# PhysicsMechanics- connected particles

#### Shah 72

##### Member
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)

#### Shah 72

##### Member
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

#### skeeter

##### Well-known member
MHB Math Helper
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

#### Shah 72

##### Member
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.