- #1

cantgetaname

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## Homework Statement

Two bodies of mass m_{1}and m_{2}are placed on a horizontal table with coefficient of friction [tex]\mu[/tex] and are joined by a spring. Initially, the spring has its natural length.

If F is the minimum force which, when applied on m_{1}, will make the other block m_{2}just move,find F. (kis the spring constant)

Here is an artistic illustration of the situation.

## Homework Equations

- Limiting Frictional Force = [tex]\mu[/tex] x Normal/reaction offered by the surface.

- Force due to spring = k * Elongation of spring (from its natural length)

## The Attempt at a Solution

If m

_{2}is to move, it needs to experience a force of [tex]\mu m_{2}g[/tex] (to oppose the friction), which can be only experienced from the spring.

This means the spring's elongation at that instant is atleast [tex]\frac{\mu m_{2}g}{k}[/tex].

An equal and opposite force acts on m

_{1}. For m

_{1}to move, net force on it must be greater than [tex]\mu m_{1}g[/tex].

That gives [tex]F=\mu m_{1}g+\mu m_{2}g[/tex], which is incorrect.

The answer is [tex]F=\mu m_{1}g+\frac{\mu m_{2}g}{2}[/tex], which I have no idea of where it could possibly come from.