Question

A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is μ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?

Answer 1

Let P be the force applied to slide the block at an angle θ.



From the free body diagram,
R + P sin θ − mg = 0
⇒ R = −P sin θ + mg (1)
μR = P cos θ (2)

From Equation (1),
μ(mg − P sin θ)−P cos θ = 0
⇒ μmg = μP sin θ + P cos θ
$\Rightarrow P=\frac{\mathrm{\mu }mg}{\mathrm{\mu }\sin \mathrm{\theta }+\cos \mathrm{\theta }}$
The applied force P should be minimum, when μ sin θ + cos θ is maximum.
Again, μ sin θ + cos θ is maximum when its derivative is zero:
$\frac{d}{d\mathrm{\theta }}\left(\mathrm{\mu }\sin \mathrm{\theta }+\cos \mathrm{\theta }\right)=0$
⇒ μ cos θ − sin θ = 0
θ = tan⁻¹ μ
So, $P=\frac{\mathrm{\mu }mg}{\mathrm{\mu }\sin \mathrm{\theta }+\cos \mathrm{\theta }}$
Dividing numerator and denominator by cos θ, we get
$=\frac{\mathrm{\mu }mg/\cos \mathrm{\theta }}{{\displaystyle \frac{\mathrm{\mu }\sin \mathrm{\theta }}{\cos \mathrm{\theta }}}+{\displaystyle \frac{\cos \mathrm{\theta }}{\cos \mathrm{\theta }}}}$

$$\begin{gathered} P=\frac{\mathrm{\mu }mg\sec \mathrm{\theta }}{1+\mathrm{\mu }\tan \mathrm{\theta }} \\ =\frac{\mathrm{\mu }mg\sec \mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}=\frac{\mathrm{\mu }mg}{\sec \mathrm{\theta }} \\ =\frac{\mathrm{\mu }mg}{\sqrt{\left(1+{\tan }^2\mathrm{\theta }\right)}}=\frac{\mathrm{\mu }mg}{\left(1+{\mathrm{\mu }}^2\right)} \end{gathered}$$(using the property $1+{\tan }^2\theta ={\sec }^2\theta$)
Therefore, the minimum force required is $\frac{\mathrm{\mu }mg}{\sqrt{\left(1+{\mathrm{\mu }}^2\right)}}$ at an angle θ = tan⁻¹ μ.