9.0 – Relative Equilibrium (Rectilinear Motion), Slides of Fluid Mechanics

Download 9.0 – Relative Equilibrium (Rectilinear Motion) and more Slides Fluid Mechanics in PDF only on Docsity!

ENGR. BON RYAN ANIBAN

• Rectilinear Translation (Moving Vessel)

✓ Horizontal Motion

✓ Inclined Motion

✓ Vertical Motion

• Rotation(Rotating Vessel)

✓ Open Cylindrical Vessel

✓ Closed Cylindrical Vessel

a

W

= mg

Reversed effective force

REF = ma

N

N

mg

ma

From force polygon shown

tan 𝜃 =

tan 𝜃 =

𝑎

𝑔

An open rectangular tank mounted on a truck is 5m long, 2 m wide and 2.5 m high is filled with water to a depth of 2m. (a)

What is the maximum horizontal acceleration can be imposed on tank without spilling any water and (b) determine the

accelerating force on the liquid mass? (c ) If the acceleration is increased to 6 m/s

2

, how much water is spilled out?

5 m

2 m

2.5 m

An open rectangular tank mounted on a truck is 5m long, 2 m wide and 2.5 m high is filled with water to a depth of 2m. (a)

What is the maximum horizontal acceleration can be imposed on tank without spilling any water and (b) determine the

accelerating force on the liquid mass? (c ) If the acceleration is increased to 6 m/s

2

, how much water is spilled out?

5 m

2 m

2.5 m

5 m

2 m

2.5 m

Solution (c )

2 m

2.5 m

5 m

tan 𝜃 =

tan 𝜃 =

An open rectangular tank mounted on a truck is 5m long, 2 m wide and 2.5 m high is filled with water to a depth of 2m. (a)

What is the maximum horizontal acceleration can be imposed on tank without spilling any water and (b) determine the

accelerating force on the liquid mass? (c ) If the acceleration is increased to 6 m/s

2

, how much water is spilled out?

5 m

2 m

2.5 m

Solution (c )

tan 𝜃 =

x

tan 𝜃 =

𝑥 = 4. 0875 m

𝑙𝑒𝑓𝑡

𝑙𝑒𝑓𝑡

= 10. 219 m

3

𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙

= 5 × 2 × 2 =^20 m

3

𝑠𝑝𝑖𝑙𝑙𝑒𝑑

𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙

𝑙𝑒𝑓𝑡

𝑠𝑝𝑖𝑙𝑙𝑒𝑑

= 20 m

3

− 10. 219 m

3

𝑠𝑝𝑖𝑙𝑙𝑒𝑑

= 9. 781 m

3

β

a

β

a β

a v

=a sinβ

a h

=a cosβ

W

= mg

REF V

= ma H

N

REF V

= ma v

N

mg +ma V

ma H

From force polygon shown

tan 𝜃 =

𝐻

𝑉

tan 𝜃 =

𝑎

𝑔 + 𝑎

β

a

β

a v

=a sinβ

a h

=a cosβ

W = mg

REF V

= ma H

N

REF V

= ma v

N

mg - ma V

ma H

From force traigle shown

tan 𝜃 =

𝐻

𝑉

tan 𝜃 =

𝑎

𝑔 − 𝑎

A vessel containing oil accelerated on a plane inclined 15° with the horizontal at 1.2 m/s

2

. Determine the inclination of the oil

surface when the motion is (a) upwards, and (b) donwwards

Solution (a )

tan 𝜃 =

𝐻

𝑉

tan 𝜃 =

1. 2 cos( 15 )

2. 81 + 1. 2 sin( 15 )

Solution (b )

tan 𝜃 =

𝐻

𝑉

tan 𝜃 =

1. 2 cos( 15 )

2. 81 − 1. 2 sin( 15 )

An open tank containing oil ( sp. gr. = 0.8) is moving vertically. Determine the pressure 3 m below the surface if the motion is

(a) upward with constant velocity, (b) upward with an acceleration of 8 m/s

2

, (c ) upward with a deceleration of 8 m/s

2

, (d)

downward with an acceleration of 8 m/s

2

, (e) downward with a deceleration of 8 m/s

2

Solution

The pressure at any depth “h” is

p = 9.81(0.8)(3)

(a) When velocity is constant , a = 0

p = 23.544 kPa

(b) For accelerating upward motion (use “+” with a = + 8 m/s

2 )

p = 42.744 kPa

p = γh ( 1 ±

a

g

p = 9.81(0.8)(3)( 1 +

(c) For decelerating upward motion (use “+” with a = - 8 m/s

2 )

p = 4.344 kPa

p = 9.81(0.8)(3)( 1 +

(d) For accelerating downward motion (use “-” with a = + 8 m/s

2

)

p = 4.344 kPa

p = 9.81(0.8)(3)( 1 −

(e) For decelerating downward motion (use “-” with a = - 8 m/s

2

)

p = 42.744 kPa

p = 9.81(0.8)(3)( 1 −