Question

Two identical uniform rectangular blocks (with longest side L) and a solid sphere of radius R are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure. If the mass of each block is M and of the sphere is $M/2$, then the maximum distance x that can be achieved is$?$

• A. $8L/15$
• B. $5L/6$
• C. $(3L/4+R)$
• D. $(7L/15+R)$

Answer 1

Answer: (A)

$8L/15$

For the system to remain stationary and un-toppled, the moments due to weights must be balanced with the normal reaction.

But, the normal reaction location is limited by the edge of the table.

$\text{Part 1 : Stability of the Upper block and Sphere}$

Let the length of the upper block, beyond the lower block be $y$.

The normal force exerted by the lower block on the upper block is $\frac{3Mg}{2}$

The farthest distance possible is when normal reaction acts at the edge.

Moment balance about the edge of the lower block gives $\frac{Mg}{2}y=Mg(\frac{L}{2}-y)$

$\therefore y=\frac{L}{3}$

$\text{Part 2 : Stability of Upper masses and lower block}$

For the farthest distance possible without toppling, normal reaction acts on the edge of the table.

Normal reaction exerted by the table is $Mg+Mg+0.5Mg=2.5Mg$

Fixing origin at the edge of the table,

Location of CG of lower block is $x-y-0.5L$

Location of CG of upper block is $x-0.5L$

Location of CG of sphere is $x$

Applying moment balance about the edge of the table, we have

$Mg(x-y-0.5L)+Mg(x-0.5L)+0.5Mg\left(x\right)=0⟺5x=2(L+y)=\frac{8L}{3}$

$\therefore x=\frac{8L}{15}$