Question

A square plate is thrown horizontally on a rough horizontal surface. Find the minimum value of coefficient of
friction so that the plate topples.​

Answer 1

The minimum value of coefficient of friction so that the plate topples is 1

Explanation:

Let the side of the square plate is a

If the plate is thrown horizontally so that it is sliding and it has to topple then the the torque due to friction at the extreme right corner of the plate should be greater than the torque due to its weight

Therefore,

$F\times \frac{a}{2}\ge Mg\times\frac{a}{2}$   (The friction force acting on the whole lowers surface of the plate can be assumed to be acting at the middle)

$\implies \mu Mg\ge Mg$

$\implies \mu\ge 1$

Therefore, the minimum value of the coefficient of friction is 1

Hope this helps.

Answer 2

Explanation:

For toppling about the extreme bottom right corner of the square plate thee torque of force should be greater than mg then only the plate can topple. Let us take the side as "a" unit.So tau(F)>tau(mg) F.a>mg.aF>mg ........ (1)To avoid sliding FMG (from (1))u>1So the answer is 1