A block of mass m is at rest on a rough inclined plane of angle of inclination $\theta$. If coefficient of friction between the block and the inclined plane is $\mu$, then the minimum value of force along the plane required to move the block on the plane is

To solve the problem of determining the minimum force required to move a block of mass $$m$$ resting on a rough inclined plane with an angle of inclination $$\theta$$ and a coefficient of friction $$\mu$$, we can follow these steps:

Step 1: Identify the forces acting on the block
1. Weight of the block: The weight $$W$$ of the block acts vertically downwards and is given by $$W=mg$$.
2. Components of weight:
- The component of weight acting parallel to the incline: $$W_{\parallel }=mg\sin \theta$$.
- The component of weight acting perpendicular to the incline: W_{\perpendicular} = mg \cos \theta .
3. Normal force: The normal force $$N$$ acts perpendicular to the inclined surface and balances the perpendicular component of the weight.

Step 2: Determine the frictional force
The frictional force $$f$$ that opposes the motion is given by:
$$f=\mu N$$
Since the normal force $$N$$ is equal to the perpendicular component of the weight, we have:
$$N=mg\cos \theta$$
Thus, the maximum frictional force can be expressed as:
$$f=\mu mg\cos \theta$$

Step 3: Set up the equilibrium condition
To move the block, the applied force $$F$$ must overcome both the component of the weight acting down the incline and the frictional force. Therefore, we can write the equilibrium condition as:
$$F+mg\sin \theta =f$$
Substituting the expression for frictional force, we get:
$$F+mg\sin \theta =\mu mg\cos \theta$$

Step 4: Solve for the applied force $$F$$
Rearranging the equation to isolate $$F$$:
$$F=\mu mg\cos \theta -mg\sin \theta$$
Factoring out $$mg$$:
$$F=mg(\mu \cos \theta -\sin \theta )$$

Final Answer
The minimum value of the force $$F$$ required to move the block along the inclined plane is:
$$F=mg(\mu \cos \theta -\sin \theta )$$