1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Forces generated by free falling objects

  1. Jul 1, 2013 #1

    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?


  2. jcsd
  3. Jul 1, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hello m.cgalloway! welcome to pf! :smile:
    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)
  4. Jul 1, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No. Why do you think the acceleration (eg during free fall) should be the same as the deceleration (eg during stopping) ?
  5. Jul 4, 2013 #4


    User Avatar
    Homework Helper

    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.
  6. Jul 4, 2013 #5


    User Avatar
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook