Calculate Tension in Cord: Elevator Accelerates

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Homework Help Overview

The discussion revolves around calculating the tension in a cord supporting a bundle in an accelerating elevator. The elevator accelerates upward, and the forces acting on the bundle include gravity and tension in the cord. The problem involves understanding the dynamics of the system and applying Newton's laws.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the bundle, including gravity and tension. They explore the relationship between these forces and the acceleration of the elevator. Questions arise about how to set up the equations of motion and the vector sum of forces.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify the roles of tension and gravitational force. Some guidance has been offered regarding the vector sum of forces and the application of Newton's second law, though no consensus has been reached on the final calculation of tension.

Contextual Notes

There is a focus on understanding the forces involved without providing a complete solution. Participants are also discussing a related multiple-choice problem involving forces and acceleration, which adds complexity to the overall discussion.

anglum
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Tension in the cord?

An elevator starts from rest with a constant acceleration upward and moves 1m in 1.8s
A passenger in the elevator is holding a 9.7kg bundle at the end of a vertical cord
Acceleration of gravity is 9.8m/s squared
What is the tension in the cord as the elevator accelerates? answer in units of N

so i calculated the acceleration to be .55555m/s squared upward

and the downward acceleration is 9.8m/s squared

and the mass is 9.7 kg

so where do i go from here
 
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What are the two forces acting on the object hanging by the cord?

you know the acceleration of this object is 0.5555m/s^2 upward.
 
and gravity is 9.8m/s^2 downward
 
so those are the 2 forces acting on the object
 
anglum said:
so those are the 2 forces acting on the object

gravity and?
 
weight of the object
 
so the tension on the rope is the difference between .5555 and 9.8*9.7kg?
 
anglum said:
weight of the object

that's gravity... you have the weight of the object, and the tension in the cord... so what is the [tex]\Sigma{F} = ma[/tex] equation for the bundle?
 
anglum said:
so the tension on the rope is the difference between .5555 and 9.8*9.7kg?

sum of all forces = ma

What goes in the left side of the above equation? What goes in the right side?
 
  • #10
the bundle force is 95.06N? and then what do i do with the acceleration?
 
  • #11
ok i think i got it...

the upward is .555555* 9.7

and the downward is 9.8*9.7

?
 
  • #12
anglum said:
ok i think i got it...

the upward is .555555* 9.7

That is not one of the individual forces acting on the cord...

and the downward is 9.8*9.7

?

Remember, the vector sum of all the individual forces = ma

There are two individual forces here. Tension and gravity. Let tension = T. So what is the vector sum of tension and gravity?
 
  • #13
the vector sum equals 0?

so the gravity is 95.06 and the tension on the cord is -95.06
 
  • #14
anglum said:
the vector sum equals 0?

so the gravity is 95.06 and the tension on the cord is -95.06

no.

just using T and mg... write out the vector sum arithmetically. don't worry about the ma for now... what is the vector sum of the tension and mg, written in terms of T and mg...
 
  • #15
vector sum = T + MG
 
  • #16
anglum said:
vector sum = T + MG

it should be T - mg. because tension acts upward... mg acts downwards...

So T - mg = ma

remember... all force problems are like this... the vector sum of all the individual forces = ma... you shouldn't think of ma as another force. All the individual forces add up to ma...
 
  • #17
ok so ... then T = MA + MG

and in this problem T = (9.7)(.5555)+(9.7)(9.8)

real fast problem that is multiple choice

imagine a Y

with the intersecting point being a box
and the downward part bein 70N and the two upward ones bein 35N

will the box experience acceleration?
a - unable to determine without the angle
b- Yes; upwards
c-Yes;downwards
d- no it is balanced

i say the answer is D they are balanced
 
Last edited:
  • #18
after taht multiple choice one i have 3 left... mind helpin me?
 
  • #19
anglum said:
ok so ... then T = MA + MG

and in this problem T = (9.7)(.5555)+(9.7)(9.8)

real fast problem that is multiple choice

imagine a Y

with the intersecting point being a box
and the downward part bein 70N and the two upward ones bein 35N

will the box experience acceleration?
a - unable to determine without the angle
b- Yes; upwards
c-Yes;downwards
d- no it is balanced

i say the answer is D they are balanced

why do you say it is balanced? What is the net force in the vertical direction?
 
  • #20
anglum said:
after taht multiple choice one i have 3 left... mind helpin me?

sure. np. my computer keeps freezing, so I may be gone for short periods of time... but I'll do my best.
 
  • #21
oooo it is not balanced since the 35N is not the vertical... we arent able to determine it without the angle then correct? its either that or downward acceleration

since it can't be balance and can't be upward
 
  • #22
i think it has to be downward becuz there is no way the vertical component can equal 70 if the angles comin of the Y are bothe givin 35N in an angle and not straight up
 
  • #23
anglum said:
oooo it is not balanced since the 35N is not the vertical... we arent able to determine it without the angle then correct? its either that or downward acceleration

since it can't be balance and can't be upward

yeah, I think it is unbalanced downward. I think that although they haven't given the angle, we can presume it is not vertical...
 

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