The force required to lift an object with weight mg must be slightly greater than mg to produce a net force upwards. There's one thing I'm struggling to understand with this concept though and would appreciate if someone can clarify it for me. If an object is resting on a table then it will have a reaction force (the normal force) also acting upwards on the object. Since the net force on this object is zero as the normal/weight forces cancel out, then shouldn't any force applied upwards to the object cause it to start moving, regardless of whether the force's magnitude is less than mg or not?
[Physics] Force required to lift an object
forces
Related Solutions
No.
If you take all your masses and and squeeze them into a tiny volume, although the density increases, the normal force is the same.
Loosely speaking consider the following:
- Density + Volume = Mass, $m = \rho V$
- Mass + Gravity = Weight, $W= m g$
- Weight + No Motion = Equal and Opposite Normal Force, $N=W$
- Normal Force + Area of Contact = Contact Pressure, $N=\iint P(\vec{r})\,{\rm d}A$
- Contact Pressure + Elasticity = Contact Deflection, $\delta = \frac{1-\nu^2}{E} \iint \frac{P(\vec{r})}{|\vec{r}|}\,{\rm d}A$
There are two points to be clarified here.
- The normal reaction force from the surface is a self-adjusting force. In particular, it can take any value so as to prevent the object in contact from penetrating. So, if an object resting on a surface has a weight $w$ then the normal reaction force would be $w$ in the upward direction. Now, if you apply an external upward force on the object (with your hand, say) of a magnitude $w/2$ then the normal reaction force from the surface would change its value to $w/2$. Now, if you apply an external force of a magnitude $w$ in the upward direction then the normal reaction force from the surface would reduce to zero.
- However, as you correctly notice, when the upward external force is exactly the same as the weight in magnitude, the object is still in perfect equilibrium. And since the initial velocity of it was zero, its velocity would still remain zero because equilibrium means no acceleration. So, there would be no movement. So, in order to actually lift the object, you do need to provide an upward force which is at least slightly greater than the weight of the object. Once you apply such a force even for a tiny amount of time, the object would pick up an upward velocity because it would have been subjected to an upward acceleration for that tiny amount of time. Once this is accomplished, you can reduce the magnitude of the upward force to be exactly the same as the magnitude of weight and the object will continue to move in the upward direction, in equilibrium, but now, with a constant velocity (that it picked up during that tiny amount of time of acceleration).
Best Answer
The normal force only acts on that object while that object is in contact with the surface it is resting on. It is also proportional to the force being applied onto the surface - which is not necessarily $mg$ but would be if there were no other forces involved. Say you went to lift the object off that surface, and you applied a force $F<mg$ upwards. What you would get is a gravitational force going down of $mg$ and a normal force of $N=mg-F$ going up and your force of $F$ going up. In the end, the situation is still balanced and the object doesn't move. The normal force does not always provide a $mg$ force going up. It's only enough to keep the object from moving down through the surface.