Sphere A is attached to the ceiling of an elevator by a string. A second sphere

Question

Sphere A is attached to the ceiling of an elevator by a string. A second sphere is attached to the first one by a second string. Both strings are of negligible mass. Here m1 = m2 = m = 3.28 kg.

(a) The elevator starts from rest and accelerates downward with a= 1.35 m/s2. What are the tensions in the two strings?

T1 =

T2 =

(b) If the elevator moves upward instead with the same acceleration what will be the tension in the two strings?

T1 =

T2 =

(c) The maximum tension the two strings can withstand is 93.9 N. What maximum upward acceleration can the elevator have without having one of the strings break?
m/s2

T1 =

T2 =

Explanation / Answer

The total force on the top sphere, taking up as positive, is

F = (T1 - T2 - mg)

so Newton's second la applied to that sphere is

T1 - T2 - mg = m a

For the second (lowest) sphere we have

T2 - m g = m a

T2 = mg + ma

T1 = 2mg + 2ma

we have m = 3.28 kg and a = - 1.35 m/s^2

A)

T1 = 2*(3.28*(9.81 - 1.35)) = 55.49 N
T2 = 27.75 N

B)

now a = +1.35 m/s^2 this gives

T1 = 2*(3.28*(9.81 + 1.35)) = 73.2 N

T2 = 36.6 N

C)

T1 = 2(mg + ma)

The maximum tension the strings can withstand is 93.9 N.

2(mg + ma) < 93.9 N

a < 93.9 / (2*m) - g

a < 93.9 /(2*3.28) - 9.81

a < 4.5 m/s^2

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