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Kainui

  • one year ago

If each box weighs 200lb , determine the least horizontal force P that the man must exert on the top box in order to cause motion. The coefficient of static friction between the boxes is μs = 0.80, and the coefficient of static friction between the box and the floor is μ′s = 0.55. https://session.masteringengineering.com/problemAsset/1526808/1/Probs.8-57_58.jpg

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  1. Kainui
    • one year ago
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    So is the whole thing going to slide, is the top box going to fall off, or is the whole thing going fall over?

  2. Kainui
    • one year ago
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    This is a question my little brother sent me last semester who's in school to become an engineer. He said it was a fun problem, and I sorta forgot about it but just found it and now it's bugging me since I don't really know what to do.

  3. ParthKohli
    • one year ago
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    Alright, just draw free-body diagrams and wish for the best.|dw:1433696058786:dw|

  4. Kainui
    • one year ago
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    I guess my confusion lies in the fact that I can sorta consider the whole thing as having a kind of pivot point at point B in the diagram, but I can also consider it to be a completely solid object that slides, or I can consider the same scenario for the box on top as either pivoting or sliding. I guess I could work out all 4 or at least get started but that just doesn't seem like the right way to think about it, mindlessly calculating.

  5. ParthKohli
    • one year ago
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    Whenever considering limiting cases, \(f_s = \mu_s N\) (actually, there is an inequality to the left but we consider the equality to anlayse limiting cases).

  6. Michele_Laino
    • one year ago
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    we have these forces: |dw:1433696128210:dw| on box A are acting these two forces: the force P and the friction force R_1 whose magnitudfe is: \[\Large {R_1} = {\mu _{SB}}mg\] where \[\Large {\mu _{SB}}\] is the static friction coefficient between boxes. On box B are acting the se two forces, -R_1 and R_2 whose magnitude is: \\Large [{R_2} = {\mu _{SF}}2mg\] where: \[\Large {\mu _{SF}}\] is the static friction coefficient between box and floor

  7. ParthKohli
    • one year ago
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    \[N_1 = 200 g\tag{top box, vertical direction}\]Where \(g\) is the gravitational constant.\[N_1 + 200 g = N_2 \tag{bottom box, vertical direction}\]

  8. ParthKohli
    • one year ago
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    Yeah, looks like my work's correct so far...

  9. Michele_Laino
    • one year ago
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    \[\Large {R_2} = {\mu _{SF}}2mg\]

  10. ParthKohli
    • one year ago
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    \[N_1 = 200 g = (200 \times 9.81) ~N\]\[N_2 = 400 g = (400 \times 9.81)~N\]\[f_1 = \mu_sN_1\]\[f_2 = \mu_s' N_2\]

  11. ParthKohli
    • one year ago
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    Again, careful about the units. I've just replaced lbs with kg because, well, SI-master-race.

  12. ParthKohli
    • one year ago
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    For the box to move, \(F = f_1 + \epsilon\). Minimise this force and you get \(F = f_1 = \mu_s N_1\).

  13. Michele_Laino
    • one year ago
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    now the equilibrium condition is given by these subsequent formulas: \[\large \left\{ \begin{gathered} {\mathbf{P}} + {{\mathbf{R}}_{\mathbf{1}}} + \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ {\mathbf{O}} \times {\mathbf{XP}} + {\mathbf{OY}} \times {{\mathbf{R}}_{\mathbf{1}}} + {\mathbf{OZ}} \times \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {\mathbf{OT}} \times {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.\]

  14. Michele_Laino
    • one year ago
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    |dw:1433696867479:dw|

  15. Michele_Laino
    • one year ago
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    oops, there is a typo: \[\large \left\{ \begin{gathered} {\mathbf{P}} + {{\mathbf{R}}_{\mathbf{1}}} + \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ {\mathbf{OX}} \times {\mathbf{P}} + {\mathbf{OY}} \times {{\mathbf{R}}_{\mathbf{1}}} + {\mathbf{OZ}} \times \left( { - {{\mathbf{R}}_{\mathbf{1}}}} \right) + {\mathbf{OT}} \times {{\mathbf{R}}_{\mathbf{2}}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.\]

  16. Michele_Laino
    • one year ago
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    the first vector equation, is equivalent to this subsequent scalar equation: \[\Large P - {R_2} = 0\]

  17. Michele_Laino
    • one year ago
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    where I have used the reference system in my drawing

  18. Kainui
    • one year ago
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    Sorry for the delay, give me a moment to read through this I'm at the kitchen table on my laptop and I'm having a conversation with my parents so it'll be a minute.

  19. Michele_Laino
    • one year ago
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    whereas the second vector equation, is equivalent to this scalar equation: \[\Large 5P - 4.5{R_1} - 4.5{R_1} = 0\]

  20. Michele_Laino
    • one year ago
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    From the firs scalar equation, we see that, motion is possible, if and only if, the subsequent condition holds: \[\Large P > {R_2} = {\mu _{SF}}2mg\]

  21. Michele_Laino
    • one year ago
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    first*

  22. ParthKohli
    • one year ago
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    Oh, I thought the question was talking about the motion of the first block only.

  23. Michele_Laino
    • one year ago
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    in that case the condition is: \[\Large {\mu _{SB}}mg < P < {\mu _{SF}}2mg\]

  24. ParthKohli
    • one year ago
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    |dw:1433698078430:dw|