Small block 1 with mass m = 3.29 kg sits on top of large block 2 of mass M = 15.

Question

Small block 1 with mass m = 3.29 kg sits on top of large block 2 of mass M = 15.3 kg, and the pair sit on a frictionless, horizontal table. Between the blocks: the coefficient of kinetic friction is ?k = 0.176, and the coefficient of static friction is ?s = 0.373.
NOTE: This is similar to a problem we did in lecture, except this time force F is applied horizontally to the upper block.

a) Find the magnitude of the maximum force applied horizontally to the upper block (block 1) that will cause the two blocks to slide together.
1

ANSWER:14.6N

b) Now, suppose the force applied to block 1 has magnitude F = 39.3 N, which is greater than the force found in part a. Find the magnitude of the acceleration of each block.
magnitude of the acceleration of block 1 (top block): 2 m/s2
magnitude of the acceleration of block 2 (bottom block): 3 m/s2

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Explanation / Answer

Ff = us m g max force applied to lower block without slipping F = (m + M) a force needed to accelerate both blocks M a = Ff acceleration of lower block due to Ff F = (m + M) * Ff / M combining last two equations F = (m + M) us m g / M = (3.29 + 15.3) * .373 * 3.29 * 9.8 / 15.3 F = 14.6 N F (upper) = 39.3 - Ff = 39.3 - uk m g net force on upper block m a = 39.3 - .176 * 3.29 * 9.8 a = 10.2 m/s^2 acceleration of upper block F (lower) = uk m g = M a solve for a of the lower block

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