We have all heard people saying that to lift an object of mass $m$, you have to apply a force $F$ equal to its weight $mg$. But isn't it getting the force equal to its weight from the surface to which it is attached to (normal force). Why it is willing to change that equilibrium state by getting the same force from us as from the surface? (Consider the situation devoid of any resistance) . I think we must be applying slightly more force to it in order to move it even with constant velocity at least at the start and balancing the force of gravity afterwards.
Forces – To Lift an Object, Does the Required Force Equal Its Weight or Greater? Insights from Newtonian Mechanics
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You are correct.
$W=mgh$ is also correct, but it brushes something under the rug. It ignores the force required to accelerate an object from rest, and it ignores the opposite force that slows the object to a stop. In between speeding up and slowing down the velocity is constant which, as you point out, implies the net force is zero. The lifting force magnitude is the same as that of gravity, and it is clear that $W=mgh$ during that interval. So what about starting and stopping? The extra vertical work needed to accelerate the object is balanced by the reduced vertical work needed to bring it to a stop. So in the end, $W=mgh$.
A normal force is simply a force that arises from contact between one object and another object or substance. There can be several normal forces acting on an object. For example, an object on a slope held in place by a wedge will have two normal forces acting on it - one from the slope, the other from the wedge. Neither one is equal to the weight of the object (or its apparent weight if the slope is in an accelerating lift).
Apparent weight is a very specific instance of a normal force - it is the normal force registered by horizontal scales on which the object rests. If the scales and the object are in an accelerating lift then the apparent weight will be greater than or less than the “true” (unaccelerated) weight of the object, depending on the acceleration of the lift.
Buoyancy is another type of normal force which acts on objects partly or fully submerged in a liquid. But it is not the same as apparent weight. If you take a submerged object that is denser than the liquid and rest it on unaccelerated horizontal scales, the normal force registered by the scales will be equal to the object’s weight in air minus the buoyancy force. This is the apparent weight of the (unaccelerated) object when submerged in the liquid.
If the scales (still in the liquid) are accelerated up or down then the apparent weight of the object (as registered by the scales) will increase or decrease. But the buoyancy force stays the same. So this also confirms that buoyancy and apparent weight are not the same - they are different instances of normal forces.
Best Answer
There are two points to be clarified here.