A body of weight W is lying at rest on a rough horizontal surface. If the angle of friction is , then the minimum force required to move the body along the surface will be

The correct answer is W sinθ.

Step 1: Understand the system and forces acting on the body
Consider a body of weight $W$ lying at rest on a rough horizontal surface.
Let a force $P$ be applied at an angle $\alpha$ with the horizontal to move the body.
The forces acting on the body are:
1. Weight $W$ acting vertically downwards.
2. Normal reaction $N$ acting vertically upwards.
3. Applied force $P$ at an angle $\alpha$ to the horizontal.
4. Frictional force $f$ acting horizontally opposite to the direction of motion.

Step 2: Set up the equations of equilibrium
Resolving the forces vertically:
$N+P\sin \left(\alpha \right)=W$
This gives the normal reaction as:
$N=W-P\sin \left(\alpha \right)$
Resolving the forces horizontally:
$f=P\cos \left(\alpha \right)$

Step 3: Relate friction and normal force
For the body to just begin moving, the frictional force must equal the limiting static friction:
$f=\mu N$
where $\mu$ is the coefficient of static friction. Substituting the expressions for $f$ and $N$:
$P\cos \left(\alpha \right)=\mu (W-P\sin (\alpha \left)\right)$
Rearranging terms to solve for $P$:
$P\cos \left(\alpha \right)+\mu P\sin \left(\alpha \right)=\mu W$
$P(\cos (\alpha )+\mu \sin (\alpha \left)\right)=\mu W$
$P=\frac{\mu W}{\cos \left(\alpha \right)+\mu \sin \left(\alpha \right)}$

Step 4: Express using the angle of friction
The relation between the coefficient of friction $\mu$ and the angle of friction $\theta$ is given by:
$\mu =\tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}$
Substituting this value of $\mu$ in the equation for $P$:
$P=\frac{\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}W}{\cos \left(\alpha \right)+\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\sin \left(\alpha \right)}$
Multiplying the numerator and denominator by $\cos \left(\theta \right)$:
$P=\frac{W\sin \left(\theta \right)}{\cos \left(\alpha \right)\cos \left(\theta \right)+\sin \left(\alpha \right)\sin \left(\theta \right)}$
Applying the trigonometric identity $\cos (A-B)=\cos \left(A\right)\cos \left(B\right)+\sin \left(A\right)\sin \left(B\right)$:
$P=\frac{W\sin \left(\theta \right)}{\cos (\alpha -\theta )}$

Step 5: Minimize the required force
For the force $P$ to be minimum, the denominator $\cos (\alpha -\theta )$ must be maximized.
The maximum value of a cosine function is 1, which occurs when its argument is zero:
$\cos (\alpha -\theta )=1\Rightarrow \alpha =\theta$
Therefore, the minimum force required to move the body is:
$P_{\text{min}}=W\sin \left(\theta \right)$