Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) / 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
Answer:
a). 8.67 x m
b).0.3011 m
c).0.0719 m
d).0.2137 N
e).1.792 N
Explanation:
Given :
Temperature of air, T = 293 K
Air Velocity, U = 5 m/s
Length of the plate is L = 6 m
Width of the plate is b = 5 m
Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X Pa-s
We know density of air is ρ = 1.21 kg /
Now we can find the Reyonld no at x = 1 m from the leading edge
Re =
Re =
Re = 332052.6
Therefore the flow is laminar.
Hence boundary layer thickness is
δ =
=
= 8.67 x m
a). Boundary layer thickness at x = 1 is δ = 8.67 X m
b). Given Re = 100000
Therefore the critical distance from the leading edge can be found by,
Re =
100000 =
x = 0.3011 m
c). Given x = 3 m from the leading edge
The Reyonld no at x = 3 m from the leading edge
Re =
Re =
Re = 996158.06
Therefore the flow is turbulent.
Therefore for a turbulent flow, boundary layer thickness is
δ =
=
= 0.0719 m
d). Distance from the leading edge upto which the flow will be laminar,
Re =
5 X =
x = 1.505 m
We know that the force acting on the plate is
=
and at x= 1.505 for a laminar flow is =
=
= 1.878 x
Therefore, =
=
= 0.2137 N
e). The flow is turbulent at the end of the plate.
Re =
=
= 1992316
Therefore =
=
= 3.95 x
Therefore =
=
= 1.792 N