Is there a universal formula for finding the apparent weight of an object in a particular situation? I heard that it is the net of all forces acting on the object in the vertical direction? Is this true? If so, then please elaborate further. (NOTE: By apparent weight, I mean the reading of a weighing scale if a body were to stand on it while the whole system was in an accelerating frame.)
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Qmechanic
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Mr. Learner
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4You've answered your own question. What a weighing scale actually shows you is the normal force on the soles of your shoes. "Sum of all forces" makes no sense. If I am standing still in my kitchen the sum of all forces is zero. – garyp Dec 31 '21 at 16:11
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Is there a universal formula for finding the apparent weight of an object in a particular situation?
No, there is no universal formula to calculate the apparent weight. It all depends on the situation and how you set the definition for the apparent weight.
The problem is that, to the best of my knowledge, there is no formal definition for the apparent weight. See related discussion: Is there a formal definition for apparent weight?
I heard that it is the net of all forces acting on the object in the vertical direction? Is this true?
Not true. The net force on the object standing still on the ground is zero. Your definition implies its’s (apparent) weight is zero, which is not true.
The most important thing is to set the definition for the apparent weight. I will discuss here three different scenarios, each with different definition. Once we set the definition, the apparent weight is easily calculated from the free-body diagram.
If so, then please elaborate further.
Let's consider scenarios when object is (i) on the floor, (ii) in free fall, and (iii) in a fluid.
Object on the floor
The apparent weight is usually defined as the force that the floor exerts on the body. This is somewhat intuitive since this is what a scale would show if the object was standing on one.
Now let's consider the most general scenario - an object in an elevator that moves (accelerates) upwards or downwards. For a person standing on the ground (inertial reference frame), the net force acting on the object in the elevator is:
$$F_\text{net} = m \cdot a \tag 1$$
where $m$ is mass of the object. Note that the mass is property of a body which does not depend on the frame of reference.
The free-body diagram would show that there are only two forces exerted on the object: (i) force that elevator floor exerts on the object $F_\text{f/o}$ in upward direction, and (ii) gravitational force that Earth exerts on the object $F_\text{e/o}$ in downward direction:
$$F_\text{f/o} - F_\text{e/o} = F_\text{net}$$
where positive acceleration $a$ is taken to be in upward direction, and the gravitational force is
$$F_\text{e/o} = m \cdot g$$
The apparent weight equals $F_\text{f/o}$ force which is
$$F_\text{f/o} = F_\text{net} + F_\text{e/o} = m \cdot (g + a)$$
When the elevator accelerates:
- upwards at $a = g$ the apparent weight is $2mg$
- downwards at $a = g$ the apparent weight is zero (free fall)
- downwards at $a = 2g$ the apparent weight is $-mg$. The negative sign implies that $F_\text{f/o}$ is in the opposite direction to our convention. In other words, the object is on the elevator ceiling and its apparent weight is $mg$.
Object in free fall
The apparent weight is usually defined such that the apparent weight of an object in free fall is zero. Note that this is very loose definition when we consider the free fall with drag!
When the body just starts falling, the velocity is zero and the free body diagram would show only one force acting on the object - gravitational force that pulls the object down. The velocity starts increasing and the drag starts acting on the object which will eventually be equal in magnitude an opposite in direction to the gravitational force. When this happens, the object's acceleration drops to zero and the object remains falling at the terminal velocity. See related discussion: https://physics.stackexchange.com/a/681282/149541
What is the apparent weight of the object when it reaches the terminal velocity? Unfortunately, the above definition does not cover this case! You could say it is zero or redefine the apparent weight.
Object in a fluid
The apparent weight is usually defined as the gravitational force minus the buoyancy. When the body floats its apparent weight is said to be zero. Note that this is the definition found in most textbooks.
The closest you can get to the intuition is to imagine holding an object submerged in a fluid. The force you need to apply to prevent the object from sinking equals gravitational force minus buoyancy.
What is the apparent weight of a sinking body? This example is very similar to the freely falling object, i.e. the free-body diagram would have three forces: gravitational force, buoyancy, and drag. Note that buoyancy and drag are much larger in fluid than in air. The definition above does not cover the sinking body case, i.e. by definition it is still gravitational force minus buoyancy. Force required to stop the body from sinking further still equals gravitational force minus buoyancy.
To conclude, when asked to find the apparent weight, you should always ask for its definition and then apply free-body diagram.
Marko Gulin
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But what about the apparent weight of a body in an elevator accelerating downwards and the apparent weight of a submerged object in a fluid? The fbd for both cases yield the force of gravity acting downwards and an upward force(the force by the elevator on the body(Fs/p)and buoyancy in the second) but the apparent weight in the first case is Fnet-mg which means ultimately it is the upward force by the elevator on the body. But the apparent weight in the second case is the real weight(mg)-buoyancy. So the upward force is NOT the apparent weight rather it is getting subtracted. – Mr. Learner Jan 01 '22 at 02:18
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2@Mr.Learner I added explanation for an elevator accelerating downwards. As for the apparent weight in a fluid with buoyancy, first define what is apparent weight. After that, the analysis is as simple as for the elevator example. – Marko Gulin Jan 01 '22 at 09:39
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As far as I know, apparent weight means the weight a body experiences when there is a net acceleration. But didn't you already give a definition yourself :"The apparent weight is the force with which person acts on the floor which is by Newton's third law the same as the force with which floor acts on the person." Could you please explain the apparent weight of a body submerged in a fluid example, too(using your own definition) ? Because I was unable to relate it to the elevator example. – Mr. Learner Jan 01 '22 at 15:08
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You said that, the force with which person acts on the floor is the same as the force with which floor acts on the person. But in all the cases we see, that they are not. In the first case the Fe/p is mg while the Fs/p is m(g+a). In the second case, Fs/p is m(g-a) and finally in the last case, it’s Fe/o - Ff/o. – Mr. Learner Jan 02 '22 at 01:55
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Also, say the object was sinking in the fluid meaning that its density is greater than that of the fluid. Then, if I were to find the apparent weight of that object like you, Fs/o + Ff/o - Fe/o= Fnet. So then the apparent weight aka Fs/o = Fnet + Fe/o - Ff/o. But we know that, even when the object is sinking the apparent weight is still Fe/o - Ff/o. So, it'd be helpful if you explained this example as well. (I actually wanted to know this example in my last comment) – Mr. Learner Jan 02 '22 at 02:07
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@Mr.Learner You are confusing action-reaction pairs. The reaction pair for $F_\text{e/p}$ is $F_\text{p/e}$, i.e. the force that person exerts on the Earth! The same applies to $F_\text{s/p}$, i.e. the person exerts force $F_\text{p/s}$ on the floor. These reaction forces are not shown in the free-body diagram of the person, i.e. only the forces exerted on the person should be used. As for the buoyancy, if the body sinks then the apparent weight is true weight minus the buoyancy, and if the body floats then the apparent weight is zero. – Marko Gulin Jan 02 '22 at 13:12
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Yes, I know that, the apparent weight of a body sinking in water is true weight minus buoyancy. But according to your formula, the apparent weight should be the net acceleration in the downwards direction times the mass of the object plus true weight minus buoyancy. Or, ma+Fe/o-Ff/o, where a is the net acceleration in the downwards direction. What you mentioned in your answer was the apparent weight of a body that floats in a fluid aka there is no vertical movement. But what i want to know is the apparent weight when there IS vertical movement(in the downwards direction(bcz it's sinking)) – Mr. Learner Jan 02 '22 at 16:05
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Not true, because net force is not part of the free-body diagram. The $m \cdot (g+a)$ force in the elevator refers to the normal force, not to the net force. But we calculated the normal force from the net force $ma$ in the elevator example. When body hangs on a spring, it will not sink! We used the spring just to define the apparent weight. – Marko Gulin Jan 02 '22 at 16:08
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@Mr.Learner I am waiting in the chat room you posted in your last comment. – Marko Gulin Jan 02 '22 at 22:09
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@ Marko Gulin, what if for the sinking object example, we attach an weighing machine at the bottom of that object instead of hanging it from a spring? – Mr. Learner Jan 04 '22 at 07:16
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Chat room tomorrow 5.1. at 9:15 AM CET, I am available only up until 10:00 AM.. – Marko Gulin Jan 04 '22 at 08:14
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@Mr.Learner I edited the answer, it might include information you need. Please check now and let me know if you have any questions. Also, if you find some of the answers useful for your original question, please consider accepting it. It does not have to be my answer, just any answer you find most useful. – Marko Gulin Jan 06 '22 at 18:27
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Amd I wanted to know what's the apparent weight when there IS vertical movement(in the downwards direction) of an object submerged in a fluid. Why is it still calculated Fe/o - Ff/o? You said that, you can't actually find the apparent weight in this case, bcz you can't attach a spring to a sinking object. Then I said, well, what if you glued a weighing machine to his feet. So, please answer these questions. I WILL accept the answer, but I have to be satisfied with the answer first, r8? – Mr. Learner Jan 07 '22 at 01:35
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@Mr.Learner I edited the section with fluids. Sinking body is the same as freely falling body, the only difference is that buoyancy and drag are much larger in fluid. You cannot think in terms of scale, rather think in terms that observer has some supernatural power that allows them to see all the forces acting on the object. This would not help much because apparent weight is not strictly defined. See discussion about freely falling object and terminal velocity. If you do not find this informative enough, I suggest you open new question, maybe someone else will provide better explanation. – Marko Gulin Jan 07 '22 at 08:00
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The apparent weight is the force of you on the "floor" which is equal and opposite to the force of the floor on you, sometimes called the normal force of constraint. The weight shown on a scale is equal in magnitude, but opposite in direction, to the normal force.
The total forces on you are the normal force, the force of gravity, and in an accelerating reference frame, fictitious forces. The total force on you is $\vec N + m \vec g + \vec F_{fict}$ where $\vec N$ is the normal force, $m \vec g$ is the force of gravity, and $\vec F_{fict}$ is the total fictitious force. Since these forces are vectors, the direction of the forces must be considered.
For example, for you standing on a scale on earth the normal force equals the force of gravity and your weight on the scale is $mg$ (ignoring the small effect of the fictitious forces due to the earth's rotation). If you are standing on a scale in an elevator in vertical free fall, the force of gravity and the fictitious force sum to zero, so the normal force is zero and you are weightless and show zero weight on the scale.
John Darby
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Your definition of apparent weight sounds good to me. Just keep in mind that the force from the scale acting on the body (which is indicated by the reading on the scale) contributes to (or opposes) any other forces which are causing the body to accelerate.
R.W. Bird
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