# Forces generated by free falling objects

• m.cgalloway
In summary: The acceleration (eg during free fall) is the rate of change of momentum, while the deceleration (eg during stopping) is the rate at which the momentum changes.
m.cgalloway
Hello:

I have quite a few questions regarding objects free falling and the forces that they generate. I have been doing some reading on this topic over the last few days but I think that I am missing some of the concepts and how they apply in the scenario that I am trying them to. I think perhaps the issue that I am running into is that I am not applying the fundamental concepts properly.

First, this is what I know, or at least I think I know: Force = Mass x Acceleration. Weight = Mass x g. Where G is the constant of 9.81m/s2. Now my questions based on this:

1. 'LBf', or pounds of force, is not the same as regular 'LB', pounds, correct? My understanding is that 'LBf' is a measure of Weight and 'LB' is a measure of mass, is this right?
2. I have seen several places refer to the units of Weight (in the equation I mentioned above) as 'LBf' (or pounds of force) and other places as 'N' (Newtons). Which is it? This confused me since from what I have been reading 1LBf is equal to ~4.4N so the units can't be interchangeable, right?
3. All other things remaining the same, When calculating the force that a free falling object generates when it is suddenly stopped I tend to think that the equation Force=Mass x Acc will give me the answer. I use 9.81m/s2 as the acceleration since it is a free falling object thus falling at the same rate as anything else in the planet would, is my assumption correct?
4. This is where I start to get confused. If my assumption in the previous question is correct (which I am aware probably isnt), then that means that an object free falling at 1 meter would generate the same force as an object falling 2meters or even 100 meters. I know this isn't right but I am not sure how to prove that. Wha I'm I missing here? When do I bring the distance fallen into the equation?

Now to get specific. the reason I am trying to understand this is because I have a scenario that I am trying to get some data from. The scenario is the following.

I am standing 3 meters at the top of a hill and I am holding a 10 meter rope with both. The other end of the rope is tied to a 10kg dead weight which is already hanging 4 meters over the edge. This means that from where I am to where the load is there are 3 meters of rope laid out horizontally and 4 meters of rope dangling vertically over the edge for a total of 7m of rope past where I am holding. If I were to quickly let go of the rope and let 1meter of rope pay out before I grabbed it again what would be the force that I would feel at my hands when I grabbed that rope again? what if I let it go for 2m instead of 1?

For this scenario take into account that the rope is rated to a ridiculous strength and does not break, that the rope does not stretch, and that the load is stopped instantaneously meaning that I would be strong enough to not be moved forward when I grab the rope to stop the falling load. I know that this is not al all realistic since in a real system there would be some shock absorption from the rope and from me but for the sake of argument let say that there isn't. How can I use the formulas above to figure out the force generated at my end of the rope?

Thanks!

MG

welcome to pf!

hello m.cgalloway! welcome to pf!
m.cgalloway said:
… If I were to quickly let go of the rope and let 1meter of rope pay out before I grabbed it again what would be the force that I would feel at my hands when I grabbed that rope again? what if I let it go for 2m instead of 1?

you're confusing force with impulse

you would feel an impulse, equal to the change in momentum

(and the impulse you would feel would be the same as the impulse felt by the ground if the weight, instead, happened to hit the ground at that point)

When calculating the force that a free falling object generates when it is suddenly stopped I tend to think that the equation Force=Mass x Acc will give me the answer. I use 9.81m/s2 as the acceleration since it is a free falling object thus falling at the same rate as anything else in the planet would, is my assumption correct?

No. Why do you think the acceleration (eg during free fall) should be the same as the deceleration (eg during stopping) ?

m.cgalloway said:
Hello:

I have quite a few questions regarding objects free falling and the forces that they generate. I have been doing some reading on this topic over the last few days but I think that I am missing some of the concepts and how they apply in the scenario that I am trying them to. I think perhaps the issue that I am running into is that I am not applying the fundamental concepts properly.

First, this is what I know, or at least I think I know: Force = Mass x Acceleration. Weight = Mass x g. Where G is the constant of 9.81m/s2. Now my questions based on this:

1. 'LBf', or pounds of force, is not the same as regular 'LB', pounds, correct? My understanding is that 'LBf' is a measure of Weight and 'LB' is a measure of mass, is this right?
2. I have seen several places refer to the units of Weight (in the equation I mentioned above) as 'LBf' (or pounds of force) and other places as 'N' (Newtons). Which is it? This confused me since from what I have been reading 1LBf is equal to ~4.4N so the units can't be interchangeable, right?
3. All other things remaining the same, When calculating the force that a free falling object generates when it is suddenly stopped I tend to think that the equation Force=Mass x Acc will give me the answer. I use 9.81m/s2 as the acceleration since it is a free falling object thus falling at the same rate as anything else in the planet would, is my assumption correct?
4. This is where I start to get confused. If my assumption in the previous question is correct (which I am aware probably isnt), then that means that an object free falling at 1 meter would generate the same force as an object falling 2meters or even 100 meters. I know this isn't right but I am not sure how to prove that. Wha I'm I missing here? When do I bring the distance fallen into the equation?

Now to get specific. the reason I am trying to understand this is because I have a scenario that I am trying to get some data from. The scenario is the following.

I am standing 3 meters at the top of a hill and I am holding a 10 meter rope with both. The other end of the rope is tied to a 10kg dead weight which is already hanging 4 meters over the edge. This means that from where I am to where the load is there are 3 meters of rope laid out horizontally and 4 meters of rope dangling vertically over the edge for a total of 7m of rope past where I am holding. If I were to quickly let go of the rope and let 1meter of rope pay out before I grabbed it again what would be the force that I would feel at my hands when I grabbed that rope again? what if I let it go for 2m instead of 1?

For this scenario take into account that the rope is rated to a ridiculous strength and does not break, that the rope does not stretch, and that the load is stopped instantaneously meaning that I would be strong enough to not be moved forward when I grab the rope to stop the falling load. I know that this is not al all realistic since in a real system there would be some shock absorption from the rope and from me but for the sake of argument let say that there isn't. How can I use the formulas above to figure out the force generated at my end of the rope?

Thanks!

MG

First part: you wanted to use a g value of 9.81m/s2. That means you will be using mass in kg, and Force in Newtons, so you don't have to worry about the various Pounds you have come across.

Second part: A falling object does not generate Force. ?

If you drop something, and then that falling object strikes something - a large force may be required for the "what happens next" scenario you get.

If you dropped a 1kg rock several metres onto a pane of glass, the rock will hardly be affected, while the glass is shattered.

If you drop the 1 kg rock from a height of 5m onto a concrete pathway, the rock will appear to stop instantly.
Compare the accelerations:
The Rock will take approx 1 second to fall the 5m - so with an acceleration of 9.81, it will have reached its final speed.
The rock then stops in a very short time - perhaps only 0.001 seconds.
If the speed change happens on 1/1000 th the time, the acceleration must have been 1000 times as large, so the average Force necessary for the change to happen is 1000 times the size as the force of gravity (weight of the rock; 9.81 N) or 9810 N.

m.cgalloway said:
3. All other things remaining the same, When calculating the force that a free falling object generates when it is suddenly stopped I tend to think that the equation Force=Mass x Acc will give me the answer. I use 9.81m/s2 as the acceleration since it is a free falling object thus falling at the same rate as anything else in the planet would, is my assumption correct?
4. This is where I start to get confused. If my assumption in the previous question is correct (which I am aware probably isnt), then that means that an object free falling at 1 meter would generate the same force as an object falling 2meters or even 100 meters. I know this isn't right but I am not sure how to prove that. Wha I'm I missing here? When do I bring the distance fallen into the equation?
There's two different things to consider. 1) The force to prevent the object from falling. This is Mass x 9.81m/s^2 And this is what you feel when you hold a book in your hand. 2) The force required to slow a moving object so that it becomes stationary. This is the force you feel when you catch a ball, for example. And this force depends on how quickly you slow the object, and the momentum the object had, just before it hit your hand.

## 1. What is the formula for calculating the force generated by a free falling object?

The formula for calculating the force generated by a free falling object is F = m x g, where F is the force in Newtons, m is the mass of the object in kilograms, and g is the acceleration due to gravity, which is typically 9.8 m/s² on Earth.

## 2. Does the weight of a free falling object affect the force it generates?

Yes, the weight of a free falling object directly affects the force it generates. The greater the weight of the object, the greater the force it will generate when falling due to the influence of gravity.

## 3. Can the shape or size of a free falling object impact the force it generates?

Yes, the shape and size of a free falling object can impact the force it generates. Objects with larger surface areas may experience more air resistance, which can decrease the force of their fall, while objects with smaller surface areas may experience less air resistance and therefore generate a greater force.

## 4. How does air resistance affect the force generated by a free falling object?

Air resistance can decrease the force generated by a free falling object by slowing down its descent. This is because air particles collide with the object, exerting a force in the opposite direction of its fall. However, in a vacuum where there is no air resistance, the force of a free falling object would be solely determined by its weight and acceleration due to gravity.

## 5. Does the height from which a free falling object is dropped impact the force it generates?

Yes, the height from which a free falling object is dropped can impact the force it generates. The higher the object is dropped from, the greater its potential energy will be, which will convert to kinetic energy as it falls. Therefore, the force generated by the object will be greater when dropped from a higher height compared to a lower height.

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