The figure from below shows a rectangular object entirely flat which is placed over two circles of radius $R$. The superficial mass density of the object is $\sigma$ and the radius $R$ of each circle makes with both contact points an angle $\phi$ with the horizontal of the reaction force.

The given alternatives are:

$\begin{array}{ll}

1.&\frac{gR\left(1-cos\phi\right)L\sigma}{\sin\phi}\cos\phi\\

2.&gR\left(1-\cos\phi\right)L\sin\phi\\

3.&gR\left(1-\cos\phi\right)L\sigma\cos\phi\\

4.&\frac{R\left(1-cos\phi\right)L\sigma}{\sin\phi}\cot\phi\\

5.&\frac{g\left(1-\cos\phi\right)L\sigma}{\sin\phi}\cos\phi\\

\end{array}$

Can somebody help me here with the vectors. Because I'm confused at exactly how to put the vectors in such a way that I can relate it with the superficial mass.

The only thing which I can remember is that the superficial mass is:

$\sigma=\frac{\textrm{mass}}{\textrm{area}}$

But in this particular situation I'm confused at exactly how to find the area of the square which would be required to get the mass and hence the force.

Can somebody help me with a drawing to find specifically how to find what it is being requested?. More importantly can somebody help me with what on earth is the

**horizontal reaction force**? :help: