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What is your weight? (in Newtons)

  1. Mar 26, 2005 #1
    Suppose that when you ride on your 7.00 kg bike the weight of you and the bike is supported equally by the two tires. If the gauge pressure in the tires is 73.5 lb/in2 and the area of contact between each tire and the road is 7.13 cm2, what is your weight? (in Newtons)

    MY WORK:



    This isn't correct. I also used the fact that the tires are round, so A=pi(d/2)^2 and used that as the area. That wasn't correct either.

    I also tried multiplying (both forms of area) by 2, since there is 2 tires. Nope, not correct again.

    And finally (all forms of force) I divided my answer by 9.81 since W does equal m*g. Still not correct.

    I have been modifying my solutions in a variety of ways, none of which are correct, summerizing (no matter how many times i reread the section) I am not understanding what is going on at all.

    Please help me.
  2. jcsd
  3. Mar 26, 2005 #2
    Dont forget to subtract the weight of the bike from the final force you get. The weight of the bike is about 70N. The total sufrace area of contact is 14.26cm^2.

    You are dividing by only he surface area of one tire.
  4. Mar 27, 2005 #3


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    Science Advisor

    From problem statement:
    {Weight of Bike} = (7 kg)*(9.81 m/sec^2) = (68.67 N)
    {Rider's Weight} = W
    {Total Tire Contact Area} = 2*(7.13 cm^2) = 2*(7.13e(-4) m^2)
    {Tire Pressure Holding Bike + Rider} = (73.5 lb/in^2) = (5.0677e(+5) N/m^2)

    Because system is at equilibrium, we have:
    {Force Applied By Tire Pressure} = {Weight Bike + Rider}
    ::: ⇒ {Tire Pressure Holding Bike + Rider}*{Total Tire Contact Area} = {Weight Bike + Rider}
    ::: ⇒ (5.0677e(+5) N/m^2)*{2*(7.13e(-4) m^2)} = W + (68.67 N)
    ::: ⇒ (722.65 N) = W + (68.67 N)
    ::: ⇒ W = (654 N) ::: (or mass of 66.6 kg = 147 lbs)

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