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Discussion Overview

The discussion revolves around calculating the tension in a lift cable when a crate is placed inside the lift, considering different scenarios of acceleration: upward, constant speed, and downward. The focus is on applying Newton's laws of motion to derive the tension in each case.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant proposes a formula for tension during upward acceleration: T - W - w = (M + m)a, where T is tension, W is the weight of the lift, and w is the weight of the crate.
  • Another participant calculates the tension for the upward acceleration scenario and arrives at a value of 3296 N, but notes discrepancies with textbook answers.
  • In the case of constant speed, one participant states that the tension should equal the total weight, leading to the equation T - 320g = 0, questioning why there is no acceleration.
  • For downward acceleration, the same participant presents the equation 320g - T = 320a, indicating the direction of acceleration affects the tension calculation.

Areas of Agreement / Disagreement

Participants express differing views on the calculations for tension, particularly in the cases of constant speed and upward acceleration. There is no consensus on the correct values or methods, as discrepancies with textbook answers are noted.

Contextual Notes

Some calculations appear to depend on assumptions about the values of g (acceleration due to gravity) and the interpretation of acceleration in different scenarios. The discussion does not resolve these assumptions or clarify the definitions used.

Shah 72
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A crate of mass 20 kg is put into a lift. The mass of the lift is 300kg. Find the tension in the lift cable.
a) when the lift accelerates upwards at 0.3m/s^2
b) when the lift travels at constant speed
c) when the lift accelerates downwards at 0.3m/s^2
For (a) is it T- W-w= ( M+m) a
Where T is tension of the cable, W is the weight of the lift and w is the weight of the mass.
I don't know how to calculate (b) and (c)
 
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Shah 72 said:
A crate of mass 20 kg is put into a lift. The mass of the lift is 300kg. Find the tension in the lift cable.
a) when the lift accelerates upwards at 0.3m/s^2
b) when the lift travels at constant speed
c) when the lift accelerates downwards at 0.3m/s^2
For (a) is it T- W-w= ( M+m) a
Where T is tension of the cable, W is the weight of the lift and w is the weight of the mass.
I don't know how to calculate (b) and (c)
For a I did T-3000-200= 320×0.3
Iam getting 3296N
b) I did T-3000-206= 320×0.3, I get 3302, textbook ans for (b) is 3200 and for (a) its 3300N
 
scalar equations ... all use the magnitude of acceleration

(a) $T - 320g = 320a$
$a$ is upward

(b) $T - 320g = 0$
no acceleration, why?

(c) $320g - T = 320a$
$a$ is downward
 
skeeter said:
scalar equations ... all use the magnitude of acceleration

(a) $T - 320g = 320a$
$a$ is upward

(b) $T - 320g = 0$
no acceleration, why?

(c) $320g - T = 320a$
$a$ is downward
Second case is constant speed so a=0.
 

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