Step 1

We are given the mass of automobile $m = 2\;{\rm{Mg}}$, the coefficient of static friction as ${\mu _s} = 0.3$, and the angle as $\theta = 30^\circ $.

We are asked to determine the towing force required to move the car.

 
Step 2

We will draw a free body diagram of the car.

Here, ${N_A}$ is the reaction force at point $A$, ${N_B}$ is the reaction force at point $B$, and $f = {\mu _s}{N_B}$ is the friction force at point $B$.

The length $GD$ is $GD = 0.6\;{\rm{m}}$.

The length $AB$ is $AB = 2.5\;{\rm{m}}$.

The length $AD$ is $AD = 1\;{\rm{m}}$.

The length $CC'$ is $CC' = 0.3\;{\rm{m}}$.

The length $AA'$ is $CA' = 0.75\;{\rm{m}}$.

 
Step 3

We will resolve the forces in the horizontal direction.

\[\begin{array}{c} \sum {{F_x}} = 0\\ f - F\cos 30^\circ = 0\\ {\mu _s}{N_B} - F\cos 30^\circ = 0\\ {N_B} = \frac{{F\cos 30^\circ }}{{{\mu _s}}} \end{array}\]

Substitute the given value in the above equation.

\[{N_B} = \frac{{F\cos 30^\circ }}{{0.3}}\]
 
Step 4

We will take the moment about point $A$.

\[\begin{array}{c} \sum {{M_A}} = 0\\ F\cos 30^\circ \times CC' + {N_B} \times AB - mg \times AD - F\sin 30^\circ \times AD = 0\\ F\left( {\cos 30^\circ \times CC' - \sin 30^\circ \times AD + \left( {\frac{{\cos 30^\circ }}{{0.3}}} \right) \times AB} \right) = mg \times AD \end{array}\]

Substitute the given value in the above equation to find the towing force.

\[\begin{array}{c} F\left( \begin{array}{l} \cos 30^\circ \times \left( {0.3\;{\rm{m}}} \right) - \\ \sin 30^\circ \times \left( {{\rm{1}}\;{\rm{m}}} \right) + \\ \left( {\frac{{\cos 30^\circ }}{{0.3}}} \right) \times \left( {2.5\;{\rm{m}}} \right) \end{array} \right) = \left( {2\;{\rm{Mg}} \times \frac{{{{10}^3}\;{\rm{kg}}}}{{1\;{\rm{Mg}}}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}\;{\rm{m}}} \right)\\ F = \frac{{\left( {2000\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}\;{\rm{m}}} \right)}}{{\left( {\cos 30^\circ \times \left( {0.3\;{\rm{m}}} \right) - \sin 30^\circ \times \left( {{\rm{1}}\;{\rm{m}}} \right) + \left( {\frac{{\cos 30^\circ }}{{0.3}}} \right) \times \left( {2.5\;{\rm{m}}} \right)} \right)}}\\ F = \left( {\frac{{19620\;{\rm{kg}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}}}{{6.976\;{\rm{m}}}}} \right) \times \left( {\frac{{{{10}^{ - 3}}\;{\rm{kN}}}}{{1\;{\rm{N}}}}} \right)\\ F = 2.81\;{\rm{kN}} \end{array}\]