Center of gravity of a stick

[diva8789]

Homework Statement

Three uniform meter sticks, each of mass m, are placed on the floor as follows: stick 1 lies along the y-axis from y = 0.500 m to y = 1.50 m, stick 2 lies along the x-axis from x = 0.110 m to x = 1.11 m, stick 3 lies along the x-axis from x = 1.34 m to x = 2.34 m. Calculate the location of the center of mass of the meter sticks. Enter the x value first and then the y value.

Homework Equations

m1*y1+m2*y2.../m1+m2...

The Attempt at a Solution

this is the problem that i have to figure out. i know the equation for the center of gravity as being basically (m1y1+m2y2...)/m1+m2+... but since these are meter sticks and they all have the same weight yet take up more than one point on the system, i don't know where to go from here...can someone please help me?

 

[Dick]

They are uniform sticks, so the center of mass is at the center of the stick.

 

[ogb p]

To calculate the center of gravity of the meter sticks, we can use the equation m1y1 + m2y2 + m3y3 / m1 + m2 + m3, where m represents the mass of each stick and y represents the distance from the origin (0,0). In this case, all three sticks have the same mass, so we can simplify the equation to y1 + y2 + y3 / 3.

To find the y values, we can use the given information that stick 1 is along the y-axis from y = 0.500 m to y = 1.50 m. This means that the midpoint of stick 1 is at y = (0.500 + 1.50) / 2 = 1.00 m. Similarly, the midpoint of stick 2 is at x = (0.110 + 1.11) / 2 = 0.610 m, and the midpoint of stick 3 is at x = (1.34 + 2.34) / 2 = 1.84 m.

Substituting these values into our equation, we get (1.00 + 0.610 + 1.84) / 3 = 1.15 m. Therefore, the center of gravity of the meter sticks is located at (0, 1.15).

It is worth noting that the x-coordinate of the center of gravity is not needed in this calculation since all three sticks are aligned along the x-axis, and therefore the center of gravity will always lie at the midpoint of the sticks. I hope this helps with your problem!