A statics problem to find minimal friction

Question

consider the following:

I need to find the minimal coeeficient of friction $\mu _{min}$ so that both recatngle boxes would remain static. The lower angle in the triangle is $2\alpha$ as indicated. I ended up with this expression: $$\mu _{min}=\frac{m_2}{2\cdot m_1 \cdot \tan \alpha}$$ but the answer in the book is rather this one: $$\frac{m_2}{(2m_1+m_2)\tan\alpha}$$ Who is correct?I know that the normal force that that the triangle is exerting on both is like this: (on both sides of course) so with geometry and soome FBD that's what I came out with.

Would like to hear your thoughts!

Answer 1

I've called $\beta$ the angle that $\alpha + \beta = 90^o$ so $\tan\alpha = 1/\tan\beta$ Since the triangle is not moving, and the situation is symmetrical: the horizontal component of $N$ will cancel out, while the vertical ones must compensate for the weight of the triangle. $$ 2N \cos \beta = m_2g$$ I'll call $F_x$ the horizontal component of $-N$ the reaction on the rectangle of $N$, $F_y$ the vertical component.

$$ F_x = N\sin \beta $$ $$ F_y = N \cos \beta $$

Friction must compensate the force on the horizontal axis $F_x$ so $$ F_{friction} = F_x \Rightarrow \mu R = F_x $$ The reaction of the table is given by the vertical component of the sum of all forces so $R = m_1g + F_y$ from the first you get $$ N = \frac{m_2g}{2\cos \beta}$$ so $$\mu = \frac{F_x}{F_y + m_1g} = \frac{N\sin\beta}{N\cos\beta + m_1g} = \frac{\frac{m_2g}{2\cos \beta}\sin\beta}{\frac{m_2g}{2\cos \beta}\cos\beta + m_1g}$$ simplify you'll get $$\mu = \frac{m_2g\tan\beta}{(2m_1 + m_2)g} = \frac{m_2}{(2m_1 + m_2)\tan\alpha}$$