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Magnitude of frictional force during terminal velocity

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data

    A parachutist, after jumping from the plane, is accelerating down with average acceleration of 1.6 m/s2 (relative to the ground). What is the force of friction slowing his motion is mass of the parachutist with his equipment is 124 kg? What is the frictional force when the parachutist reaches terminal velocity?



    2. Relevant equations

    F = ma



    3. The attempt at a solution

    F=ma
    F=124 kg x 1.6 m/s2
    F= 199.36 N [down]

    Would the force of friction also be 199.36 N except in [up] direction because for every action there is an equal and opposite reaction?

    What happens when he reaches terminal velocity? Wouldn't acceleration = 0 and cause force unbalance to equal to 0. Doesn't this mean the two forces of gravity and friction are equal but in opposite direction? So how do I find the frictional force upwards during terminal velocity?
     
  2. jcsd
  3. Nov 9, 2012 #2
    Note that when F = ma is used the symbol F represents the RESULTANT force on the system.
    Then when he is accelerating, resultant force (downwards) = weight - frictional force
    and when he is moving at terminal speed, resultant force = 0.
     
  4. Nov 9, 2012 #3

    PhanthomJay

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    This is the net force acting down. There are 2 forces acting on the 'chutist. The net force is the vector sum of both.

    No.
    yes
    yes
    solve your last question!
     
  5. Nov 9, 2012 #4
    Well....

    Fg=ma
    =124 kg x9.8m/s2 [down]
    =1215.2 N[down]

    This is force of gravity on the parachutist.

    FNET = 199.36 N [down]

    FNET = Fg + Ff
    Ff = FNET - Fg
    = 199.36 N[down] - 1215.2 N [down]​
    = -1015.84 N [down]​

    ∴ Frictional force during Vterminal = 1015.84 N [up]
     
  6. Nov 9, 2012 #5
    ?

    This however is not right because frictional should be equal to gravity?

    Am I overthinking everything and the frictional force at V_terminal is simply 1215.2 N [up] (magnitude of his weight).



    EDIT: sorry for double posting, I accidentally clicked quote instead of edit and I don't know how to delete this
     
  7. Nov 9, 2012 #6

    PeterO

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    Yes - at terminal velocity friction must exactly "balance" weight force.
     
  8. Nov 9, 2012 #7
    Okay, thank you.
     
  9. Nov 9, 2012 #8

    PhanthomJay

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    Say Ace. it appears that you made an error in determining the direction of the friction force when the chutist is accelerating downward. One must be very careful in the use and interpretation of the minus sign. Please give this important topic some thought.
     
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