p09_008.pdf

8. The centers of mass (with centimeters understood) for each of the ﬁve sides are as follows:

( x 1 ,y 1 ,z 1 )=(0 , 20 , 20) for the side in the yz plane

( x 2 ,y 2 ,z 2 )=(20 , 0 , 20) for the side in the xz plane

( x 3 ,y 3 ,z 3 )=(20 , 20 , 0) for the side in the xy plane

( x 4 ,y 4 ,z 4 )=(40 , 20 , 20) for the remaining side parallel to side 1

( x 5 ,y 5 ,z 5 )=(20 , 40 , 20) for the remaining side parallel to side 2

Recognizing that all sides have the same mass m , we plug these into Eq. 9-5 to obtain the results (the

ﬁrst two being expected based on the symmetry of the problem).

(a)

x com = mx 1 + mx 2 + mx 3 + mx 4 + mx 5

5 m

= 0+20+20+40+20

=20cm

(b)

y com = my 1 + my 2 + my 3 + my 4 + my 5

5 m

= 20+0+20+20+40

=20cm

(c)

z com = mz 1 + mz 2 + mz 3 + mz 4 + mz 5

5 m

= 20+20+0+20+20

=16cm