Answer 1:

It is easier to move any body by pulling it than by pushing it. A body starts to move when the applied force is more than the frictional force. (Which acts in a direction opposite to that of the motion of the body). This frictional force is proportional to the weight of the body under which it acts. (W) When we move a body shorter than us, (a lawn mover in this case) the force applied is not horizontal. When a force (F) is applied to a body at an angle `Ø', it is equivalent to applying a horizontal force of F CosØ and a vertical force of F SinØ. Where `Ø' is the angle of the line of action of the force to the horizontal. This is termed as resolving the forces). F CosØ is the horizontal component & F SinØ is called the vertical component.

When we push a body, the vertical component acts downward and adds up to the weight of the body. So, pushing a body by applying a force `F' at an angle `Ø' is equivalent to moving a body a body of weight W + (F SinØ), with a horizontal force F CosØ. Where `W' is the weight of the body. When we pull a body, the vertical component acts upward. So pulling a body of weight `W' by applying the same forces `F' at an angle `Ø', is equivalent to moving a body of weight W — (F SinØ) with a horizontal force F CosØ.

So it is possible to pull a body by applying a force less than that is required to push it.

When we push a body, we also press it down. And when we pull a body, we also apply a lifting force to that body, which makes it feel lighter. The force required to pull and push a body will be equal if it is applied horizontally (Ø=0).

Answer 2:

It is simply because there would be more friction against motion of a lawn roller in case of pushing than in case of pulling. When a lawn roller is pulled by a certain force (F), the vertical component (FSinØ) of that pulling force decreases the weight (mg) of the roller which inturn results in decrease of normal reaction (R=mg-F SinØ).

Since frictional force (F{-f}=ìR) pulling is directly proportional to the normal reaction, friction is less here.

When the lawn roller is pushed by the same force (F), the vertical component of the pushing force now increases the weight (w=mg) of the roller which inturn results in increase of normal reaction (R{+1}=mg+F SinØ). Therefore here the friction (F{-f} =ìR{+1}) is more.

*Published in The Hindu on Jan 03, 2002.*