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johnweldon1993

  • one year ago

Question with Dynamics?

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  1. johnweldon1993
    • one year ago
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    So I'm going to use a circle for my FBD First thing is...without the brakes being applied...would I not have a friction force? |dw:1436399040735:dw| Meaning for impulse \[\large mv_1 + \Sigma \int_{0}^{0.06} F_{wall} = mv_2\] \[\large (\frac{2700lb}{32.2ft/s^2})(4ft/s) + F_{wall}(0.06s) = 0\] \[\large F_{wall} = \frac{337.5lb/s}{0.06s} = 5625lb\]

  2. johnweldon1993
    • one year ago
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    I mean, that doesnt make sense to me at all, there is always friction, otherwise there would be no traction...regardless since the coefficient of kinetic friction wasn't introduced until the second part of the problem...it's the only thing I can think of

  3. johnweldon1993
    • one year ago
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    So now with the friction force added in because of the applied breaks we would have \[\large F_y = ma_y\] \[\large W = N\] \[\large N = 2700lb\] \[\large F_f = \mu_k N = 0.3(2700lb) = 810lb\] Impulse again \[\large mv_1 + \Sigma \int_{0}^{0.06}Fdt = 0\] \[\large (\frac{2700lb}{32.2 ft/s^2})(4ft/s) + F_{wall}(0.06s) - 810(0.06s) = 0\] \[\large F_{wall} = 4780.06lb\]

  4. johnweldon1993
    • one year ago
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    Just review so reply with whatever you can correct me on because again I'm not even 75% sure on this :)

  5. Vincent-Lyon.Fr
    • one year ago
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    The friction force, if any, would be negligible compared to the force by the wall.

  6. anonymous
    • one year ago
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    The correct units in line 3 is -337.5 lb.seconds / .06s = -5625 lb. negative total Force -5625lb (total) = -810 lb (brakes) + -4815 lb (wall) You _should_ apply the brakes _before_ hitting the wall, so their Force lasts longer.

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