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 = 3.35 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 = N T2 = N (b) If the elevator moves upward instead with the same acceleration what will be the tension in the two strings? T1 = N T2 = N (c) The maximum tension the two strings can withstand is 93.1 N. What maximum upward acceleration can the elevator have without having one of the strings break? m/s2

Explanation / Answer

a)

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

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

So

T2 = mg + ma

T1 = 2mg + 2ma

With m=3.35 kg and a= -1.35 m/s^2 we have

T1 = 2( 3.35kg * (9.81m/s^2 - 1.35m/s^2)) = 56.682N

T2 = 28.341N

b)

In this case: a= +1.35 m/s^2 this gives

T1 = 2( 3.35kg * (9.81m/s^2 + 1.35m/s^2)) = 74.772N

T2 = 37.386N

c)

T1 = 2(mg + ma)

2m(g + a) <93.1N

a< 93.1N/(2*3.35kg) - 9.81m/s^2

a< 4.09 m/s^2

So the max a = 4.09 m/s2

Hope this helps :)

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