A misconception in an application of Newton's laws

When a body is kept in ground, and is at rest, the downward gravitational pull is balanced exactly by the Normal Reaction if we assume the earth to be an inertial frame. But this would mean that any external force provided to that body would lead to its motion. if this was the case, moving things both heavy and light would be same. what is the external force that abstains it from happening?

• How about the friction? Jul 7 '15 at 13:22
• Suppose that the ground surface is frictionless for the moment and I'm pulling the body upwards Jul 7 '15 at 13:26
• Then you'll have to win the battle with gravity Jul 7 '15 at 13:29
• when the body is at rest, normal Reaction does it for me Jul 7 '15 at 13:30
• No the moment you pull the weight there is no normal force anymore. Jul 7 '15 at 13:30

I think the apparent paradox is driven by a misunderstanding of the very nature of reaction forces. A reaction force is a force $F_N$ perpendicular to a surface which counterbalance the gravitational pull, or any other force perpendicular to the surface. This is because reaction force are basically what prevent two bodies to occupy the same volume, and their microscopically origin is the electrostatic (Coulomb) repulsion. They are always normal to the surface and with outward direction.

Imagine now that the body on the ground has mass $m$ and one applies an external force $F_E$ perpendicular to the ground, pulling upwards. We have $$F_E-m g+F_N=m a,$$ where we choose our $x$ axis perpendicular to the ground and directed upwards. The force $F_N$ is perpendicular to the ground (outward direction), i.e., $F_M\ge0$ for our axis choice.

Now there are two cases:

1) $F_E<mg$ in this case the normal reaction is $F_N=mg - F_E>0$, and the body does not move ($a=0$). Intuitively, the force $F_E$ is not enough to lift the body.

2) $F_E>mg$ in this case one would have $F_N=mg - F_E<0$, i.e., the normal force would push the body to the ground. But this contradicts the assumption since $F_N<0$. Therefore in this case $F_N=0$ and $F_E-m g=m a\neq 0$, i.e., the body has a finite acceleration.

Therefore, the conclusion is very intuitive: in order to lift a body of mass $m$ from the ground, one must provide an external force $F_E\ge m g$.

• This describes the response to lateral forces, however in a comment to his question the OP makes it clear he is thinking about normal forces. Jul 7 '15 at 15:43
• I think I misunderstood the question... Jul 7 '15 at 15:47
• Yes, the question was not initially clear. Jul 7 '15 at 15:48
• I update the answer. I hope that it addresses the question now. Jul 7 '15 at 16:29

As Tom mentioned, the normal force has changed.

Draw a free body diagram. You should have three forces displayed, assuming no horizontal influences, such as friction.

Sintetico discusses horizontal motion below. Considering vertical motion only, as you said, the body is still at rest (and not accelerating), thus $\vec{a}=0$. Newton's second law then says $\sum\vec{F}_i=0$. So, "your" upward force and the normal force balance the gravitational force.

When a body is kept in ground, and is at rest, the downward gravitational pull is balanced exactly by the Normal Reaction if we assume the earth to be an inertial frame.

This is not correct. The normal force is a constraint force that acts in one direction (upward). Suppose you exert an upward force of five newtons on an object that weighs eight newtons and another object that weighs four newtons. In the case of the eight newton object, the upward normal force exerted by the ground will decrease from eight newtons to three newtons as you apply the force. In the case of the four newton object, the object accelerates upwards. The normal force vanishes the moment the object leaves the surface. The normal force is not glue; it does not switch from an upward force that keeps objects from sinking into the Earth to a downward force. It acts in one direction, normal to and away from the surface.