MHB Mechanics- connected particles

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SUMMARY

The discussion focuses on calculating the tension in a lift cable under various conditions involving a crate of mass 20 kg and a lift of mass 300 kg. For upward acceleration at 0.3 m/s², the tension is calculated as T = (M + m) * a + W + w, resulting in 3300 N. When the lift travels at constant speed, the tension equals the total weight, yielding 3200 N. For downward acceleration at 0.3 m/s², the tension is determined using T = 320g - 320a, leading to a tension of 3296 N.

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  • Understanding of Newton's Second Law of Motion
  • Basic knowledge of forces, specifically weight and tension
  • Ability to perform calculations involving acceleration and mass
  • Familiarity with scalar equations in physics
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  • Study the derivation of tension in varying acceleration scenarios
  • Learn about the implications of constant speed on force calculations
  • Explore real-world applications of tension in mechanical systems
  • Investigate the effects of varying mass on tension in lift systems
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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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