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
AI Thread 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
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...

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
Back
Top