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Solving problems involving forces

  1. Jul 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A loaded barge has a mass of 1.5exponent6 kg and its travelling at 3 metres per second. If a tugboat applies an opposing force of 1.2exponent4 N for 10s

    What is it final velocity?
    How long does it take to stop

    2. Relevant equations
    F = ma
    F = m(v-u)/t


    3. The attempt at a solution
     
  2. jcsd
  3. Jul 14, 2016 #2

    CWatters

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    Please post your attempt at a solution.
     
  4. Jul 14, 2016 #3
    Ive subtituted all the values given in the formula. From then, i couldnt figure out the next step
     
  5. Jul 14, 2016 #4

    CWatters

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    It's best NOT to substitute values at the outset. The problem asks for the final velocity. Try rearranging the second equation to give an equation for the final velocity. Show your working.
     
  6. Jul 14, 2016 #5
    So can i substitute F on the second equation with the value of N given in the question. It says that it is an opposing force. So i have to do F-N = m(v-u)/t. Right? Then i will have 2 unknowns F and V. What now?
     
  7. Jul 14, 2016 #6

    CWatters

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    No it's easier than that. The "N" in the question just stands for Newtons the unit of force not another variable.

    |F| = 1.2 * 104 Newtons

    Actually you need to think about the sign.
     
  8. Jul 14, 2016 #7

    haruspex

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    The second question is a bit unclear. I think it means, how long will it take to stop if the same force continues?
     
  9. Jul 14, 2016 #8
    I think haruspex is right that the question is asking how long it will take to stop if the same opposing force from the tugboat continues. The reason is because I think this problem is intended to be equivalent to an object sliding on a frictionless surface. In reality, a barge and a tugboat is not that situation because of the varying resistance of the water. So I think a barge and tugboat is really not a good choice for a problem of this nature.

    I also had to look up the equation F = m(v - u)/t because I didn't know what it meant. Maybe I am revealing my age, but I remember V = Vo + at, which would rearrange to F = m(V-Vo)/t. But afifbaha, please use the equation that you listed and are familiar with: F = m(v-u)/t

    So as CWatters said, try rearranging the equation to solve for the unknown.
    In the first part of the problem, the final velocity is the unknown. In the second part of the problem, the final velocity is given and time is the unknown.

    And like CWatters pointed out, 'N' is not a variable; it is a symbol for Newtons - a unit of force, like 'lb' for pounds.

    Sorry to be so wordy, but this is my very first post after my introduction.
    ~ inept new guy
     
  10. Jul 14, 2016 #9
    I suggest to use momentum & impuls equation
     
  11. Jul 14, 2016 #10
    Actually your second equation is the simple impuls equation,
    You can solve the first question with your second equation, just subtitute and solve...
     
  12. Jul 15, 2016 #11

    CWatters

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    I think we should let the OP attempt the problem again offering more help.
     
  13. Jul 16, 2016 #12
    I knew that N is newtons. What i was saying was for F in F=ma, the F is net force which means it has to be subtracted with any other forces. In this case it has to be subtracted with the opposing force right? That is what i meant.
     
  14. Jul 16, 2016 #13

    CWatters

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    Ok sorry about that.

    As far as I can see the tug is the only force acting on the barge - at least no other forces are stated in the problem.
     
  15. Jul 16, 2016 #14
    Ya thats what i was figuring out. Anybody else could help?
     
  16. Jul 16, 2016 #15

    haruspex

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    CWatters is saying (and I agree) that you should consider the tug's arresting pull as the only horizontal force on the barge. Of course, in reality, there would be drag from the water, but that would be very small in comparison.
     
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