Question

Calculate the tension in a rope with a diameter of 2 inches and a hanging load of 1000 lbs in: a. lbs/sq. in. b. mega pascal

Answer 1

a)
`T=318.3\ lb/in^2`

b)T= 2.1 MPa

Step-by-step explanation:

Given that

Diameter ,d= 2 in

Load ,P= 1000 lbs

We know that

1 lb = 4.45 N

1 in = 0.0254 m

P= 4448.22 N

So tension in rope T

`T=(P)/((\pi d^2)/(4))`

a)

`T=(P)/((\pi d^2)/(4))`

`T=( 1000)/((\pi * 2^2)/(4))`

`T=318.3\ lb/in^2`

b)

We know that

`1\ lb/in^2=6894.76\ N/m^2`

`318.3\ lb/in^2=2.1* 10^6\ N/m^2`

`1\ MPa=10^6\ Pa`

So T= 2.1 MPa

Answer 2

(a)
`318.471lbs/inch^2`

(b)
`2.196mega\ pascal`

Step-by-step explanation:

We have given load = 1000 lbs

And diameter of the rope d = 2 inches

So radius
`r=(d)/(2)=(2)/(2)=1inch`

Area
`a=\pi r^2=3.14* 1^2=3.14inch^2`

(a) Tension
`T=(load)/(area )=(1000)/(3.14)=318.471lbs/inch^2`

(b) We know that 1 lbs = 4.4482 Newton

So 1000 lbs = 1000×4.4482 = 4448.8 N

And
`1inch^2=0.000645m^2`

So 3.14
`inch^2` = 3.14×0.000645 =
`2.0253* 10^(-3)m^2`

So
`T=(load)/(area )=(4448.8N)/(2.0253* 10^(-3)m^2)=2196.61* 10^3N/m^2=2196.61kPa` =
`2.196mega\ pascal`