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 = 3.14 kg.(a) The elevator starts from rest and accelerates downward with a = 1.35 m/s2. What are the tensions in the two strings in newtons?T1 =________ NT2 =________N.(b) If the elevator moves upward instead with the same acceleration what will be the tension in the two strings in newtons?T1 = _______NT2 = _________N.(c) The maximum tension the two strings can withstand is 92.6 N. What maximum upward acceleration (in m/s2) can the elevator have without having one of the strings break?_______m/s2.

Answer 1

Part a)

`T_1 = 53.13 N`

`T_2 = 26.6 N`

Part b)

`T_1 = 70.1 N`

`T_2 = 35 N`

Part c)

`a = 4.62 m/s^2`

Step-by-step explanation:

Part a)

As we know that the elevator is accelerating downwards

so we have force equation for sphere A given as

`m_ag + T_2 - T_1 = m_a a`

also for second sphere we have

`m_bg - T_2 = m_b a`

from above equations we have

`(m_a + m_b)g - T_1 = (m_a + m_b) a`

`(2m)g - T_1 = (2m)a`

so we have

`T_1 = (2m)(g - a)`

`T_1 = (2* 3.14)(9.81 - 1.35)`

`T_1 = 53.13 N`

Now from other equation we have

`T_2 = m_2(g - a)`

`T_2 = 3.14(9.81 - 1.35)`

`T_2 = 26.6 N`

Part b)

Now the elevator is accelerating upwards

so we have force equation for sphere A given as

`T_1 - (m_ag + T_2) = m_a a`

also for second sphere we have

`T_2 - m_b g = m_b a`

from above equations we have

`T_1 - (m_a + m_b)g = (m_a + m_b) a`

`T_1 - 2mg = (2m)a`

so we have

`T_1 = (2m)(g + a)`

`T_1 = (2* 3.14)(9.81 + 1.35)`

`T_1 = 70.1 N`

Now from other equation we have

`T_2 = m_2(g + a)`

`T_2 = 3.14(9.81 + 1.35)`

`T_2 = 35 N`

Part c)

Now we know that maximum possible tension in the string is

T = 92.6 N

so we have

`T_1 = (2m)(g + a)`

`92.6 = 2(3.14)(9.81 + a)`

`a = 4.62 m/s^2`