### Video Transcript

The table shows the distribution of a system of masses on a uniform lamina. Which of the following is the center of mass of this system?

We are given five options (A) to (E), which each contain expressions for the 𝑥- and 𝑦-coordinate of the center of mass. We can answer this question by recalling the negative mass method. In the table, we are told that we have a lamina with mass 𝑚 one. A mass 𝑚 two is added to the lamina. And a mass 𝑚 three is removed. This means that we have two positive mass bodies, 𝑚 one and 𝑚 two, together with a negative mass body, 𝑚 three.

We are also given in the table the centers of mass of 𝑚 one, 𝑚 two, and 𝑚 three. These lie at the coordinates 𝑥 one, 𝑦 one; 𝑥 two, 𝑦 two; and 𝑥 three, 𝑦 three, respectively. We recall that the 𝑥-coordinate of the center of mass of the system, denoted 𝑥 bar, is equal to the sum from 𝑖 equals one to 𝑛 of 𝑚 sub 𝑖 multiplied by 𝑥 sub 𝑖 divided by the sum from 𝑖 equals one to 𝑛 of 𝑚 sub 𝑖.

In this question, the numerator will be equal to 𝑚 one 𝑥 one plus 𝑚 two 𝑥 two minus 𝑚 three 𝑥 three. And the denominator will be equal to 𝑚 one plus 𝑚 two minus 𝑚 three. This corresponds to the 𝑥-coordinate in option (A). Options (B) and (D) cannot be correct as the 𝑥-coordinate here contains 𝑦-values. Option (C) is incorrect as we have added the mass 𝑚 three and not removed it. And option (D) is not correct as the signs of 𝑚 two and 𝑚 three are incorrect. This expression corresponds to removing 𝑚 two and adding 𝑚 three.

We can repeat this process to find the 𝑦-coordinate of the center of mass 𝑦 bar as shown. This is equal to 𝑚 one 𝑦 one plus 𝑚 two 𝑦 two minus 𝑚 three 𝑦 three all divided by 𝑚 one plus 𝑚 two minus 𝑚 three. This also corresponds to the expression for the 𝑦-coordinate given in option (A). The center of mass of the system in the table is 𝑚 one 𝑥 one plus 𝑚 two 𝑥 two minus 𝑚 three 𝑥 three divided by 𝑚 one plus 𝑚 two minus 𝑚 three, 𝑚 one 𝑦 one plus 𝑚 two 𝑦 two minus 𝑚 three 𝑦 three divided by 𝑚 one plus 𝑚 two minus 𝑚 three.