step instructions with solution Three objects are located at the following positions in atwo-dimensional (x, y) coordinate system: m1 at (2.3 m, 4.4 m), m2at (12.2 m, 89.4 m), and m3 at    (74.1 m, -0.8m).
If the masses are 1 kg, 1.5 kg, and 2.1 kg, respectively, what isthe position of the center of mass of this system? Please give step by step instructions with solution

Explanation / Answer

Good way to handle this is treating the x and y coordinatesseperately. We'll just go to what we'll apply here to reduce the clutter of theformula derivations. (M1*2.3 + M2*12.2 +M3*74.1)/(M1+M2+M3) for our x coordinate of CoM All we did we sum the multiples of each mass*coordinate and divideit by the sum of the masses. If you run this through your calc with m1 m2 and m3 being 1, 1.5,2.1 you end up with a x coordinate of 38.307 Intuitive balancing of the 3 objects with 1 and 1.5 on theleft and 2.1 on the right, 38 sounds about right. now we do the same with our y component (M1*4.4 + M2*89.4 +M3*-0.8)/(M1+M2+M3)   notice how our last component will turn the addition into anegative. We end up with a Y component of29.743               so CoM is      (38.307, 29.743) in the Y case our 1.5 mass was very high, but we had so much weightbelow 4.4 (3.1kg) to bring the com of the Y components downto 30ish.            I think most dont have much trouble visualizing the COM, its just amatter of knowing to sum the x and y coordinates with their massmultiples and divide by the sum of the masses. So justremember             (X1*M1 + X2*M2 + ......+ Xn*Mn)/(M1 + M2 +..... + Mn)       for either your x or y component and you should be fine.

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