Question

the maximum stress that can be applied to the material of a wire used to suspend an elevator is 1.3×10^8 N/m square.if mass of the elevator is 900 kg .and it moves up eith an acceleration of 2.2m/s square what is the minimum diameter of the wire​

Answer 1

Answer:

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Answer 2

Answer:

S

max

=

π

3

×10

8

N/m

2

m=900kg

ma=T−mg

900×2.2=S

m

×A−900×10

S

m

.A=900×12.2

A

max

=

π

3

×10

8

900×12.2

πr

min

2

=

π

3

×10

8

900×12.2

r

min

2

=12.2×3×10

−6

r

min

=6×10

−3

r

min

=6mm