MHB Mechanics- connected particles

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To find the tension in the lift cable for a crate and lift system, the calculations vary based on the lift's motion. When the lift accelerates upwards at 0.3 m/s², the tension is calculated using T - (weight of lift + weight of crate) = (total mass) * acceleration, resulting in approximately 3300 N. For constant speed, the tension equals the total weight of the lift and crate, which is around 3200 N, as there is no acceleration. When the lift accelerates downwards at 0.3 m/s², the equation reverses, leading to a different tension value. Understanding these dynamics is crucial for accurate calculations in mechanics involving connected particles.
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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