hw5_solns

# hw5_solns - Answer Key – Homework 5 – David McIntyre...

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Unformatted text preview: Answer, Key – Homework 5 – David McIntyre – 45123 – Mar 25, 2004 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. Chapter 5 problems. 001 (part 1 of 1) 10 points Assume: All surfaces, wheels, and pulley are frictionless. The inextensible cord and pulley are massless. m 1 m 2 F M Given M = 29 . 8 kg, m 1 = 4 . 29 kg, m 2 = 3 . 38 kg, and g = 9 . 8 m / s 2 . What horizontal force must be applied to the cart shown in the figure in order for the blocks to remain stationary relative to the cart? Correct answer: 289 . 314 N. Explanation: Note: The blocks m 1 and m 2 being sta- tionary relative to the cart M means that they have the same non-zero horizontal accel- eration a relative to the ground. Consider the free-body diagrams. N 1 m 1 g T m 2 g T N 2 Applying Newton’s second law to the hori- zontal motion of the m 1 block yields m 1 : X F x = T = m 1 a, (1) while for the vertical motion of the m 2 block we have m 2 : X F y = T- m 2 g = 0 (2) because it accelerates horizontally but not vertically. Combining the above two Eqs. (1) & (2), and solving for the a yields a = m 2 m 1 g . (3) On the other hand, a is related to the ex- ternal force F via the equation of horizontal motionforthewholecart-plus-two-blockssys- tem: N total ( M + m 1 + m 2 ) g F In light of the above whole-system diagram, the horizontal equation is simply X F x = F = ( M + m 1 + m 2 ) a (4) and hence F = ( M + m 1 + m 2 ) × m 2 m 1 a = 289 . 314 N ....
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hw5_solns - Answer Key – Homework 5 – David McIntyre...

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