Question

Is it easier to push or pull a body over a rough surface? Explain your answer.

Answer 1

1. The normal force acts on a level plane due to the weight of the body. We also know that a vector quantity has components, and force is a vector quantity.
2. When we pull an object on the horizontal plane, then the vertical component of the force will act vertically downward and will increase the weight of the object leading to an increase in friction. If the friction will be then it will be difficult to move the object as friction acts in the opposite direction of the force.
3. As a result, when the friction force is lower, it is easier to move the body.
4. On the other hand when we pull an object, however, the vertical component works in the upward direction, lessening the friction force. As a result, when the friction force is lower, it is easier to move the body.

Mathematically explanation of the above demonstration

$F_{net}=ma..........\left(i\right)[where{F}_{net}==netforceontheobject,m=massoftheobject,a=acceleationoftheobject]$

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

$F_f=\mu N.........\left(ii\right)[where\mu =coefficientofsttaicofkineticfriction,N=normalforce]$

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 $F_1=\mu N_1$

Substituting the value of $N_1$, $F_1=\mu (F\cos \theta +mg)..............\left(iii\right)$

Let us now apply Newton’s second law of motion to the object being pulled in the vertical direction.

$$\begin{gathered} N_2+F\cos \theta =mg \\ {N}_2=mg-F\cos \theta \end{gathered}$$

Hence, the frictional force acting on this object according to equation (2) is $F_2=\mu N_2=\mu (mg-F\cos \theta ).........\left(iv\right)$

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.

Hence, it is easier to pull than to push a body.