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Please Help / Desperate, Feedback needed by tomorrow.

  1. Dec 18, 2006 #1
    1. The problem statement, all variables and given/known data

    A 98.6 kg box is resting on a horizontal surface. The coefficients of friction are uk= .1602 and us= .6027 for the surface to box. The box is attached to a 59.1 kg rock by a cord that passes over a frictionless pulley.

    What is the acceleration of the 59.1 kg mass? Make sure the direction is correct.

    What is the tension in the cord?


    2. Relevant equations

    Some equations I think may help with this problem (keep in mind I am clueless in Physics atm) are:

    F=ma
    (force of kinetic friction) Fkf = uk*Fn
    (force of static friction) Fsf= us*Fn
    Ft= mg+ma

    3. The attempt at a solution

    I have not enough time to type the 1/2 pg. of BS I started with.
    My grade is slipping in this class and I need your help. Teacher doesn't give enough of it when I ask. I have one or more problems that I need help with similar to this, but will refrain from typing them incase I don't end up with any responses, as this is my first time using this forum.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 18, 2006 #2

    berkeman

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    What does your free body diagram (FBD) look like? What are the forces on the box and on the rock due to the cord? What directions are those forces pointed in? If the forces are different, what does that tell you about the motion of the box and rock? Does it look like you should use uk or us? Why?
     
  4. Dec 18, 2006 #3
    My FBD for the box shows a small friction in the left direction and a longer arrow going to the right as the tension. For the rock's FBD, there is the force of gravity downwards and the force of tension upwards.
    I would assume we would have to use both uk and us. But for the acceleration I'd guess we use Uk to find it, because the box would be moving (kinetic).
    would the tension in the cord be:
    Ft = mg + ma, we obviously would have needed to first find A which I am unsure of how to do, I'd further be unsure of what masses to use in the Ft equation. both mass of the rock and box or just one of them?
     
  5. Dec 18, 2006 #4

    berkeman

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    Good start! You're close on the tension, but be careful to use separate mass symbols for the box and rock. And the tension on the rope from the box has two components -- some force from the acceleration of the box and some tension from the resistance of the friction. First determine if the whole system is moving or not. Assume no movement and see if the static frictional force will be overpowered or not....
     
  6. Dec 18, 2006 #5
    So you include both objects mass in the tension equation?

    I think it would be best to answer the questions in order. So, we are looking for the acceleration of the 59.1kg rock. I tried something and got 9.8 m/s2 which I think is wrong since that is just the force of gravity. What I did was this:

    A=?
    m = 59.1kg
    Fn (normal force) = mg
    Fn = 59.1kg*9.8 m/s2
    Fn= 579.18 N

    F=ma
    579.18N = 59.1kg*a
    divided both sides by 59.1kg to get a by itself, and got 9.8 m/s2.

    What did I do wrong?
    Also, if you have AIM let me know.
     
  7. Dec 18, 2006 #6

    berkeman

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    Well, the first thing you should do is multiply Fn by the statuc mu (us). What do you get? Why is understanding this force important to do first?
     
  8. Dec 18, 2006 #7
    That would give me 349.071 N but what does that represent? the minimum force it takes to get the box moving? how do I find the rock's acceleration?
     
  9. Dec 18, 2006 #8

    berkeman

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    And what is the initial force supplied by the hanging rock via the rope and pulley? What does that tell you about what the box does in the beginning?
     
  10. Dec 18, 2006 #9
    The inital force supplied by the hanging rock is the rock's mass times gravity (579.18 N)? I don't know what that tells you about the box in the beginning. It is at rest? I feel like you're not answering my questions
     
  11. Dec 18, 2006 #10

    berkeman

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    I'm not directly answering questions that you should be able to answer yourself, given proper hints. Now go back and re-calculate the force due to static friction at the box. I don't get 349N for that. Compare the static friction force available to the weight of the rock.
     
  12. Dec 18, 2006 #11
    If you didn't get 349 N, tell me how you came to your answer, because obviously I don't know how to find the force due to static friction at the box. I tried, now help because I don't know what to do. I got 349 N by using:

    Ffs = Us*Fn
    Ffs = Us*mg
    Ffs = .3027*59.1kg*9.8m/s2
    = 349
     
  13. Dec 18, 2006 #12

    berkeman

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    The box is 98.6kg, and the rock is 59.1kg.
     
  14. Dec 18, 2006 #13
    Are you saying I just used the wrong mass?
    If so, it would be 582 N.
     
  15. Dec 18, 2006 #14

    berkeman

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    And re-checking your post #9 tells you what? Look back at your FBD if that helps. Looks like you're done!

    Well, except for the obvious two quiz questions I'll ask when you post the answer....
     
  16. Dec 18, 2006 #15
    I'm really confused, how can I be done? I don't know what 582 N is. It's the force needed to move the box? How does that answer the question What is the acceleration of the 59.1kg mass/rock?
     
  17. Dec 18, 2006 #16

    berkeman

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    Okay, to summarize some of the steps you've taken, fill in the following:

    -- Mass of box, force of static friction available before box starts slipping.

    -- Mass of rock, force of tension available by rock pulling down on rope, force of tension available to start the box slipping.

    -- Conclusion about acceleration of the box and rock.

    Ready for the two quiz questions now?
     
  18. Dec 18, 2006 #17
    I wish you'd stick around longer and help me until I'm actually finished, this has been really prolonged and I have a ton of homework, I do appreciate your help so far though
     
  19. Dec 18, 2006 #18

    berkeman

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    Please fill in the blanks, and I think you're done.
     
  20. Dec 18, 2006 #19
    98.6 kg is the mass of the box.
    582 N is the force of static friction avail. before box starts slipping.
    59.1 kg is the mass of rock.
    I don't know how to find the force of tension available ??
    I haven't even found the acceleration of the rock !
     
  21. Dec 18, 2006 #20

    berkeman

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    The initial tension before anything starts to move is the weight of the rock. What is that. And then what happens?
     
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