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A cubical block of mass $m$ and edge length $a$ slides down a rough inclined plane of inclination $\alpha$ radian with a uniform speed. Find the torque of the normal force acting on the block about its center.

On the back of the book the answer is $\frac{1}{2} mga \sin(\alpha)$.

I have done the basic steps like the force of friction $f=mg\sin(\alpha)$ and the normal reaction is $n=mg\cos(\alpha)$. Also as the object is only slipping so net torque $=0$ or $T(N)+T(F)=0$ now if we calculate $T(F)$ question will be solved and even though I have determined the force of friction but what distance should I take and why?

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    $\begingroup$ Have you asked your instructor for help? $\endgroup$
    – Bill N
    Commented Oct 12, 2015 at 17:43

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enter image description hereInclined system
The block moves down the inclined with uniform speed. Therefore, $$ F_f = mgsin\alpha\space, $$ Due to symmetry as you said above gravity will produce no torque.
Now the only left-over forces are Normal and Friction which will produce torque. $$ T\space_n\space_e\space_t=T_N+T_F\space, $$ But the block doesn't roll which means there is no net torque to provide angular acceleration . Therefore, $$ T\space_n\space_e\space_t=T_N+T_F\space=0, $$ $$ T_N=-T_F $$ $$ T_N=-F(a/2) $$ $$ T_N=-(a/2)mgsin\alpha\space\space(i.e\space clockwise) $$ $$ Magnitude\space of\space T_N=(a/2)mgsin\alpha $$

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  • $\begingroup$ Can you elaborate a bit more on how you got the value of T(F) $\endgroup$
    – Freelancer
    Commented Oct 12, 2015 at 16:55
  • $\begingroup$ @Freelancer We have already found that $$ Frictional\space Force \space=mgsin\alpha $$ and it is given that the block slides down with uniform speed.Therefore we can say that the forces are balanced i.e the gravitational and frictional force. $$ mgsin\alpha=\mu mgcos\alpha=F_f $$ $\endgroup$
    – Sathyaram
    Commented Oct 12, 2015 at 17:04
  • $\begingroup$ I meant how you got a/2 here. $\endgroup$
    – Freelancer
    Commented Oct 12, 2015 at 17:29
  • $\begingroup$ @Freelancer Did you see the pic above, it is the position of COM in a cube, To find torque simply do F X R. i.e. F(a/2) $\endgroup$
    – Sathyaram
    Commented Oct 12, 2015 at 17:31
  • $\begingroup$ But was frictional force only acting on the particles near the edge !! I think it was acting on all the pertickes but still you are calculating torque about centre and taking distance as a/2 $\endgroup$
    – Freelancer
    Commented Oct 12, 2015 at 17:37

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