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# examf_sol - PHY 3221 Mechanics I Fall Term 2010 Final Exam...

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Unformatted text preview: PHY 3221: Mechanics I Fall Term 2010 Final Exam, December 14, 2010 • This is a closed book exam lasting 90 minutes. • Since calculators are not allowed on this test, if the problem asks for a numerical answer, answering 2 + 2 is as good as 4, and √ 2 is as good as 1 . 4142 ... . • There are six equally weighted problems worth a total of 30 points. The problems appear on the second and third page of this test. Begin each problem on a fresh sheet of paper. Use only one side of the paper. Avoid microscopic handwriting. • Put your name, the problem number and the page number in the upper right-hand corner of each sheet. • To receive partial credit you must explain what you are doing. Carefully labelled figures are important! Randomly scrawled equations are not helpful. • Draw a box around important results (or at least results which you think might be im- portant). • Good luck! 1 Problem 1. [5 pts] You use a rope to pull a crate across a level floor. The maximum tension that the rope can have without breaking is T max , and the coefficient of kinetic friction is μ k . Suppose that you tug on the rope so that it makes an angle θ with the horizontal (refer to Fig. 1). Find the maximum possible mass M max of the crate that you can pull at constant speed without breaking the rope, and the optimal angle θ opt that will allow you to do that. Hint: make sure to draw a force diagram and clearly label all forces. M θ vectorg μ k x y Figure 1: An illustration for the crate pulling problem. Solution. The fources which act on the crate are: gravity G = Mg , tension T , normal reaction N and friction F = μ k N . When we pull the crate at constant speed, the net force is zero. In components: x : − F + T x = − μ k N + T cos θ = 0 , y : − G + T y + N = − Mg + T sin θ + N = 0 . Eliminate N N = T cos θ μ k − Mg + T sin θ + T cos θ μ k = 0 and solve for M : M = T μ k g (cos θ + μ k sin θ ) Obviously, the mass will be maximum when T = T max . To find the corresponding optimum angle θ opt , find the extremum of M ( θ ): dM dθ = 0 = ⇒ − sin θ opt + μ k cos θ opt = 0 = ⇒ tan θ opt = μ k = ⇒ θ opt = tan − 1 μ k Explicitly: sin θ opt = μ k radicalBig 1 + μ 2 k , cos θ opt = 1 radicalBig 1 + μ 2 k 2 and M max = T max μ k g (cos θ opt + μ k sin θ opt ) = T max μ k g radicalBig 1 + μ 2 k Problem 2. [5 pts.] Obtain the Fourier expansion of the periodic function F ( t ) = +1 , if −...
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## This note was uploaded on 12/15/2011 for the course PHY 3221 taught by Professor Chan during the Spring '08 term at University of Florida.

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examf_sol - PHY 3221 Mechanics I Fall Term 2010 Final Exam...

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