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Suppose we have a very heavy object resting on a frictionless floor. If we pull upwards without moving it, what happens to the force we are applying?

I am a bit confused regarding normal forces in this case too. If someone could help me out by clearing this doubt, specially including normal terms, I would really be grateful.

I understand the sum would be zero, but in which direction does the normal force act. If you consider pushing the object downward then normal force by the floor would act on the object upward. (which would be equal to the weight of the object and the force we applied), due to action reaction force. I want to know,if we are pulling the object up, in which direction does the normal force act? Downwards probably, but how can that be.?? A free body diagram would help.

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    $\begingroup$ The object will elongate {deform} a bit and that is caused by the applied force. $\endgroup$ – Farcher Mar 18 '18 at 12:28
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Since both the person acting and the box are on earth's surface, both the box and the person cause a normal force at the connection with the ground. The location and distribution of this normal force may vary due to different actions by the person on the box. The figure below illustrates that:

  1. When the box is at rest on the floor, the normal force is $N_B = m_{\small B}g$ onto the floor under the box, equal to the weight $W_B$ of the box against the floor, and total normal force on the feet of the person is $N_P = W_P = m_{\small P}g$. In these expressions, $g$ is gravitational acceleration, subscripts ${\small B}$ and ${\small P}$ refer to the box and person respectively.
  2. When the person exerts a downward push on the box, he/she is transferring some of his/her weight onto the box. The normal force under the box is equal to the effective box weight: $N_B = W_B = m_{\small B}g + F$ and the total normal force under the person's feet is equal to his/her effective weight $N_P = W_P = m_{\small P}g - F$.
  3. When the person exerts a lifting force on the box, they transfer some of the weight of the box onto their feet, and under the box, the normal force is $N_B = W_B = m_{\small B}g - F$. Then, the total normal force under the person's feet is $N_P = W_P = m_{\small P}g + F$.
  4. When the person pushes the box against the ceiling (which is connected to the ground as usual), there is no contact between the box and the floor. The box pushes on the ceiling with force F causing a normal force downwards $N = F$. The total normal force under the person's feet is $N_P = W_P = m_{\small P}g + m_{\small B}g + F$:
    Normal force on box and person
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    $\begingroup$ Wow, now that's what i had needed, thank you very much. It helped. $\endgroup$ – 123IR Mar 21 '18 at 3:25
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The normal force is a contact force and comes into play only when the object and floor are in contact with each other.it is a self-adjusting force and is there to explain Newton's third law of motion.As we force the object in upward direction and the object doesn't move ,the normal force self-adjusts itself such that the sum of the applied force , normal force and gravitational force becomes zero. If the object moves in upward direction then the object is out of contact with floor hence there is no such thing called normal force.

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  • $\begingroup$ I understand the sum would be zero, but in which direction does the normal force act. If you consider pushing the object downward then normal force by the floor would act on the object upward. (which would be equal to the weight of the object and the force we applied), due to action reaction force. I want to know,if we are pulling the object up, in which direction does the normal force act? Downwards probably, but how can that be.?? $\endgroup$ – 123IR Mar 19 '18 at 2:49
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Force, unlike energy or mass, is not conserved. There is no cause to suppose that a force 'goes' anywhere.

A force applied to a motionless chair can change its shape, and alter the force applied by/to the floor beneath. As long as the chair doesn't accelerate, the sum of forces on it is zero, so the chair is responding to any applied force by small elastic deformation, perhaps applying stress in different direction and magnitude at its points of floor contact, to null the sum of forces.

Similarly, a butcher ought not lean his thumb on the scale pan when he weighs out your roast... in that case, the force goes from your food budget.

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  • $\begingroup$ Thanks, it helped. Could you also give some explanation on normal/net forces acting in the system? A free body diagram would help. $\endgroup$ – 123IR Mar 19 '18 at 5:19
  • $\begingroup$ @123IR The 'normal' force at the floor is a component of all the adhesive or compressed-spring interaction that acts on (for instance) the feet of the hypothetical object-on-a-floor; in addition to normal force there's lateral forces from friction (and adhesive forces) that act parallel to (and not normal :== perpendicular) to the floor surface. Net force just means the sum of all forces. In a wind, or on board a rolling ship, or when a magnet passes near a steel chair, the 'diagram' is just a useful way of symbolizing... all of it. $\endgroup$ – Whit3rd Mar 19 '18 at 5:23
  • $\begingroup$ I think, in this case normal force on the floor is just due to the weight and the force with which we were pulling did not affect that force. Since normal force can be also called contact force, so there would be a normal force on us by the body only at the contact parts. Correct me if i am wring, i got that the net forces would be zero. Thanks for your help. $\endgroup$ – 123IR Mar 19 '18 at 12:50
  • $\begingroup$ @123IR : Contact force means applied at a contact point, but otherwise is unspecific (friction, adhesion, pressure, are all included). If the force diagram is to be useful, it ought not distinguish 'contact force'' and the directional part 'normal force' . Those 'force' parts are not distinct one from the other. It's like describing a population as partly adults, and partly female; those parts overlap yet don't include everyone. $\endgroup$ – Whit3rd Mar 20 '18 at 2:02

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