UT PHY 317K - HW02-solutions (7 pages)

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HW02-solutions



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HW02-solutions

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7
School:
University of Texas at Austin
Course:
Phy 317k - General Physics I
General Physics I Documents
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gilbert hlg574 HW02 tsoi 55730 This print out should have 18 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 part 1 of 2 10 0 points An elevator accelerates upward at 1 2 m s2 The acceleration of gravity is 9 8 m s2 What is the upward force exerted by the floor of the elevator on a n 65 kg passenger Correct answer 715 N Explanation 1 590 N 1 73 0N 601 N a What is the angle 1 which the righthand cable makes with respect to the ceiling Round up to four significant digits Express unit as degrees Correct answer 38 4808 degrees a Explanation Observe the free body diagram below N mg When the elevator is accelerating upward Fnet m a N m g N m a m g 002 part 2 of 2 10 0 points If the same elevator accelerates downwards with an acceleration of 1 2 m s2 what is the upward force exerted by the elevator floor on the passenger Correct answer 559 N Explanation a N mg When the elevator is accelerating downward Fnet m a m g N2 N2 m g m a 003 part 1 of 2 10 0 points Consider the 601 N weight held by two cables shown below The left hand cable had tension 590 N and makes an angle of with the wall The right hand cable had tension 730 N and makes an angle of 1 with the ceiling 2 F2 1 F1 Wg Note The sum of the x and y components of F1 F2 and Wg are equal to zero Given Wg 601 N F1 730 N 1 38 4808 2 90 Basic Concepts X Fx 0 F1x F1 cos 1 F12 cos2 1 X and Fy and F2x F2 cos 2 F22 cos2 2 1 2 0 F1y F2y F3y 0 F1 sin 1 F2 sin 2 F3 0 F1 sin 1 F2 sin 2 F3 F12 sin2 1 F22 sin2 2 2 F2 F3 sin 2 F32 since 3 F3 sin 3 F3 sin 270 F3 and F3 cos 3 F3 cos 270 0 gilbert hlg574 HW02 tsoi 55730 2 Solution Since sin2 cos2 1 and adding Eqs 2 and 3 we have F22 F12 2 F1 F3 sin 1 F32 F 2 F32 F22 sin 1 1 2 F1 F3 2 F3 F12 F22 1 arcsin 2 F1 F3 601 N 2 730 N 2 arcsin 2 730 N 601 N 590 N 2 2 730 N 601 N 38 4808 Why doesn t the rope break when he is supported as shown at the left above To answer this find the tension in the rope Correct answer 355 N 004 part 2 of 2 10 0 points b What is the angle which the left hand cable makes with respect to the wall Round up to four significant digits Express unit as degrees Correct answer 75 2391 degrees Explanation Using Eq 1 we have F1 cos 1 F2 F1 2 arccos cos 1 F2 730 N cos 38 4808 arccos 590 N 14 7609 90 2 90 14 7609 75 2391 cos 2 005 part 1 of 2 10 0 points Harry the painter swings year after year from his bosun s chair His weight is 710 N and the rope unknown to him has a breaking point of 405 N Explanation In the left figure Harry is supported by two strands of rope that share his weight so each strand supports only 355 N below the breaking point Total upward force supplied by the ropes equals weight acting downward giving a net force of zero and no acceleration 006 part 2 of 2 10 0 points One day Harry is painting near a flagpole and for a change he ties the free end of the rope to the flagpole instead of to his chair as shown at the right Why did Harry end up taking his vacation early To answer this find the tension in the rope Correct answer 710 N Explanation In the right figure Harry is now supported by one strand requiring the tension to be 710 N Since this is above the breaking point of the rope it breaks 007 part 1 of 5 10 0 points Given The pulley is massless and frictionless and attached to the ceiling A massless inextensible string is attached to the masses Mb and Ma where Mb Ma The tensions Ty Tz Tx and the gravitational constant g are magnitudes T Tz 3 T gilbert hlg574 HW02 tsoi 55730 Tx Mb Ma Mb g Ty Mb Ma g a Ma Consider the following possibilities and select the correct choice a R Since the larger mass will move down and the smaller mass up we can take motion downward as positive for Mb and motion upward as positive for Ma Apply Newton s second law to Ma and Mb respectively and then combine the results we have for mass Ma X Fx T Ma g Ma a 2 1 Ty Mb g and Tx Ma g 2 Ty Mb g and Tx Ma g 3 Ty Mb g and Tx Ma g 4 Ty Mb g and Tx Ma g 5 Ty Mb g and Tx Ma g correct 6 Ty Mb g and Tx Ma g 7 Ty Mb g and Tx Ma g 8 Ty Mb g and Tx Ma g 1 Ty Tx 9 Ty Mb g and Tx Ma g 2 Ty Tx correct T Ma g a or T Ma g and for mass Mb X Fy Mb g T Mb a 3 T Mb g a or T Mb g 008 part 2 of 5 10 0 points Consider the following possibilities and select the correct choice 3 Ty Tx Explanation The magnitude of the tension in the string is the same at all points along the string Explanation See Eq 1 in Part 1 009 part 3 of 5 10 0 points Consider the following possibilities and select the correct choice T Ty Tx 1 1 Tz Mb g Ma g correct 2 Tz Mb g Ma g Consider the free body diagrams 3 Tz Mb g Ma g gilbert hlg574 HW02 tsoi 55730 Explanation Using Eq 1 in Part 1 we have Two masses m1 and m2 are connected in the manner shown T1 Tz Ty Tx Mb g a Ma g a Mb g Ma g Mb Ma a Mb g Ma g 010 part 4 of 5 10 0 points Consider the following possibilities and select the correct choice 1 Ty Tx Tz 2 Ty Tx Tz 3 Ty Tx Tz correct Explanation Visualize a free body diagram for the pulley The forces are in equilibrium so Tz Ty Tx 2 T 4 m1 a T2 g 2 m2 The system is accelerating downward with g acceleration of magnitude 2 Determine T2 1 1 T2 m2 m1 g 2 1 2 T2 m2 m1 g 2 1 3 T2 m2 m1 g 2 3 4 T2 m2 g 2 1 5 T2 m2 g correct 2 6 T2 m2 …


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