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I had a problem in my book stating:

A boy lifts a ball with mass $m$ with force of constant magnitude $F$ to a height $h$.
Calculate the force magnitude $F$ acting on the ball .

I know this is a very silly question. But I am screwing up in understanding the answer in my book. Please help me! Here,the answer given in my book is $mg$. But I think the force must be greater than that.

To explain my answer, I have taken the scenario as follows: If the boy is holding the ball in his hand, then the force of gravity and the force by his hand counter balance each other as a result of which ball stays stationary. Now, if the boy has to move the ball to a greater height, he must apply an additional force along with the force that keeps the ball stationary. Thus, my answer.

So, can the force being equal to the weight raise the ball up to a height $h$, or does the force need to be larger than this?

BioPhysicist
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Shogo Makishima
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    I have edited the post to read less like a "check my work" post, which is off topic, and to focus more on your conceptual question. Please edit anything if I have misinterpreted the question. – BioPhysicist Oct 31 '21 at 11:29
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    Since the question does not state the initial conditions the force may very well be negative:If the ball is already flying upwards at high velocity it must be braked or it would overshoot (if we impose the end condition "at rest", whatever that means on Earth). – Peter - Reinstate Monica Nov 01 '21 at 01:25
  • Note also that if, when you start observing, at $t_0$, the ball is already moving with constant velocity, and remains at that velocity all the way up, and has the same velocity at $t_1$ when it reaches $h$, then the force $\vec F$ will be $-m\vec g$ - you don't need a net force to keep the ball moving, only to change its speed/direction. Which is probably what this poorly worded problem statement was getting at. – Filip Milovanović Nov 01 '21 at 21:55
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    Likely a duplicate of this: Do I have to apply more force than gravity to lift my leg above the ground? But see also on how the force arises when holding something: https://physics.stackexchange.com/a/668342/247642 – Roger V. Nov 02 '21 at 15:31

5 Answers5

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This is a horribly written question (the exercise, not your post).

You are correct in your reasoning. If the ball starts at rest, the force $F$ needs to be larger than $mg$ in order for it to begin moving upwards.

However, if the ball already starts with some initial upwards velocity, a force $F=mg$ would be enough to get the ball to keep moving upwards, and any force larger than $mg$ still gets the job done. With a sufficiently large initial velocity, the force $F$ could actually be smaller than $mg$, since all we require is the ball move to a height $h$ rather than move to a height $h$ and then still keep moving upwards.

So, to summarize, $F>mg$ always works, and $0<F\leq mg$ works for a sufficiently large initial upward velocity$^*$.

The issue with the problem is that it doesn't specify the initial conditions of the ball, and even if they had been specified there is not a unique answer; there is just a minimal force needed to raise the ball a height $h$. So yeah, ignore the fact that you didn't get the answer and instead focus on the concepts that you have correct. $F=mg$ means no acceleration, and if the ball starts at rest this means, in this case, it could not reach height $h$ with only $F$ and the weight acting on it.

$^*$ F could technically be negative with a sufficiently large initial upward velocity, but since we are told the ball is being lifted we will assume the force has to be upwards.

BioPhysicist
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  • However, as you note, the extra force you need to put in to get the ball moving is exactly the same as the lower force necessary at the end. Every bit of force you put in to get the ball moving returns back to you if you slow down to a stop at the end. – Luaan Oct 31 '21 at 21:45
  • @Luaan Impulse, not force. You might be speeding it up and slowing it down with different forces over different timeframes. – Bergi Oct 31 '21 at 22:36
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    @Luaan I'm not assuming it needs to end at rest. $F$ is constant, so it cannot come to rest for a finite time once it is moving. – BioPhysicist Nov 01 '21 at 00:07
  • @BioPhysicist Good point, I missed that the force is supposed to be constant! In that case, the only possible force that could be the correct answer is exactly mg, and there's no ambiguity in the question. It's still an annoying way to put the question, especially if it comes from a text book, though :D – Luaan Nov 01 '21 at 09:10
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    @Luaan As my answer points out, no, the only possible correct answer is not $mg$. By saying that is "the answer" you are assuming the ball starts with some initial vertical velocity. But if you do assume this, then $mg$ is not a unique answer. Any force larger than $mg$ would work, and for a sufficiently large initial velocity, forces less than $mg$ could also work. – BioPhysicist Nov 01 '21 at 11:21
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Yes, a very poorly worded question. First of all, we must make some assumptions. So let’s assume

  • the force $F$ is applied directly to the ball
  • the ball is lifted vertically through a height $h$
  • the ball is stationary at the start and the end of its motion
  • there are no friction or drag forces

Note that none of this is stated explicitly in the question.

With these assumptions then, as you say, to get the ball moving in the first place the initial force applied to the ball must be greater than $mg$. And if we want the ball to be stationary again at height $h$ then at some point in the motion we need to reduce the force to less than $mg$ so that the ball slows down again. In this case all we can say, even with these assumptions, is that the average force applied to ball as it is raised is $mg$.

And, of course, if the ball does not need to be stationary at height $h$ then we can raise it with any direct force we like that is greater than $mg$.

Further more, if the force is not applied directly to the ball or the ball is not lifted vertically, then the boy may raise the ball with a force less than $mg$ if he applies it over a greater distance, using, for example, a lever or an inclined plane.

So the main educational benefit of this question seems to lie in thinking of all the ways in which the given answer can be wrong !

gandalf61
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I think the key here is that the question states that the force is 'of constant magnitude', and yet it moves the ball.

Whenever you accelerate an object away from you you increase the force with which you press on it, and whenever you let it decelerate you decrease the force. As the force is stated to be of constant magnitude we must assume that the acceleration is so small that we can ignore it when calculating the force. If we did not ignore acceleration the force would be higher at the start as the ball is accelerated upwards and lower at the end as it is decelerated.

Or alternatively maybe the acceleration is so brief that we can ignore it.

Once we've ignored acceleration, the force the boy applies to the ball must exactly match the force the earth applies to it as gravity in the opposite direction, so the magnitude is mg.

bdsl
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The questioner stated that "But I think the force must be greater than that." which is of course the correct answer.

The basic equation which defines no movement of the ball would be F=mg. Without complicating the question any vertical upward component of applied force greater than mg will move the ball upwards. If this vertical component of applied force is a constant the ball will keep accelerating upwards as long as F>mg and so this simple equation F>mg will satisfy all of the terms included in the question.

Denis
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  • Rather than "no movement", it would be more correct to say "no change in acceleration" (Newton's 1st law, inertia). With this change, you may find that the rest of your answer changes, in order to address show the question the OP was answering gives insufficient information about the initial inertial state of the ball. – Dewi Morgan Nov 01 '21 at 08:52
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    @DewiMorgan you mean no change in velocity – user253751 Nov 01 '21 at 12:00
  • @user253751 I'm fairly sure I mean acceleration, because f=ma. Any given force will establish the acceleration of a mass, not its velocity. Of course, when acceleration changes, velocity changes as well, but you would also need values for "time" and "initial velocity" to get a velocity from mass and force. – Dewi Morgan Nov 02 '21 at 03:07
  • @DewiMorgan yes F=ma so F=0 means a=0 not Δa=0 – user253751 Nov 02 '21 at 09:25
  • @user253751 a in the expression is acceleration, not velocity. – Dewi Morgan Nov 03 '21 at 04:23
  • @DewiMorgan Yes. F=0 means a=0 not Δa=0 – user253751 Nov 03 '21 at 09:19
  • @user253751 Ah, I see, you're being hypercorrect about my phrasing "it would be more correct to say "no change in acceleration"". Very droll. Ho, ho. – Dewi Morgan Nov 05 '21 at 21:49
  • @DewiMorgan That's not hypercorrect - it's simply correct... – user253751 Nov 06 '21 at 12:26
  • @user253751 Clarity fail on my part, I agree. I did not make it clear that I was talking about setting a value for f. Setting f is a change of f, which causes a change to a. Having constant f would mean constant a, I agree. – Dewi Morgan Nov 06 '21 at 22:13
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As the time interval in which the force is applied is also important, I will reword the question as follows.

An object of mass $m$ moving vertically upward and decelerating due to gravity $g$ has an initial velocity $u\ge 0$ at $t=0$. If the object must reach a point $h>0$ above it at $0< t \le T$, find the minimum additional upward force $F$ needed.

Case 1: $u=0$

\begin{align} F-mg &=ma\\ h&=\frac{1}{2}aT^2 \end{align} The minimum force is $F=m\left(\frac{2h}{T^2}+g\right)$.

Case 2: $u>0$ and $T\ge\frac{u}{g}$ and $h\le\frac{u^2}{2g}$

The minimum force is $F=0$. Let the object continue moving and it is guaranteed that it will reach the target point.

Case 3: Otherwise

\begin{align} F-mg &=ma\\ h&=uT + \frac{1}{2}aT^2 \end{align} The minimum force is $F=m\left(\frac{2(h-uT)}{T^2}+g\right)$.

Display Name
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  • I think case 1 and case 3 can be combined into a single case. I did not realize that at the time this answer was written. – Display Name Nov 02 '21 at 02:08