Explain why pulling is easier than pushing with a neat sketch on a rough horizontal plane.

Hint:Check which forces act on the object when it is pulled or pushed in the same direction. Draw the free body diagram of the object being pulled and being pushed. Use the formula for the frictional force acting on the object. From the free body diagram, determine the value of the normal forces acting on both of these objects and substitute these values in the formula for frictional forces. Hence, determine whether pulling is easy or pushing is easy.

Formula used:
The expression for Newton’s second law of motion is
\[{F_{net}} = ma\] …… (1)
Here, \[{F_{net}}\] is the net force acting on the object, \[m\] is mass of the object and \[a\] is acceleration of the object.

Complete step by step answer:
When an object placed on a horizontal surface is pulled or pushed, there is a force of friction between the object and the surface and this frictional force opposes the motion of the particle. We know that the frictional force \[{F_f}\] is given by
\[{F_f} = \mu N\] …… (2)
Here, \[\mu \] is the coefficient of static or kinetic friction and \[N\] is the normal force exerted by the surface on the object.

Let us draw a free body diagram of an object being pushed and being pulled as follows:

In the above free body diagram, \[{F_1}\] and \[{F_2}\] are the frictional forces acting on the objects of mass while pushing and pulling the objects respectively, \[mg\] is the weight of both the object, \[{N_1}\] and \[{N_2}\] are the normal forces acting on the objects while pushing and pulling respectively. The components of the pushing and pulling forces \[F\] are shown in the above diagram.

Let us first apply Newton’s second law of motion to the object being pushed in the vertical direction.
\[{N_1} = F\cos \theta + mg\]
Hence, the frictional force acting on this object according to equation (2) is
\[{F_1} = \mu {N_1}\]
Substitute \[F\cos \theta + mg\] for \[{N_1}\] in the above equation.
\[{F_1} = \mu \left( {F\cos \theta + mg} \right)\] …… (3)

Let us now apply Newton’s second law of motion to the object being pulled in the vertical direction.
\[{N_2} + F\cos \theta = mg\]
\[ \Rightarrow {N_2} = mg - F\cos \theta \]
Hence, the frictional force acting on this object according to equation (2) is
\[{F_2} = \mu {N_2}\]
Substitute \[mg - F\cos \theta \] for \[{N_2}\] in the above equation.
\[\therefore{F_2} = \mu \left( {mg - F\cos \theta } \right)\] …… (4)
From equations (3) and (4), we can conclude that the frictional force on the object being pushed is more than the frictional force on the object being pushed.

This makes pulling more easier than pushing.

Note:The students should keep in mind that the above criteria which makes pulling an object more easier than pushing an object is only applicable when the horizontal surface over which the object is being pulled or pushed is not smooth. Because if there is no friction between the object and the surface then equal force is required to pull or push the object.