# hw4b - HW#4b Problem 1 A 100 kg person stands on a scale...

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HW#4b Page 1 of 6 Problem 1. A 100 kg person stands on a scale. a.) What would be the scale readout? (How does this compare with the person's weight?) b.) If the person stands on the scale in an elevator accelerating upwards at 5 m/s 2 , what is the scale readout? (How does this compare with the person's weight?) c.) If the person stands on the scale in an elevator accelerating downwards at 5 m/s 2 , what is the scale readout? (How does this compare with the person's weight?) d.) If the person stands on the scale in an elevator accelerating downwards at 9.8 m/s 2 , what is the scale readout? (How does this compare with the person's weight?) e.) Explain why your result from part d makes sense. Is the actual weight of the person any different? What does the person perceive his weight to be? f.) If the person stands on the scale in an elevator moving up at a constant speed of 100 m/s, what is the scale readout? (How does this compare with the person's weight?) Explain why during most of the time during an elevator ride, a person will perceive nothing unusual. I’ll use m = 100 kg, for parts b-c: accelerates upwards, downwards at 5 m/s 2 A) Scale reading is the same as person’s weight (mg). B) Scale reading (1480 N) is larger than person’s actual weight (980 N). Person will feel heavier (feel more support) than they really are (because the floor is pushing up on them with a force larger than their weight). C) Scale reading (480 N) is less than person’s actual weight (980 N). Person will feel lighter than they really are (because floor is pushing up on them with a force smaller than their weight).

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HW#4b Page 2 of 6 D) Scale reading (0 N) is less than person’s actual weight (980 N). E) The scale reads zero. Accordingly, the person will “feel” weightless. It’s not that the actual weight (mg = 980 N) of the person is any different from normal. Gravity still exerts the same force on them. But now, since a y = -9.8 m/s 2 , the person and elevator are in freefall. They will fall together, and there is no danger of the person smashing the surface. (Normal force is only as large as it needs to be, so it is zero here.) The floor falls out from underneath the person as fast as the person can fall towards it, so they maintain the same distance apart without actually pressing against each other. Then N can be 0. F) Scale reading is same as person’s weight. This looks identitcal to part A, since both at rest and constant velocity are equivalent states according to Newton’s 1 st Law. For both, F net = 0. Person will “feel” their weight just as if standing at rest (not feel heavier or lighter). For most of an elevator ride the elevator is going at a constant velocity, so the person will not feel anything unusual. It is only during the short periods when the elevator is accelerating (getting up to speed from rest, and slowing to a stop) that the person feels lighter or heavier than normal.
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