Given data

• Mass of body: $m = 1\,\text{kg}$
• Coefficient of static friction: $\mu_s = \dfrac{1}{\sqrt{3}}$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

We need the minimum force required to just make the body move.

Idea

If a force $F$ is applied at an angle $\theta$ above the horizontal, then:

• Horizontal component: $F\cos\theta$
• Vertical upward component: $F\sin\theta$

The upward component reduces the normal reaction, hence reduces friction. That is why the least force is obtained by applying the force at some angle above the horizontal.

Normal reaction

Since the body is still on the floor, $N = mg - F\sin\theta$

Maximum static friction is $f_{\max} = \mu_s N = \mu_s (mg - F\sin\theta)$

Condition for impending motion

For the body to just start moving, $F\cos\theta = \mu_s (mg - F\sin\theta)$

Rearranging, $F\cos\theta + \mu_s F\sin\theta = \mu_s mg$ $F(\cos\theta + \mu_s \sin\theta) = \mu_s mg$

Hence, $F = \frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta}$

Minimizing $F$

To minimize $F$, we must maximize $\cos\theta + \mu_s \sin\theta$

The maximum value of $a\cos\theta + b\sin\theta$ is $\sqrt{a^2+b^2}$.

Here, $a=1$ and $b=\mu_s$, so maximum is $\sqrt{1+\mu_s^2}$

Therefore, $F_{\min} = \frac{\mu_s mg}{\sqrt{1+\mu_s^2}}$

Substitute values

Since $\mu_s = \frac{1}{\sqrt{3}}, \quad m=1, \quad g=10$

$F_{\min} = \frac{\left(\frac{1}{\sqrt{3}}\right)(10)}{\sqrt{1+\frac{1}{3}}}$

$= \frac{10/\sqrt{3}}{\sqrt{4/3}}$

$= \frac{10/\sqrt{3}}{2/\sqrt{3}} = 5\,\text{N}$

Rounded to nearest integer

$5$

Comparison with stored answer

Stored correct answer = $5$

Our derived answer matches it.