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pathosdebater

  • 2 years ago

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

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  1. JakeV8
    • 2 years ago
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    |dw:1354921431255:dw|

  2. JakeV8
    • 2 years ago
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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"

  3. pathosdebater
    • 2 years ago
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    So E, none of the above.

  4. JakeV8
    • 2 years ago
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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.

  5. pathosdebater
    • 2 years ago
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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?

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

  7. pathosdebater
    • 2 years ago
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    yes

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

  9. JakeV8
    • 2 years ago
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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.

  10. JakeV8
    • 2 years ago
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    Were you given a coefficient of friction, μ?

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

  12. pathosdebater
    • 2 years ago
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    It is already moving as a constant velocity.

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

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

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

  16. JakeV8
    • 2 years ago
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    good idea :)

  17. pathosdebater
    • 2 years ago
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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?

  18. JakeV8
    • 2 years ago
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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.

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

  20. JakeV8
    • 2 years ago
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    |dw:1354922360538:dw|

  21. JakeV8
    • 2 years ago
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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).

  22. pathosdebater
    • 2 years ago
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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?

  23. JakeV8
    • 2 years ago
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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!!)

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