Question

What will be the value of maximum acceleration of the truck in the forward direction so that the block kept on the back does not topple $?$

• A. $Ag/h$
• B. $hg/h$
• C. $Ag/2h$
• D. $h/bg$

Answer 1

The correct option is (A) $Ag/h$

We know,

When the truck moves forward with acceleration$\left(\alpha \right)$ the block gets a pseudo force in the backward direction which acts at the center of the block and this force gives an anticlockwise torque which must be balanced by the clockwise torque provided by the weight of the block.

So,

The torque due to pseudo force $=$ Torque due to weight of the block

Putting all the values

Taking toppling condition:

$m\times \alpha \times \frac{h}{2}=m\times g\times \frac{A}{2}$

After solving

$\alpha =\frac{Ag}{h}$

Hence the maximum acceleration is $\frac{Ag}{h}$