a thin rod of length 3l is bent at aright angle at adistance of l of on the x axis and at adistance of 2l on the y axis locate the center of mass with respect to the corner tale and l is1.2

0.2i*+0.8j*

To locate the center of mass of the thin rod, we need to find the coordinates of the center of mass (x_cm, y_cm) with respect to the corner of the table.

Let's break down the problem into two parts: the horizontal part (along the x-axis) and the vertical part (along the y-axis).

1. Horizontal Part:
The rod is bent at a right angle at a distance of l on the x-axis. This means that the center of mass on the x-axis will be located at a distance of l/2 from the bend point.

Since the length of the rod is 3l, the horizontal distance from the corner of the table to the center of mass is:
(x_cm_horizontal) = (distance from corner of table to bend point) + (distance from bend point to center of mass)

(x_cm_horizontal) = l + (l/2) = (3/2)l

2. Vertical Part:
The rod extends 2l on the y-axis. This means that the center of mass on the y-axis will be located at a distance of l from the bend point.

In this case, the vertical distance from the corner of the table to the center of mass is:
(y_cm_vertical) = (distance from corner of table to bend point) + (distance from bend point to center of mass)

(y_cm_vertical) = 2l + l = 3l

Therefore, the center of mass with respect to the corner of the table is located at coordinates (x_cm, y_cm):
(x_cm, y_cm) = ((3/2)l, 3l)

Since you have given the value of l as 1.2, we can substitute it into the equation to find the precise coordinates:
(x_cm, y_cm) = ((3/2) * 1.2, 3 * 1.2) = (1.8, 3.6)

So, the center of mass with respect to the corner of the table is located at (1.8, 3.6).