Finding torque due to normal force [closed]

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?

• Have you asked your instructor for help? Oct 12 '15 at 17:43  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, $$T\space_n\space_e\space_t=T_N+T_F\space=0,$$ $$T_N=-T_F$$ $$T_N=-F(a/2)$$ $$T_N=-(a/2)mgsin\alpha\space\space(i.e\space clockwise)$$ $$Magnitude\space of\space T_N=(a/2)mgsin\alpha$$
• @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. $$mgsin\alpha=\mu mgcos\alpha=F_f$$ Oct 12 '15 at 17:04