Respuesta :
Answer:
(a) [tex]318.471lbs/inch^2[/tex]
(b) [tex]2.196mega\ pascal[/tex]
Explanation:
We have given load = 1000 lbs
And diameter of the rope d = 2 inches
So radius [tex]r=\frac{d}{2}=\frac{2}{2}=1inch[/tex]
Area [tex]a=\pi r^2=3.14\times 1^2=3.14inch^2[/tex]
(a) Tension [tex]T=\frac{load}{area }=\frac{1000}{3.14}=318.471lbs/inch^2[/tex]
(b) We know that 1 lbs = 4.4482 Newton
So 1000 lbs = 1000×4.4482 = 4448.8 N
And [tex]1inch^2=0.000645m^2[/tex]
So 3.14[tex]inch^2[/tex] = 3.14×0.000645 =[tex]2.0253\times 10^{-3}m^2[/tex]
So [tex]T=\frac{load}{area }=\frac{4448.8N}{2.0253\times 10^{-3}m^2}=2196.61\times 10^3N/m^2=2196.61kPa[/tex] =[tex]2.196mega\ pascal[/tex]
Answer:
a)[tex]T=318.3\ lb/in^2[/tex]
b)T= 2.1 MPa
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
[tex]T=\dfrac{P}{\dfrac{\pi d^2}{4}}[/tex]
a)
[tex]T=\dfrac{P}{\dfrac{\pi d^2}{4}}[/tex]
[tex]T=\dfrac{ 1000}{\dfrac{\pi \times 2^2}{4}}[/tex]
[tex]T=318.3\ lb/in^2[/tex]
b)
We know that
[tex]1\ lb/in^2=6894.76\ N/m^2 [/tex]
[tex]318.3\ lb/in^2=2.1\times 10^6\ N/m^2 [/tex]
[tex]1\ MPa=10^6\ Pa[/tex]
So T= 2.1 MPa
