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

Answer:

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)

$a = 4.62 m/s^2$

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

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

$T_1 = (2\times 3.14)(9.81 - 1.35)$

$T_1 = 53.13 N$

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

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

$T_1 = (2\times 3.14)(9.81 + 1.35)$

$T_1 = 70.1 N$

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

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

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

$a = 4.62 m/s^2$