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F = ma, net force on shopping carts

  1. Aug 4, 2011 #1
    In a supermarket parking lot, an employee is pushing ten empty shopping carts, lined up in a straight line. The acceleration of the carts is 0.05 m/s^2. The ground is level, and each cart has a mass of 26 kg.

    a. What is the net force acting on any one of the carts?
    b. Assuming friction is negligible, what is the force exerted by the fifth cart on the sixth cart?

    Answers:
    a. 1.3 newtons
    b. 6.5 newtons

    I most definitely understand the concept behind B and how to reach the answer of 6.5 newtons.

    I know that to get answer A, you do F = ma = 26(0.05) =1.3 newtons....However, I am confused as to why part A is not 13 newtons.

    Isn't the net force on any single cart going to be 13 newtons? My reasoning was that there are ten carts total, for a total mass of 260 kilograms.

    Total mass = 10(26) = 260 kg

    Fnet = total mass x acceleration = 260 x 0.05 = 13 newtons.

    Don't the masses of the carts add up to give the net force on any cart??
     
  2. jcsd
  3. Aug 4, 2011 #2

    PeterO

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    Homework Helper

    If you can work out how strongly Cart 5 pushes on Cart 6, try cart 3 on cart 4 and all the way back to the employee pushing Cart 1.
    Then consider the difference between the net force on the bunch of carts vs the net force on each of them.
     
  4. Aug 4, 2011 #3
    Newton's second law states that the net force on an object is equal to the rate of change of momentum (F=ma for this case). Does (a) make sense then?
     
  5. Aug 4, 2011 #4
    I do not know what momentum is. I have just started physics a month ago. I've been looking at this problem for 30 minutes and I still do not understand.
     
  6. Aug 4, 2011 #5
    Well, you might want to study out of a proper textbook in that case. (a) is a straight forward application of Newton's second law. Given the mass m=26 kg and the acceleration a=0.05 m/s2, the net force is given by (ma) or 1.3 N. The net force comes out from the forward force due to the previous carts and the reaction force due to the carts in front.
     
  7. Aug 4, 2011 #6

    PeterO

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    Imagine instead you were holding a long pole, with 10 separate strings attached to the 10 carts which were lined up beside each other.

    You then accelerate them at 0.05 m/s^2

    Can you now see the difference between the total force you apply, as compared to the force on each of the trollies.

    In each case you apply a force to the set of trollies, and each trolley is accelerated.
     
  8. Aug 4, 2011 #7
    Pi-bond: No need to be rude...If I could afford a proper textbook, don't you think I would get one? I am 13 years old, unable to get a job just yet, and my mom is a heroine-addicted, single mother working minimum wage. I can't afford a textbook that is $100+... so I rely on borrowing my friend's free internet to learn and when I can get there, I go to the library to use the limited textbooks available...and I can only use these INSIDE the public library during business hours... I get help where I can, and I'm not afraid to ask for help when I need it. EFF THIS SITE. I'm out.
     
  9. Aug 4, 2011 #8
    Thank you PeterO, I think I am just over-thinking and over-analyzing the question. I do that a lot..
     
  10. Aug 4, 2011 #9
    your second example just clicked :D thank you again :)))))) -jehan
     
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