MHB What are the tensions in the strings of an accelerating elevator?

  • Thread starter Thread starter cbarker1
  • Start date Start date
  • Tags Tags
    Elevator Tension
Click For Summary
In an accelerating elevator scenario with two spheres connected by strings, the tensions in the strings depend on the direction and magnitude of the elevator's acceleration. When the elevator accelerates downward at 1.45 m/s², the equations of motion reveal the tensions in the strings can be calculated using the gravitational force and the net acceleration. Conversely, when the elevator accelerates upward with the same acceleration, different equations yield the new tensions. Additionally, the maximum tension the strings can withstand is 80.0 N, which limits the maximum upward acceleration of the elevator to prevent breaking the strings. Understanding these dynamics is crucial for solving the tension equations effectively.
cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
Dear Everyone,

A sphere 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 $m_1=m_2=m=3.52\text{ kg}$.

4-p-064.gif


(a) The elevator starts from rest and accelerates downward with $a=1.45\,\dfrac{\text{m}}{\text{s}^2}$. What are the tensions in the two strings?

(b) If the elevator starts from rest and accelerates upward with the same acceleration, what will be the tension in the two strings?

(c) The maximum tension the two strings can withstand is 80.0 N. What maximum upward acceleration can the elevator have without having one of the strings break?

I would need to some help to setup and find the value of Tension of the cable one.

Thanks,

Cbarker1
 
Mathematics news on Phys.org
(a) ...

$m_2g - T_2 = m_2 a$

$m_1g+T_2-T_1 = m_1 a$

------------------------------ sum the two scalar equations ...

$(m_1+m_2)g - T_1 = (m_1+m_2)a$

$(m_1+m_2)g - (m_1+m_2)a = T_1$

evaluate $T_1$, then determine $T_2$(b) ...

$T_1 - (m_1g + T_2) = m_1a$

$T_2 - m_2g = m_2a$

same drill ...

I'll leave part (c) for you to try
 

Similar threads

Replies
2
Views
2K
Replies
1
Views
2K
Replies
12
Views
1K
Replies
10
Views
4K
Replies
10
Views
2K
Replies
10
Views
6K
Replies
8
Views
2K