Comparing Net Forces: Hammer and Feather in a Vacuum

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SUMMARY

The discussion centers on the comparison of net forces acting on a hammer and a feather dropped in a vacuum. The hammer, having 100 times the mass of the feather, experiences a net force that is also 100 times greater due to the equation F=ma. Despite this difference in force, both objects accelerate at the same rate of 9.8 m/s² because acceleration is independent of mass. The gravitational force acting on each object is derived from the equation F=GMm/r², confirming that while the net forces differ, the acceleration remains constant for both objects in a vacuum.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with gravitational force equations (F=GMm/r²)
  • Concept of acceleration due to gravity (9.8 m/s²)
  • Basic knowledge of mass and its relationship to force
NEXT STEPS
  • Explore the implications of Newton's Second Law in different gravitational fields
  • Study the derivation and applications of the gravitational force equation F=GMm/r²
  • Investigate the effects of air resistance on falling objects
  • Learn about the concept of free fall and its relation to mass and acceleration
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Physics students, educators, and anyone interested in understanding the principles of motion and gravitational forces in a vacuum.

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A hammer and a feather drops together in a vacuum space. The hammer has 100x the mass of the feather.

How do their net force compare?

I am thinking that the hammer's force is equal to F= m( a). So the hammer would have a 100x force than the feather's. But shouldn't their net force equal to their acceleration which is 9.8?
 
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BadSkittles said:
But shouldn't their net force equal to their acceleration which is 9.8?

Force and acceleration have different units, they can never be equal! But, one is related to the other by F=ma as you posted. This equation, F=ma, relates force to mass and acceleration but it doesn't tell you the force in this case. Consider the force due to gravty, F=GmM/r2. Does that look familiar? With that equation you can get a force due to gravity and then use that value to calculate "a" via F=ma. Protip - don't actually plug in any numbers at first, try to do it with the letters/symbols only. You will get better understanding that way.
 
It might be useful to imagine the situation where you drop each of these objects on Earth and ignore air resistance. Then we have two expressions for force:

$$F = m_{o}a$$
$$F = \frac{GMm_{o}}{r^{2}}$$

Here I have used the symbol ##m_{o}## to represent the mass of an object.By setting these two expressions equal to each other we notice something quite interesting. Can you figure out what this is? (Note: You should consider these expressions both for the same object. That is, when you equate them they are both talking about the same object - say, a ball, for example.)
 
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Tsunoyukami said:
It might be useful to imagine the situation where you drop each of these objects on Earth and ignore air resistance. Then we have two expressions for force:

$$F = m_{o}a$$
$$F = \frac{GMm_{o}}{r^{2}}$$

Here I have used the symbol ##m_{o}## to represent the mass of an object.By setting these two expressions equal to each other we notice something quite interesting. Can you figure out what this is? (Note: You should consider these expressions both for the same object. That is, when you equate them they are both talking about the same object - say, a ball, for example.)

ModusPwnd said:
Force and acceleration have different units, they can never be equal! But, one is related to the other by F=ma as you posted. This equation, F=ma, relates force to mass and acceleration but it doesn't tell you the force in this case. Consider the force due to gravty, F=GmM/r2. Does that look familiar? With that equation you can get a force due to gravity and then use that value to calculate "a" via F=ma. Protip - don't actually plug in any numbers at first, try to do it with the letters/symbols only. You will get better understanding that way.

So in that F = GmM/r^2 equation. Does that mean the mass of the two objects is irrelevant compared to the mass and the radius of the earth. So they would have the same Force?

I can see how acceleration cannot equal to force. But that equation just shows me otherwise ?
 
I think you should review the force of gravity equation and what the different pieces mean. "r" is not necessarily the radius of the Earth and the two masses are not the feather and the hammer (because we are not considering their attraction to each other, we are considering their individual attraction to the earth).

Did you do what Tsunoyukami suggested? "By setting these two expressions equal to each other we notice something quite interesting."
 
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ModusPwnd said:
I think you should review the force of gravity equation and what the different pieces mean. "r" is not necessarily the radius of the Earth and the two masses are not the feather and the hammer (because we are not considering their attraction to each other, we are considering their individual attraction to the earth).

Did you do what Tsunoyukami suggested? "By setting these two expressions equal to each other we notice something quite interesting."

So the final equation I got is

G (M of Hammer) (M of Earth) / (r ^2) = G (M of feather ) (M of Earth) / (r^2)

Canceling out similar variables.

(100x) = ( 1x)

Oh is that how you get there
 
This material should be familiar to you from class or the textbook but perhaps I should have been slightly more explicit.

The expression $$F = \frac{GMm_{o}}{r^{2}}$$ is the general expression for force between any two objects of masses ##m_{o}## and ##M## separated by a distance of ##r##.

You should set them equal separately and notice something interesting. Explicitly what I mean is apply the two expressions I mentioned previously to BOTH objects. That is,

$$F = \frac{GMm_{o}}{r^{2}} = m_{o}a = F$$
$$\frac{GMm_{o}}{r^{2}} = m_{o}a$$

Assume "o" is the hammer and finish the problem. Afterwards, assume "o" is the feather. You should notice something very interesting.EDIT: To make this even MORE obvious - here I am talking about ##M## as the mass of the Earth - not of the "other object". You can write ##M_{e}## if that helps for clarity.
 
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Tsunoyukami said:
This material should be familiar to you from class or the textbook but perhaps I should have been slightly more explicit.

The expression $$F = \frac{GMm_{o}}{r^{2}}$$ is the general expression for force between any two objects of masses ##m_{o}## and ##M## separated by a distance of ##r##.

You should set them equal separately and notice something interesting. Explicitly what I mean is apply the two expressions I mentioned previously to BOTH objects. That is,

$$F = \frac{GMm_{o}}{r^{2}} = m_{o}a = F$$
$$\frac{GMm_{o}}{r^{2}} = m_{o}a$$

Assume "o" is the hammer and finish the problem. Afterwards, assume "o" is the feather. You should notice something very interesting.EDIT: To make this even MORE obvious - here I am talking about ##M## as the mass of the Earth - not of the "other object". You can write ##M_{e}## if that helps for clarity.

So Force is independent of mass? It would mean they had the same force, because you can cross out m of o.
 
You can cross out m0, but you don't have an equation for the force anymore! What do you have an equation for?
 
  • #10
BadSkittles said:
So Force is independent of mass? It would mean they had the same force, because you can cross out m of o.

Not quite, but very close.

Something is independent of the mass of the object ##m_{o}##, but it's not force.
 
  • #11
wrong quote
 
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  • #12
Office_Shredder said:
You can cross out m0, but you don't have an equation for the force anymore! What do you have an equation for?

Tsunoyukami said:
Not quite, but very close.

Something is independent of the mass of the object ##m_{o}##, but it's not force.

You see that the GM/r^2 = a = 9.8 = F. That would mean mass is independent of acceleration

Is the final answer "equal force" on the feather and the hammer? !?

I thought acceleration and force couldn't be equal?
 
  • #13
That would mean mass is independent of acceleration

Yes: acceleration is independent of the object. What this means is that both the hammer and feather (and any other object for that matter) will accelerate at the same rate when dropped in such a (gravitational) force field.
BUT - you have something wrong with what you've written.

You see that the GM/r^2 = a = 9.8 = F. That would mean mass is independent of acceleration

Here you should have:

$$F = \frac{GMm_{o}}{r^{2}} = m_{o}a = F$$
$$\frac{GMm_{o}}{r^{2}} = m_{o}a$$
$$\frac{GM}{r^{2}} = a$$
$$F \neq \frac{GM}{r^{2}} = a$$

Is that clear?So you have a surprising result (at least I was surprised when I first saw this). All objects accelerate at the same rate? Yes. Even a feather and a hammer? Yep. Pretty interesting, right?
However, your question asks you to consider the (net) force on each of these objects falling in a vacuum. I'll jump right to the punchline here and say that you kind-of right in your first post: the force on the hammer will be 100 times that of the force on the feather.Can you explain why? Try thinking about what happens as you apply more and more force to an object.
 
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  • #14
The force on the hammer is 100x higher than on the feather, but the mass of the hammer is 100x higher than the mass of the feather. So the two accelerations are the same.
 
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  • #15
Yes, the acceleration is the same for both objects but the problem asks to compare the net force on each of the objects.
 
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  • #16
Got it guys! So essentially, the hammer does have a higher net force. m*a= F. Their acceleration is the same as proved by Gm1M2/ r^2 = m*a and is independent of mass. Therefore, the only variable that is really cared for is mass.
 
  • #17
That's correct.

Sorry for the rather roundabout approach but I thought you might find it interesting and enlightening to really see why.

Essentially the force of gravity acting on the hammer is larger than the force of gravity acting on the feather because more force is needed to accelerate the hammer than is needed to accelerate the feather. :)
 

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