A block ofmass m 1 = 5.30 kgsits on top of a second block ofmass m 2 = 16.4 kg,

Question

A block ofmass m1 = 5.30 kgsits on top of a second block ofmass m2 = 16.4 kg, which in turn is on a horizontal table.The coefficients of friction between the two blocks areµs = 0.300 andµk = 0.100. The coefficients offriction between the lower block and the rough table areµs = 0.500 andµk = 0.400. You apply a constanthorizontal force to the lower block, just large enough to make thisblock start sliding out from between the upper block and the table. (a) Draw a free-body diagram of each block, naming the forces oneach. (Do this on paper. Your instructor may ask you to turn inthis work.)

(b) Determine the magnitude of each force on each block at theinstant when you have started pushing but motion has not yetstarted. In particular, what force must you apply?
N
(c) Determine the acceleration you measure for each block.
m/s2 (m1)
m/s2 (m2) (a) Draw a free-body diagram of each block, naming the forces oneach. (Do this on paper. Your instructor may ask you to turn inthis work.)

(b) Determine the magnitude of each force on each block at theinstant when you have started pushing but motion has not yetstarted. In particular, what force must you apply?
N
(c) Determine the acceleration you measure for each block.
m/s2 (m1)
m/s2 (m2)

Explanation / Answer

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So your applied force must be:

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F - kinetic friction from floor = (m1+ m2) a

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   F = 85.064 + 21.7 *2.94 =   148.86 N

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Now... if your force exceeds this value, then the upper blockwill slip. Let’s assume your force is 149 N, just slightlygreater than necessary.

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Then the upper block experiences only kinetic friction and itsacceleration is...

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   upper block acceleration = force ofkinetic friction / mass = 5.194 /5.30 =  0.980m/s 2

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and   lower block acceleration = (F - f1 - f2) / mass = (149 - 5.194– 85.064) / 16.4 =

.

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