# How can I find the horizontal contact force between two circles?

#### Chemist116

The problem is as follows:

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: #### skeeter

Math Team
Please state the question in the text of the post. Thank you.

$\sigma = \dfrac{m}{LW}$

$W = 2R(1-\cos{\phi}) \implies mg = 2gR(1-\cos{\phi})L\sigma$

I get #1 ... hope someone checks me.

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#### Chemist116

Please state the question in the text of the post. Thank you.

$\sigma = \dfrac{m}{LW}$

$W = 2R(1-\cos{\phi}) \implies mg = 2gR(1-\cos{\phi})L\sigma$

I get #1 ... hope someone checks me.
My apologies for that. I did wrote this in a rush as I'm reviewing before my exam. The question mentioned at the end. Find the horizontal component of the contact force in the system shown. Your answer is correct according to the answers sheet in my book.

I'm stuck at how did you get the value for $W$ in terms of the radius of the circle?.

There is still the fact on how did you get to $1$?. Can you show the steps that you omitted? :help:

I get:

$F_{x}= r \cos \phi$

$r\sin\phi=mg$

$r=\frac{mg}{\sin\phi}$

Therefore:

$F_{x}= \frac{mg}{\sin\phi} \cos \phi$

Then:

$F_{x}= \frac{2gR(1-\cos{\phi})L\sigma}{\sin\phi} \cos \phi$

But what's up with 2?. I can't find a way to cancel. I'm confused about the direction of the reaction. Is pointing downwards the same as the weight of the flat surface. Would be negative?.

Can you tell which part I got lost?. :help:

#### skeeter

Math Team
$r\sin{\phi} = \dfrac{mg}{2}$