Find the height of the center of mass of the equilateral triangular lamina from the side AB as shown in figure. Given, side of the triangular lamina is 3m.
A
√3m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3√2m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
√32m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
√23m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C√32m Given, side of the equilateral triangular lamina is 3m.
In this case, triangular lamina is symmetric about y− axis. So, x− co-ordinate of COM of the triangular lamina is 0.
We know that, y-coordinate of COM is at 13rd of height from the base i.e 13×√3a2
∴ Position of COM of the triangular lamina is given by (xCOM,yCOM)=(0,a2√3)
So, required height (H) of COM above AB is equal to a2√3 H=a2√3=32√3m=√32m.