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Unformatted text preview: Solution for Long-answer Homework 14 Force problems Solution to Long-answer Homework Problem 14.1(Force Pairs) Problem: You are seated in a small airplane. The child sitting next to you had been getting a bit loud, so you are letting him sit in your lap so he will be quiet. Your inertia is 60kg , the childs inertia is 15kg , and the planes inertia is 10000kg . All of a sudden, the plane hits an air pocket and begins to drop at | a | = 1 . 5m / s 2 . (Only for a short time, everyone gets out of this okay.) (a)Draw free-body diagrams for the vertical forces on the child, yourself, and the airplane, labelling the forces using the F type by,on notation introduced in the reading and in lab. Make sure the lengths of the force vectors are appropriate for the magnitudes of the forces. Make a notation if you must draw one to a different scale. (b)Identify all the forces that form interaction pairs. (c)Find the magnitudes of the forces. Show your work, or explain briefly how you obtained these magnitudes. (d)When would you feel heavier, when the plane is accelerating downward, or after it has levelled off again? Briefly explain your answer. Solution to part (a) (a) Since the child is accelerating downward at 1 . 5m / s 2 , | F c | = 15kg 1 . 5m / s 2 = 22 . 5N The size of the force of gravity is always found from: F g ec = m c g = 147N Now we put what we know into the net force equation, sum of pointy bits =ma, getting our signs from the directions on the free body diagram,- 22 . 5N = | F c yc | - | m c g | | vector F c yc | = m c g- 22 . 5N = 124 . 5N (Note that the scales of the diagrams are different. The relative scale can be determined by the forces we know should be the same size because they are pairs. The relative sizes of the vectors indicated on the diagrams are also not exact, but it is emphasized which are larger and which are smaller.) F c yc F g ec a 1 (b) Forces on You: You are accelerating downward at 1 . 5m / s 2 , so | F y | = 60kg 1 . 5m / s 2 = 90N The size of the force of gravity is always found from: | vector F g ey | = m y g = 588N By Newtons third law, vector F c cy and vector F c yc form an interaction pair, so | vector F c cy | = 124 . 5N Putting what we know into the net force equation, getting our signs from the directions on the free body diagram,- 90N = | F c py | - | F c cy | - | F g ey | | vector F c py | = | vector F c cy | + | vector F g ey | - 90N | vector F c py | = 622 . 5N a F c py F g ey F c cy (c) Forces on Plane: The plane is accelerating downward at 1 . 5m / s 2 , so |...
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