First slide

Methods to locate centre of mass

Question

Two identical bricks of length L are piled one on top of the other on a table as shown in the figure. The maximum distance S the top brick can overhang the table with the system still balanced is:

Moderate

A

$\frac{1}{2} L$
B

$\frac{2}{3} L$
C

$\frac{3}{4} L$
D

$\frac{7}{8} L$

Solution

Maximum upper block can hange on upper $(\frac{l}{2})$
Maximum lower can hang on edge $(\frac{l}{4})$
$\begin{array}{l} S = \frac{l}{2} + \frac{l}{4} \\ = \frac{3 l}{4} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App