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It is inclined to the vertical at an angle $\theta$. The coefficient of friction is $u$? The block can be pushed along the surface only if_______ enter image description here

Clearly, the total normal force on the block will be $$N=Mg+Mg\cos\theta$$Then friction would be $$f=uMg(1+\cos\theta)$$ So $$F\sin\theta \ge f$$ $$Mg\sin\theta \ge uMg(1+\cos\theta)$$ $$\sin\theta \ge u(1+\cos\theta)$$ I can’t go further, the answer is $$\tan \theta /2 \ge u$$

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  • $\begingroup$ That looks good. I think giving the condition as you have is enough. Another way to go is to solve for the critical angle which is $2\arctan(u)$. To check if your result looks reasonable: note that if $u=0$ then the block will move as long as $\theta > 0$ which is correct (no friction so any force in the $x$ direction will do). If $\theta = \pi$ then it will not move no matter how small the friction is (also reasonable as this time the force points stright up). $\endgroup$
    – Winther
    Nov 19 '19 at 15:45
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Note that $$\tan\frac{\theta}{2} = \frac{\sin\theta}{1+\cos\theta}.$$ Now, you will be able to push the block if: $$Mg\sin\theta \ge u(Mg+Mg\cos\theta) \Rightarrow \sin\theta \ge u(1+\cos\theta) \Rightarrow \frac{\sin\theta}{1+\cos\theta}=\tan\frac{\theta}{2} \ge u$$

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