Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with Force?

  1. Jul 30, 2008 #1
    I'm trying to calculate the amount of force I will need to move a 9.6g object at an average speed of 450 feet/ second. So when I do this I use Force=Mass*Acceleration. Since acceleration is change in speed/change in time (both of which are initially 0) I come out with 450*9.6. This yeilds 4320. Now, either i'm doing something terrible wrong, or I just have my units incorrect. I am not even sure what unit this answer is in, so if someone could explain this to me it would be amazing... Also this is my first time ever using the forum so if I have posted it in the wrong spot could you please redirect me.

  2. jcsd
  3. Jul 30, 2008 #2


    User Avatar

    Staff: Mentor

    You'll need to pay more attention to units because you are mixing English with Metric. The typical way to use the equation is using kilograms as the mass and meters per second as the acceleration, giving a force in Newtons.

    So you could use .0096 kg and 137 m/s/s, getting 1.3 N. But what does that mean?

    You've just calculated the force required to accelerate 9.6N to 137 m/s in 1s. But you didn't say anything about time. You could use 68.5 m/s/s for 2 seconds... So you've just discovered that there isn't any one required force to accelerate an object if there is no friction involved. You can use any force you want. Any force will provide an acceleration.

    And if there is friction involved, then the answer is slighty over the friction force and you'll need to calculate that via other means.
    Last edited: Jul 30, 2008
  4. Jul 30, 2008 #3
    I have an extremely limitted knowledge of physics. Could you please explain more about the time...I don't understand. Thank you for explaining that I mixed systems of measurement, I knever knew that gram and feet were not in the same one. However I do not see why you divide the time by 2 if the time elapsed became 2..."You've just calculated the force required to accelerate 9.6N to 137 m/s in 1s. But you didn't say anything about time. You could use 68.5 m/s/s for 2 seconds..." Could you please explain this as well?
  5. Jul 30, 2008 #4


    User Avatar

    Staff: Mentor

    The equation is f=ma. Force = mass times acceleration. There's nothing in there about speed or how long it takes to get up to accelerate to a certain speed.
    That's just a random example to illustrate the point above. The point is that you can pick any acceleration you want. The equation relating acceleration and velocity is v=a*t. You want an average velocity of 450 ft/s. With constant acceleration, you'll then need a final speed of 900 ft/s. So plugging in 900=a*t, you can pick any acceleration or time you want and calculate the other.

    Regarding units, when doing math, you do the math to the units just like you do fractions. Ie, for a*t=v --- m/s^2 * s = m/s. The only caveat is that a Newton is defined as kg-m/s^2.
  6. Aug 3, 2008 #5
    I'll get a word in edgeways:
    Newton had just told us the force equals the change rate of the momentum,which writes:
    F∝mv/t,Or F=kmv/t.
    In Metric, k =1.However,you can make k≠1 in other measurement metrics.That's where the floor host's mistake is,I thought.
  7. Aug 3, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    F=ma is not the place you want to start from to do this sort of problem.

    In a elementray Physics course you will be presented with a set of equations that must be memorized. It is one of these that you need to use.

    V = at + V 0

    Here V is the final velocity, a is the acceleration, t is time, and V 0 is the velocity at time =0.

    If at time = 0 the velocity is 0 this becomes, simply:


    Now since F=ma or a = F/m you can write:

    [tex] V = \frac {Ft} m [/tex]

    To find the force rearrange this to get:

    [tex] F = \frac {Vm} t [/tex]

    From this it is clear that given a final velocity and a mass you must also specify a time to arrive at a force.

    Last but not least, to get a useful number all of the above quanities must be in a consistent set of units.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook