A block of mass m=2kg is kept in position on the wall by applying an oblique force 'F' at 60° to the vertical as shown . calculate the minimum value of 'F'(the coafficient of friction between the wall and the block = 0.9) (Take 'g'=10ms^-2)?

Jul 7, 2018

Let the force $\vec{F}$ be applied on the block obliquely at an angle of $\theta = {60}^{\circ}$ with the vertical as shown in the figure above.

The resolved part of the applied force $\vec{F}$ perpendicular to the wall will be $\vec{F} \sin \theta$ and the resolved part parallel to the wall in upward direction is $\vec{F} \cos \theta$

So static fictional force in upward direction is

${F}_{s} \uparrow = \text{coefficient of friction} \left({\mu}_{s}\right) \times F \sin \theta$

$= 0.9 \vec{F} \sin \theta$

So considering the equilibrium of forces we have

$\vec{F} \cos \theta + 0.9 \cdot \vec{F} \sin \theta = m g$

$\implies \vec{F} \cos 60 + 0.9 \cdot \vec{F} \sin 60 = 2 \cdot 10$

$\implies \vec{F} = \frac{20}{\frac{1}{2} + \frac{0.9 \sqrt{3}}{2}} \approx 15.6 N$

If the applied force $\vec{F}$ acts obliquely downward making an angle $\theta = {60}^{\circ}$ with the vertical then the component of $\vec{F}$ in vertical direction will be downward and equilibrium of force will give the following relation.

$0.9 \cdot \vec{F} \sin \theta = m g + \vec{F} \cos \theta$

$\implies - \vec{F} \cos 60 + 0.9 \cdot \vec{F} \sin 60 = 2 \cdot 10$

$\implies \vec{F} = \frac{20}{- \frac{1}{2} + \frac{0.9 \sqrt{3}}{2}} \approx 71.6 N$