PS2-3

# PS2-3 - Problem Set 2: Problem 3. Problem: At time t = 0,...

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Unformatted text preview: Problem Set 2: Problem 3. Problem: At time t = 0, Block A begins sliding down an incline of angle φ with constant acceleration a and Block B begins moving to the right with constant acceleration 5 a relative to Block A. Determine the 4 absolute acceleration of Block B, aB and its speed, |vB |, as a function of a, t and φ. If the acceleration of Block A is a = 10 cm/sec2 and φ = 30o , what is |vB | when t = 5 sec? y ................. . .. . .. . ..... . . . ............. ... ........... . ................. .................. .. ...... ........ .. . ...................... . .. ... .................. ...................... . ...................... . . . .. ..................... . . ...................... . .. . . ...................... ..... ......... . ..................... ................ . ...................... . ................. . ..................... . ..................... ............ .............. . . . . . . . . . ................................. . . . . . ... ......................... . ...... .. . .............. .. .................... . . ........................................................................ ............................ .................. . . ..... . . . . . . . . . ......... . ................................................................................................... ............... . . .. .......................................................................................................................... . . .. ................................................................................................................................ . ... . . . . ... .. . ........ . .. . . ............................................................................................................. . . . ............................................................................................................. . .. . . . ...................................................................... .... ................ . ............................................................................................................ . . .................................................................................................. . . . . .... ..... . . .......................................................................................................... . . .. . . .. .... .... . ............................................................................................. . .............................................................. .. ......... ................................................................................................................... ... .. . . . ................................................. ... ........... . . ..................................................................... . .... . . . ...................................................................... . . ........................................................ .. . .. . ... ... ... . . ............................................................. . . . . ............................................................. . . ... . ... ..... . . ............. .......... .. .......................................... .. . . . ............................................... . . .............. . .. . . ............................. . . . . ..... .. ... ................................... . ........................ ..................................... . . . .. . .. .... ... . ....... .... . . ..... ............ . ..................... . ................... ...................... . ..................... . .................... . .................. . .................. ..................... ..................... . . . ............................. . . ..... . .. . . . . ............... ...................... . . ............................ . . ..... . ... ................... .............................................................................. . ................................................................................................... . . .. . . ... .. . . ........ .... B A φ x Also, Block B’s acceleration vector, relative to Block A, is aB/A = Thus, the absolute acceleration of Block B is Solution: Since Block A slides down the incline, its acceleration vector is tangent to the incline. Hence, we conclude that aA = a(−i cos φ − j sin φ) 5 ai 4 5 aB = aA + aB/A = a(−i cos φ − j sin φ) + a i 4 Rearranging terms, we find 5 − cos φ a i − a sin φ j 4 Thus, since the velocity vector is vB = aB t, the speed at time t is aB = |vB | = at = at = at = at Therefore, Block B’s speed at time t is |vB | = at 4 41 − 40 cos φ 5 − cos φ 4 2 + sin2 φ 25 5 − cos φ + cos2 φ + sin2 φ 16 2 25 5 − cos φ + 1 16 2 41 5 − cos φ 16 2 We are given a = 10 cm/sec2 and φ = 30o , what is |vB | when t = 5 sec. Substituting into the equation above, we have the following. √ 1 cm cm |vB | = (5 sec) 41 − 40 cos 30o = 31.5 10 2 4 sec sec ...
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## This note was uploaded on 10/12/2009 for the course AME 301 at USC.

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