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rvc

  • one year ago

The coefficient of friction are µs = 0.4 and µK = 0.3 between all surfaces in contact. Determine the smallest force P required to start the 30 kg block moving if cable AB (a) is attached as shown (b) is removed

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  1. rvc
    • one year ago
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    need help with these three questions

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  2. Michele_Laino
    • one year ago
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    sorry, my word has crashed, can you make a sketch of your exercise, please?

  3. rvc
    • one year ago
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    wait please

  4. rvc
    • one year ago
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  5. Michele_Laino
    • one year ago
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    part (b) the cable is removed In this case total friction force, acting on composed block is: \[{\mu _s}\left( {{m_A} + {m_B}} \right)g\] where \(\mu_s\) is the friction coefficient between the composed block and the plane, whereas \(m_A\) and \(m_B\) are the masses of the two blocks

  6. Michele_Laino
    • one year ago
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    |dw:1443977990481:dw|

  7. rvc
    • one year ago
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    hmm

  8. Michele_Laino
    • one year ago
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    more precisely, what are your friction coefficients \(\mu_S\) and \(\mu_K\)?

  9. rvc
    • one year ago
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    0.4 and 0.3 respectively

  10. Michele_Laino
    • one year ago
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    yes! I know, can you indicate the surfaces to which those corfficients are referring to, please?

  11. Michele_Laino
    • one year ago
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    when the cable is removed, then we have two blocks which form a composed block, and the friction force, at beginning of motion, is: \[\Large R = {\mu _S}\left( {{m_A} + {m_B}} \right)g\]

  12. anonymous
    • one year ago
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    Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble A was initially moving at a velocity of 0.5 m/s, but after the collision it has a velocity of -0.1 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer. (5 points)

  13. IrishBoy123
    • one year ago
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    |dw:1443981768532:dw| |dw:1443981937384:dw|

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