Center of Mass (easy)

WhoCares357 · #1 $Old$ November 1st, 2009, 11:05 AM

Edit: Nevermind. I figured out that I was using the wrong coords and the wrong mass of the second object. The correct answer was -.25m.

Quote:

A uniform square plate 6. m on a side has had a square piece 2. m on a side cut out of it (see the figure). The center of that piece is at x=2 m, y = 0. The center of the square plate is at x = y = 0. Find the coordinates of the center of mass of the remaining piece. x-component ( m)?

The formula for center of mass is $r_x=\frac{x_1*m_1+x_2*m_2+x_3*m_3}{m_1+m_2+m_3}$
$m=(density)Ah$ Where A is area and h is height of the object.

This is a very easy topic and a very easy question. I'm just having trouble seeing where the coords are. In my attempt I had the following values:
$x_1=x_2=4$
$x_3=1$
$m_1=m_2=2^2*1*1=4$
$m_3=6^2*1*1=36$
My final answer was 1.54m. The online system tells me this is wrong. Does anyone see where I went wrong?

Thanks.

earboth · #2 $Old$ November 1st, 2009, 01:01 PM

Quote:

Originally Posted by WhoCares357 $View Post$

Edit: Nevermind. I figured out that I was using the wrong coords and the wrong mass of the second object. The correct answer was -.25m.

The formula for center of mass is $r_x=\frac{x_1*m_1+x_2*m_2+x_3*m_3}{m_1+m_2+m_3}$
$m=(density)Ah$ Where A is area and h is height of the object.

This is a very easy topic and a very easy question. I'm just having trouble seeing where the coords are. In my attempt I had the following values:
$x_1=x_2=4$ <<<<<< here: x_1 = x_2 = 0
$x_3=1$ <<<<< x_3 = -1
$m_1=m_2=2^2*1*1=4$
$m_3=6^2*1*1=36$
My final answer was 1.54m. The online system tells me this is wrong. Does anyone see where I went wrong?

Thanks.

The mass and the area of the partitions of the plate have the same measures. m_1 and m_2 are the masses of the top and bottom rectangles. The centers of mass of these rectangles must be at x = 0.

m_3 is the mass of the reduced rectangle. The center of mass of this rectangle must be at x = -1

Now using your formula you'll get:

$r_x=\dfrac{-1 \cdot 8}{12+12+8}=-\dfrac8{32} = -0.25$

I've modified you sketch a little bit to illustrate my considerations.