Question

. You are holding a shopping basket at the grocery store with two 0.62-kg cartons of cereal at the left end of the basket. The basket is 0.61 m long. Where should you place a 1.9-kg half gallon of milk, relative to the left end of the basket, so that the center of mass of your groceries is at the center of the basket?

Answer 1

This is a variation from the normal center of mass problem.

We have mass 1 (M1) at the extreme left, a balance point in the dead center of a stick, and a mass 2 (M2) at a variable position near the right side. We're given the values:

M1 = 1.24 kg (because you have 2 * (0.62 kg) cartons of cereal)

M2 = 1.9 kg

R = 0.61 m

c = 0.61/2 = 0.305 m

First, sum the masses:

MT = M1 + M2

MT = (1.24 kg) + (1.9 kg)

MT = 3.14 kg

Second, We know that M1 is on the extreme left, and M2 is an unknown variable distance within the total length R of the basket. The right end of the basket of total distance R is the datum. Multiply each mass by its distance from the datum

M1D = (1.24 kg) * (0.61 m)

M1D = 0.7564 kg-m

M2D = (1.9 kg) * (x)

Third, knowing that the center of mass = 0.305 m, you divide the M-D sum by the MT sum and solve:

[ M1D + M2D ] / MT = c

Do a little isolation of variables before punching in numbers leaves us with M2D which we need to break down

M2D = [ c * MT ] - M1D

Now you can add in the numbers and find your final value:

(1.9 kg) * (x) = [ (0.305 m) * (3.14 kg) ] - [ 0.7564 kg-m ]

x = 0.106 m

This is the distance from the datum, which was the extreme right side as defined (0.61 m), so we just subtract that to get the distance relative to the left end of the basket

d = (0.61 m) - (0.106 m)

d = 0.504 m