How to Calculate and Solve for Pouring Speed | Design of Gating System

The image above represents pouring speed.

To compute for poring speed, three essential parameters are needed and these parameters are Co-efficient for friction (μ), Acceleration due to Gravity (g) and Rated head static (Hs).

The formula for calculating pouring speed:

v = μ√(2gHs)

Where:

v = Pouring Speed
μ = Co-efficient for Friction
g = Acceleration due to Gravity
Hs = Rated Static Head

Let’s solve an example;
Find the pouring speed when the co-efficient for friction is 14, acceleration due to gravity is 9 and rated static head is 16.

This implies that;

μ = Co-efficient for Friction = 14
g = Acceleration due to Gravity = 9
Hs = Rated Static Head = 16

v = μ√(2gHs)
v = 14 x √(2 x 9 x 16)
v = 14 x √(288)
v = 14 x 16.97
v = 237.58

Therefore, the pouring speed is 237.58 m/s.

Calculating the Co-efficient for Friction when the Pouring Speed, the Acceleration due to Gravity and the Rated Static Head is Given.

μ = v / √(2gHs)

Where;

μ = Co-efficient for Friction
v = Pouring Speed
g = Acceleration due to Gravity
Hs = Rated Static Head

Let’s solve an example;
Find the co-efficient for friction when the pouring speed is 24, the acceleration due to gravity is 9 and the rated static head is 11.

This implies that;

v = Pouring Speed = 24
g = Acceleration due to Gravity = 9
Hs = Rated Static Head = 11

μ = v / √(2gHs)
μ = 24 / √(2 x 9 x 11)
μ = 24 / √198
μ = 24 / 14.07
μ = 1.70

Therefore, the co-efficient for friction is 1.70.

Continue reading How to Calculate and Solve for Pouring Speed | Design of Gating System

How to Calculate and Solve for Reaction: Lift Falls Freely | Motion

The image above represents reaction: lift falls freely.

To compute for reaction: lift falls freely, two essential parameters are needed and these parameters are mass (m) and acceleration due to gravity (g).

The formula for calculating reaction: lift falls freely:

R = mg

Where;

R = Reaction
m = Mass
g = acceleration due to gravity

Let’s solve an example;
Find the reaction when the mass is 30 and acceleration due to gravity is 9.8.

This implies that;

m = Mass = 30
g = acceleration due to gravity = 9.8

R = mg
R = 30 x 9.8
R = 294

Therefore, the reaction is 294 N.

Calculating the Mass when the Reaction and the Acceleration due to Gravity is Given.

m = R / g

Where;

m = Mass
R = Reaction
g = acceleration due to gravity

Let’s solve an example;
Find the mass when the reaction is 40 and the acceleration due to gravity is 9.8

This implies that;

R = Reaction = 40
g = acceleration due to gravity = 9.8

m = R / g
m = 40 / 9.8
m = 40 / 9.8
m = 4.08

Therefore, the mass is 4.08.

Continue reading How to Calculate and Solve for Reaction: Lift Falls Freely | Motion

How to Calculate and Solve for Reaction: Lift Moves Down | Motion

The image above represents reaction: lift moves down.

To compute for reaction: lift moves down, three essential parameters are needed and these parameters are mass (m), acceleration (a) and acceleration due to gravity (g).

The formula for calculating reaction: lift moves down:

R = m(g – a)

Where;

R = Reaction
m = Mass
g = acceleration due to gravity
a = Acceleration

Let’s solve an example;
Find the reaction when the mass is 28, acceleration is 9 and acceleration due to gravity is 9.8.

This implies that;

m = Mass = 28
g = acceleration due to gravity = 9.8
a = Acceleration = 9

R = m(g – a)
R = 28(9.8 – 9)
R = 28(0.80)
R = 22.4

Therefore, the reaction is 22.4 N.

Calculating the Mass when the Reaction, the Acceleration and the Acceleration due to Gravity is Given.

m = R / g – a

Where;

m = Mass
R = Reaction
g = acceleration due to gravity
a = Acceleration

Let’s solve an example;
Find the mass when the reaction is 42, the acceleration is 8 and the acceleration due to gravity is 9.8

This implies that;

R = Reaction = 42
g = acceleration due to gravity = 9.8
a = Acceleration = 8

m = R / g – a
m = 42 / 9.8 + 8
m = 42 / 17.8
m = 2.359

Therefore, the mass is 2.359.

Continue reading How to Calculate and Solve for Reaction: Lift Moves Down | Motion

How to Calculate and Solve for Reaction: Lift Moves Up | Motion

The image above represents reaction: lift moves up.

To compute for reaction: lift moves up, three essential parameters are needed and these parameters are mass (m), acceleration (a) and acceleration due to gravity (g).

The formula for calculating reaction: lift moves up:

R = m(a + g)

Where;

R = Reaction
m = Mass
g = acceleration due to gravity
a = Acceleration

Let’s solve an example;
Find the reaction when the mass is 18, acceleration is 21 and acceleration due to gravity is 9.8.

This implies that;

m = Mass = 18
g = acceleration due to gravity = 21
a = Acceleration = 9.8

R = m(a + g)
R = 18(21 + 9.8)
R = 18(30.8)
R = 554.4

Therefore, the reaction is 554.4 N.

Calculating the Mass when the Reaction, the Acceleration and the Acceleration due to Gravity is Given.

m = R / a + g

Where;

m = Mass
R = Reaction
g = acceleration due to gravity
a = Acceleration

Let’s solve an example;
Find the mass when the reaction is 42, the acceleration is 21 and the acceleration due to gravity is 9.8

This implies that;

R = Reaction = 42
g = acceleration due to gravity = 9.8
a = Acceleration = 21

m = R / a + g
m = 42 / 21 + 9.8
m = 42 / 30.8
m = 1.36

Therefore, the mass is 1.36.

Continue reading How to Calculate and Solve for Reaction: Lift Moves Up | Motion

How to Calculate and Solve for Period | Motion

The image above represents period.

To compute for period, two essential parameters are needed and these parameters are length (l) and acceleration due to gravity (g).

The formula for calculating period:

T = 2π(√(l / g))

Where;

T = Period
l = Length
g = Acceleration due to gravity

Let’s solve an example;
Find the period when the length is 4 and the acceleration due to gravity is 9.8.

This implies that;

l = Length = 4
g = Acceleration due to gravity = 9.8

T = 2π(√(l / g))
T = 2π(√(4 / 9.8))
T = 2π(√(0.408))
T = 2π(0.6388)
T = 6.28 x 0.6388
T = 4.014

Therefore, the period is 4.014 s.

Calculating the Length when the Period and the Acceleration due to Gravity is Given.

l = (T / )2 x g

Where;

l = Length
T = Period
g = Acceleration due to gravity

Let’s solve an example;
Find the length when the period is 28 and the acceleration due to gravity is 9.8.

This implies that;

T = Period = 28
g = Acceleration due to gravity = 9.8

l = (T / )2 x g
l = (28 / 6.28)2 x 9.8
l = (4.458)2 x 9.8
l = 19.87 x 9.8
l = 194.7

Therefore, the length is 194.7.

Continue reading How to Calculate and Solve for Period | Motion

How to Calculate and Solve for Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image above represents the maximum velocity to avoid overturning of a vehicle moving along a level circular path.

To compute for the maximum velocity, four essential parameters are needed and these parameters are Acceleration due to Gravity (g), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the maximum velocity:

vmax = √(gra / h)

Where:
vmax = Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path
g = Acceleration due to Gravity
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the maximum velocity when the Acceleration due to Gravity (g) is 10.2, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 14, Radius of Circular Path (r) is 22 and Half of the Distance between the Centre Lines of the Wheel (a) is 32.

This implies that;
g = Acceleration due to Gravity = 10.2
h = Height of Centre of Gravity of the Vehicle from Ground Level = 14
r = Radius of Circular Path = 22
a = Half of the Distance between the Centre Lines of the Wheel = 32

vmax = √(gra / h)
vmax = √((10.2)(22)(32)/14)
vmax = √((7180.79)/14)
vmax = √(512.91)
vmax = 22.647

Therefore, the maximum velocity to avoid Overturning of a Vehicle moving along a Level Circular Path is 22.647 m/s.

Continue reading How to Calculate and Solve for Maximum Velocity to avoid Overturning of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for the Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

The image represents reaction at the inner wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the inner wheel of a vehicle moving along a level circular path:

RA = mg / 2[1 – v²h / gra]

Where:
RA = Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 13, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 11, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 5, Radius of Circular Path (r) is 7 and Half of the Distance between the Centre Lines of the Wheel (a) is 3.

This implies that;
m = Mass of the Vechicle = 13
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 11
h = Height of Centre of Gravity of the Vehicle from Ground Level = 5
r = Radius of Circular Path = 7
a = Half of the Distance between the Centre Lines of the Wheel = 3

RA = mg / 2[1 – v²h / gra]
RA = 13(9.8) / 2[1 – (11)²(5) / (9.8)(7)(3)]
RA = 127.4 / 2[1 – (121)(5) / 205.8]
RA = 63.7[1 – 605 / 205.8]
RA = 63.7[1 – 2.939]
RA = 63.7[-1.939]
RA = -123.56

Therefore, the reaction at the inner wheel of a vehicle moving along a level of circular path is -123.56 N.

Continue reading How to Calculate and Solve for the Reaction at the Inner Wheel of a Vehicle moving along a Level Circular Path | The Calculator Encyclopedia

How to Calculate and Solve for the Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path | Nickzom Calculator

The image represents reaction at the outer wheel of a vehicle moving along a level circular path.

To compute for the reaction, six essential parameters are needed and these parameters are Mass of the Vechicle (m), Acceleration due to Gravity (g), Velocity of the Vehicle (v), Height of Centre of Gravity of the Vehicle from Ground Level (h), Radius of Circular Path (r) and Half of the Distance between the Centre Lines of the Wheel (a).

The formula for calculating the reaction at the outer wheel of a vehicle moving along a level circular path:

RB = mg / 2[1 + v²h/gra]

Where;
RB = Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path
m = Mass of the Vechicle
g = Acceleration due to Gravity
v = Velocity of the Vehicle
h = Height of Centre of Gravity of the Vehicle from Ground Level
r = Radius of Circular Path
a = Half of the Distance between the Centre Lines of the Wheel

Let’s solve an example;
Find the reaction when Mass of the Vechicle (m) is 12, Acceleration due to Gravity (g) is 9.8, Velocity of the Vehicle (v) is 28, Height of Centre of Gravity of the Vehicle from Ground Level (h) is 16, Radius of Circular Path (r) is 8 and Half of the Distance between the Centre Lines of the Wheel (a) is 4.

This implies that;
m = Mass of the Vechicle = 12
g = Acceleration due to Gravity = 9.8
v = Velocity of the Vehicle = 28
h = Height of Centre of Gravity of the Vehicle from Ground Level = 16
r = Radius of Circular Path = 8
a = Half of the Distance between the Centre Lines of the Wheel = 4

RB = mg / 2[1 + v²h/gra]
RB = 12 x 9.8 / 2[1 + 28² x 16/9.8 x 8 x 4]
RB = 117.60 / 2[1 + 784 x 16/313.6]
RB = 58.80[1 + 40]
RB = 58.80[41]
RB = 58.80[41]
RB = 2410.8

Therefore, the reaction of the outer wheel of a vehicle moving along a level circular path is 2410.8 N.

Continue reading How to Calculate and Solve for the Reaction at the Outer Wheel of a Vehicle moving along a Level Circular Path | Nickzom Calculator

How to Calculate and Solve for Mass, Height and Potential Energy | The Calculator Encyclopedia

The image above represents potential energy.

To compute for the potential energy, three essential parameters are needed and these parameters are mass (m), height (h) and acceleration due to gravity (g).

The formula for calculating the potential energy:

P.E = mgh

Where;
P.E. = Potential Energy
m = Mass
g = acceleration due to gravity
h = Height

Let’s solve an example;
Find the potential energy when the mass is 12 with a height of 24 and acceleration due to gravity of 9.8.

This implies that;
m = Mass = 12
g = acceleration due to gravity = 9.8
h = Height = 24

P.E = mgh
P.E = 12 x 9.8 x 24
P.E = 2822.4

Therefore, the potential energy is 2822.4 Joules (J).

Calculating the Mass when Potential Energy, Height and Acceleration due to Gravity is Given.

m = P.E / gh

Where;
m = Mass
P.E. = Potential Energy
g = acceleration due to gravity
h = Height

Let’s solve an example;
Find the Mass when potential energy is 450 with a height of 30 and acceleration due to gravity of 10.

This implies that;
P.E. = Potential Energy = 450
g = acceleration due to gravity = 10
h = Height = 30

m = P.E / gh
m = 450 / 10 x 30
m = 450 / 300
m = 1.5

Therefore, the mass is 1.5.

Continue reading How to Calculate and Solve for Mass, Height and Potential Energy | The Calculator Encyclopedia

How to Calculate and Solve for Escape Velocity | The Calculator Encyclopedia

The image above represents the escape velocity.

To compute the escape velocity of a field, two essential parameters are needed and the parameters are acceleration due to gravity (g) and radius (r).

The formula for calculating the escape velocity:

V = √(2gR)

Where;
V = Escape velocity
g = Acceleration due to gravity
R = Radius

Let’s solve an example;
Find the escape velocity of a field when the acceleration due to gravity is 12 and the radius is 24 cm.

This implies that;
g = Acceleration due to gravity = 12
r = Radius = 24 cm

V = √(2gR)
V = √(2 x 12 x 24)
V = √(576)
V = 24

Therefore, the escape velocity is 24 m/s.

Continue reading How to Calculate and Solve for Escape Velocity | The Calculator Encyclopedia