Locate the center of mass x y z of the homogeneous block ass

Locate the center of mass (x, y, z) of the homogeneous block assembly. Please show equations used and calculations thank you for your help!!!

250 mm 200 mm 150 mm 100 mm 150 mm 150 mm

Solution

For the centre of mass of a system, we know that: Xcom = dmx_i / dm

Similarly for Y and Z directions,we have: Ycom = dmy_i / dm

and Zcom = dmz_i / dm

Now, for the given assembly we need to indentify the smaller pieces for which the centre of mass is known and then put them into the relations above to find the co-ordinates for the centre of mass of the system.

Also, it needs to be understood that for a cuboid, the centre of mass lies at the centre of the cuboid, and for an isosceles triangle, it lies on the perpendicular bisector of the base.

Let us assume that the density of the material is p, hence we have:

For X direction,

Xcom = [150x150 x 550p x 75 + 150 x 200 x 150p x 225 + 37.5 x 0.5 x 150 x 150 x 100p] / [0.5 x 150 x 150 x 100p + 150 x 200 x 150p + 150x150 x 550p]

or, Xcom = [5.5p x 75 + 2p x 225 + 37.5 x 0.5p] / [0.5p + 2p + 5.5p] = 110.15625 mm

For Ydirection:

Ycom = [150x150 x 550p x 275 + 150 x 200 x 150p x 450 + 0.5 x 150 x 150 x 100p x 50] / [0.5 x 150 x 150 x 100p + 150 x 200 x 150p + 150x150 x 550p]

or, Ycom = [5.5p x 275 + 2p x 450 + 0.5p x 50] / [0.5p + 2p + 5.5p]

or, Ycom = 304.6875 mm

For Z direction:

Zcom = [150x150 x 550p x 75 + 150 x 200 x 150p x 75 + 0.5 x 150 x 150 x 100p x 37.5 ] / [0.5 x 150 x 150 x 100p + 150 x 200 x 150p + 150x150 x 550p]

or, Zcom = [5.5p x 75 + 2p x 75 + 0.5p x 37.5 ] / [0.5p + 2p + 5.5p]

or, Zcom = 72.65625 mm

Therefore the required centre of mass is at (110.15625, 304.6875, 72.65625)