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Posted by on Wednesday, August 28, 2013 at 10:15pm.

A block of mass
M1 = 2.7 kg
rests on top of a second block of mass
M2 = 4.7 kg,
and the second block sits on a surface that is so slippery that the friction can be assumed to be zero (see the figure below).

(a) If the coefficient of static friction between the blocks is
¦ÌS = 0.21,
how much force can be applied to the top block without the blocks slipping apart?

Incorrect: Your answer is incorrect.
N

(b) How much force can be applied to the bottom block for the same result?

  • math - , Friday, September 12, 2014 at 2:50am

    (a)
    Top block:
    F-friction=M1a
    F=2.7a+(0.21*2.7*9.8)

    From bottom block:
    friction=M2a
    a=(0.21*2.7*9.8)/4.7

    F=(0.21*2.7*9.8*2.7)/4.7+(0.21*2.7*9.8)=8.749N


    (b) Bottom block:
    F-friction=M2a
    F=4.7a+(0.21*2.7*9.8)

    Top block: 0.21*2.7*9.8=2.7a
    a=0.21*9.8

    F=(0.21*4.7*9.8)+(0.21*2.7*9.8)=15.23N

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