Answer:
as answered in the attached file.
Explanation:
The detailed steps, derivation and appropriate differentiation is as shown in the attachment
Answer:
The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.
<u>Explanation:</u>
To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is
, we know ρ=density of fluid=1000
,
g=acceleration due to gravity=9.81
, h=height of the pool=2 m and l=length of the pool=8 m.
hydrostatic force on each wall=
= 313920 N.
<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool.
</em>
At 1 m from the ground this hydrostatic force is acted on each wall.
The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation =
=
= 1255680 N. This is 4 times 313920 N.
Answer:
Explanation:
Given data in question
mean stress = 50 MPa
amplitude stress = 225 MPa
to find out
maximum stress, stress ratio, magnitude of the stress range.
solution
we will find first maximum stress and minimum stress
and stress will be sum of (maximum +minimum stress) / 2
so for stress 50 MPa and 225 MPa
=
+
/ 2
50 =
+
/ 2 ...........1
and
225 =
+
/ 2 ...........2
from eqution 1 and 2 we get maximum and minimum stress
= 275 MPa ............3
and
= -175 MPa ............4
In 2nd part we stress ratio is will compute by ratio of equation 3 and 4
we get ratio =
/
ratio = -175 / 227
ratio = -0.64
now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.
magnitude =
-
magnitude = 275 - (-175) = 450 MPa