Apparent Weight.? A 550-N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 850 kg. As the elevator starts moving, the scale reads 450 N. (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads 670 N? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?
Solution 20E Step 1: Weight of the student on a bathroom scale W 1 = 550 N Combined mass of student and elevator M = 850 kg Problem (a) As the elevator starts moving, find the acceleration of the elevator Step 1: W Mass of the student (m) = g 1 = 59.8 = 56.12 kg Step 2: R = 450 N R = 450 N < W = 550 N 1 Apparent weight of the student is less than the actual weight Hence the elevator moves downward The reaction R = m(g - a ) Step 3: a = g - R m a = 9.8 - 450 56.12 2 Magnitude of acceleration is (a) = 1.78 m/s Direction is downward Problem (b) Step 1: R = 670 N R = 670 N > W 1 = 550 N Apparent weight of the student is greater than the actual weight Hence the elevator moves upward The reaction R = m(g + a ) Step 2: a = R - g m a = 670 - 9.8 56.12 2 Magnitude of acceleration is (a) = 2.13 m/s Direction is upward Problem (c) If the scale reads zero, R = 0 N This shows that a = g, the elevator moves at the rate of free fall, hence the student should worry about this. In this case the the student feels the weightlessness. Problem (d) Step 1: Tension in the cable when a = 1.78 m/s 2 Tension ( T ) = M ( a + g ) M = 850 kg T = 850 ( 1.78 + 9.8 ) T = 9843 N Step 2: Tension in the cable when a = g Here the acceleration is downward, therefore a = -9.8 m/s 2 T = 850 ( -9.8 + 9.8 ) T = 0 The tension in the cable is zero when the elevator moves downward at the acceleration = g. Problem (d) Tension in the cable Step 1: 2 When a = -1.78 m/s T = M ( a+g) Where M is combined mass T = 850 * (-1.78 + 9.8) T = 6817 N