Can anyone please solve this......An elevator and it's load weigh a total of 1600 lb. Find the tension T in the supporting Cable when the elevator,originally moving downwards at 20ft/sec,is brought to rest with constant acceleration in a distance of 50ft.
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Answered by
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Here,
m = 800 kg
u = 20 ms
v = 0 ms
s = 50 m
g = 9.8 ms-2
u2 = -2as
=> 202 = -2a×50
=> a = - 4 ms-2
Now the acceleration acting upwards by the cable = 4 ms-2
So, the elevator experiences an downward acceleration = 4
Net acceleration downwards = 9.8 + 4 = 13.8 ms-2
Tension in the cable will be 800×13.8 = 11040 N
@skb
m = 800 kg
u = 20 ms
v = 0 ms
s = 50 m
g = 9.8 ms-2
u2 = -2as
=> 202 = -2a×50
=> a = - 4 ms-2
Now the acceleration acting upwards by the cable = 4 ms-2
So, the elevator experiences an downward acceleration = 4
Net acceleration downwards = 9.8 + 4 = 13.8 ms-2
Tension in the cable will be 800×13.8 = 11040 N
@skb
Answered by
0
Answer:
T=1800lb
Explanation:
V=0
U=20 ft/s
a=?
Δx=50ft
V²=U²-2aΔx
0=20²+2a(50)
a=-4ft/s²
T-mg=-ma
T=m(g-a)
T=1600(32-(-4))
T=1800lb
answer is T=1800lb
#SPJ2
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