A 568-N physics student stands on a bathroom scale in an 884-kg (including the s
ID: 1594483 • Letter: A
Question
A 568-N physics student stands on a bathroom scale in an 884-kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 467 N .
1) Find the magnitude of the acceleration of the elevator.
2) Find the direction of the acceleration of the elevator.
3) What is the acceleration if the scale reads 601 N ?
4) If the scale reads zero, should the student worry? Explain.
5) What is the tension in the cable in part A?
6) What is the tension in the cable in part D?
Explanation / Answer
we know, the elevator always gives normal force reading.
mass of the student, m = W/g
= 568/9.8
= 57.96 kg
1) N = m*g - m*a
==> m*a = m*g - N
a = g - N/m
= 9.8 - 467/57.96
= 1.74 m/s^2
2) downward
3) N = m*g + m*a
==> m*a = N - m*g
a = N/m - g
= 601/57.96 - 9.8
= 0.57 m/s^2
4) upward
5) T = (m + M)*(g-a)
= (57.96 + 884)*(9.8 - 1.74)
= 7592.2 N
6) T = (m + M)*(g+a)
= (57.96 + 884)*(9.8 + 1.74)
= 10870.2 N
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