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I did not understand the following steps regarding calculation of minimum force F required to topple object about point A.

Total Torque = Fb - mga/2 If torque>0 then body topples.

What i did not understand is that how he got the term 'mga/2'. Also at the time when the body is about to topple the normal reaction will be acting at the edge A and so, wont the torque of this force about point A be zero?

Could someone please help me out.

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You should also note that static friction acts on the body when it is about to topple. Friction acts along the ground so it's torque about A is also 0.

Torque due to F is clockwise and torque due to mg is counter-clockwise. Since torque= lever arm * force (or in vector form $\vec{\tau}=\vec{r}x\vec{F}$), torque due to mg is $mg\frac{a}{2}$. So net torque is $Fb-mg\frac{a}{2}$ in the clockwise about A.

A more detailed analysis:

Since the block is right now in equilibrium, balancing forces along the horizontal and vertical :$F=f, N=mg$.

Now balancing the torque about the center of mass of the body, torque due to $mg$ is 0 (moment arm is 0), due to $N$ is $N\frac{a}{2}$ and due to $F$ and $f$ is $F\frac{b}{2}$ and $f\frac{b}{2}$ respectively. Using the relations mentioned above, you can see that the expression for the net torque remains the same.

You can try this yourself for any point you wish to consider. (Since the net force on the block is 0, the expression for torque will be same for all points).

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  • $\begingroup$ But isn't the normal reaction acting at the point of consideration itself so that torque due to it is 0 $\endgroup$ – physics123 Jan 23 '16 at 15:57
  • $\begingroup$ yes, it's true. In the expression, you don't see the torque due to the normal since it's 0. $\endgroup$ – Akshit Jan 24 '16 at 0:56
  • $\begingroup$ the torque due to normal about point A is 0. $\endgroup$ – Akshit Jan 24 '16 at 1:09
  • $\begingroup$ Does 'f' stand for static frictional force? $\endgroup$ – physics123 Jan 24 '16 at 6:58
  • $\begingroup$ @physics123 yes. since the block is not sliding, the friction is static instead of kinetic. $\endgroup$ – Akshit Jan 24 '16 at 12:26

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