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pathosdebater

Please help! (: Two objects have different weights. The first is three times as heavy as the second. If the objects are stacked (with the second block on the bottom and placed on a surface that exhibits friction), how much would the friction be? (Select the BEST answer). a) Same as block 1 alone b) Same as block 2 alone c) Three times the amount of the second block alone d) 8 times the amount of second block alone e) None of the above

  • one year ago
  • one year ago

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  1. JakeV8
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    |dw:1354921431255:dw|

    • one year ago
  2. JakeV8
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    If the friction depends on the total mass "pushing down", then the friction from the combination of blocks is proportional to the total mass "4x"

    • one year ago
  3. pathosdebater
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    So E, none of the above.

    • one year ago
  4. JakeV8
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    E is what it appears to me. You have to be a little careful to make sure they don't sneak an answer in that might work, but it just comes down to comparing 4x versus each of the other single blocks.

    • one year ago
  5. pathosdebater
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    Now if 4x = 1,451 pounds, what would the friction force being exerted on the box be? Or how would I find it?

    • one year ago
  6. JakeV8
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    Is the total mass of the two blocks equal to 1451 pounds?

    • one year ago
  7. pathosdebater
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    yes

    • one year ago
  8. JakeV8
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    Like your other problem, Force of friction = "mu" times the normal force.\[F _{friction} = \mu N\]

    • one year ago
  9. JakeV8
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    So the normal force is the force pressing up against the bottom of the blocks... since they gave it to you in pounds, that is a measure of mass times acceleration already... in other words, the normal force is equal to the weight, 1451 pounds.

    • one year ago
  10. JakeV8
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    Were you given a coefficient of friction, μ?

    • one year ago
  11. pathosdebater
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    They said it is being pushed across a floor at a constant velocity.

    • one year ago
  12. pathosdebater
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    It is already moving as a constant velocity.

    • one year ago
  13. JakeV8
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    ok, but that doesn't tell you much about the friction force.

    • one year ago
  14. pathosdebater
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    I'm just going to type the entire question out.

    • one year ago
  15. JakeV8
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    Unless they tell you what the force that is pushing it horizontally is.

    • one year ago
  16. JakeV8
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    good idea :)

    • one year ago
  17. pathosdebater
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    A mover is pushing a 1,451 pound box across a garage floor. Assuming that it is already moving at a constant velocity, what is the friction being exerted on the box?

    • one year ago
  18. JakeV8
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    That's not enough to be able to solve and get a number, but you can solve in terms of that coefficient of friction. To see why, realize that the garage floor could be almost as slick as ice... so very little friction exerted, and the mover just pushes lightly to maintain constant velocity.... OR.... a very rough floor, requiring the mover to push hard to maintain constant velocity against the much larger force of friction exerted. The problem gives you no way to know the amount of roughness... that's what the μ is.

    • one year ago
  19. pathosdebater
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    Dang... my physics prof never explains anything, so I'm not surprised that his exams are flawed.

    • one year ago
  20. JakeV8
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    |dw:1354922360538:dw|

    • one year ago
  21. JakeV8
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    So, the mover just has to push equal to the force of friction exerted, because as the problem says, the box is at constant velocity (i.e. no acceleration, therefore no net horizontal force).

    • one year ago
  22. pathosdebater
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    Can you figure out what force must the mover push in order to keep the box moving at a constant velocity? (Probably not) Or, How much force does the mover have to exert to start moving the box?

    • one year ago
  23. JakeV8
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    To keep it moving at constant velocity, the mover's push must equal the force of friction which is mu * N, or here, 1451 * mu, but since you don't know mu, you can't solve fully. I can't remember how to handle the static versus dynamic friction part... you have to push harder to start, due to static coefficient of friction, then it becomes easier after it's moving, but I don't recall how to set up the problem (sorry!!)

    • one year ago
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