**What is the Velocity Ratio (V.R) ? **

**Velocity ratio is the ratio of the distance covered by an effort to the distance covered by the load in the same time interval.** We also refer to the velocity ratio of an inclined plane as vr of inclined plane.

## What is the Velocity Ratio of an Inclined Plane?

The velocity ratio of an inclined plane is the ratio of the effort’s velocity to the load’s velocity, allowing easier movement of loads along the slope compared to vertical lifting (along an inclined plane).

The velocity ratio of an inclined plane represents the advantage gained in moving a load along its length compared to lifting it vertically. It is the ratio of input force (effort) to output force (load) velocities, making it easier to move heavy objects by applying a smaller force over a longer distance.

Additionally, the velocity ratio of an inclined plane does not have a unit. It is a dimensionless quantity that expresses the mechanical advantage gained in using the plane to move a load compared to lifting it vertically.

## Velocity Ratio of Inclined Plane Formula

We can express the velocity ratio (v.r) formula in mathematical form as

Velocity Ratio (V.R) **=** Distance of the Effort **/** Distance of the Load

V.R rely on the geometry of the machine

For an inclined plane, the machine will assist us in pushing bodies up an inclined plane with ease. We apply an inclined plane to push heavy objects up a slope.

Velocity ratio (V.R) = length of plane / height of a plane = L/h

Thus

Sin θ = h/L

Therefore, we can say that

**V.R = 1 / Sin θ**

## How to Find Velocity Ratio of an Inclined Plane

To find the velocity ratio of an inclined plane:

- Measure the length of the inclined plane (effort distance) and the vertical height it covers (load height).
- Calculate the ratio: Velocity Ratio = Effort Distance / Load Height.
- The result indicates how much easier it is to move a load along the plane compared to lifting it vertically.

## Solved Problems: How to Calculate the Velocity Ratio of an Inclined Plane

Here are a few problems to help us understand how to calculate the velocity ratio of an inclined plane:

### Problem 1

Calculate the velocity ratio of an inclined plane which makes an angle of 60^{0} with the horizontal.

**Solution**:

Data:

From the question above, we are to calculate the V.R which is unknown to us

Therefore V.R = ?

Angle (θ) = 60^{0}

and the mathematical expression for V.R = 1 / Sin θ

and θ = 60^{0}

Which implies

V.R = 1 / Sin 60^{0}

and Sin 60^{0} = 0.866

This implies that

V.R = 1 / 0.866

Thus,

V.R = 1.15

Therefore, the velocity ratio of an inclined plane which makes an angle of 60^{0} to horizontal is 1.15

This is a very simple example of how to calculate the velocity ratio of an inclined plane

### Problem 2

A floor of a lorry is 2.0 m high. A plank 5 m long is used as an inclined plane to raise some load up the lorry. If the efficiency of this machine is 50% what is the minimum effort (applied parallel to the plane) required to raise 200 N load up the plane?

Solution

Data:

We need to gather our data to be able to solve the problem

Firstly, we can see from the question that the effort required to raise the load = length of the plank = 5 m

Additionally, the distance moved by the load = floor of the lorry = 2.0 m

Efficiency = 50%

Load = 200 N

From the above data, we apply the formula

Velocity Ratio (V.R) = 1 / Sin θ = Distance moved by effort = Distance moved by load

V.R = 5 / 2 = 2.5

Also

Efficiency = ( M.A / V.R ) x 100%

By substituting our data into the above equation and the solution for V.R we have

50% = ( M.A / 2.5 ) x 100%

by making M.A the subject of the formula, we have

( 50% / 100% ) x 2.5 = M.A

Therefore

M.A = 0.5 x 2.5 = 1.25

Now we can write

M.A = Load / Effort

To find Effort, we make it the subject of the formula from the above equation

Effort = Load / M.A

This implies

Effort = 200 / 1.25 = 160 N

### Problem 3

A block of mass 2 kg is placed on an inclined plane whose angle of inclination is increased until the block is just about to slide down the plane. If the angle of inclination of the plane is 60^{0} at this point, calculate the force, parallel to the plane which will just move the block up the inclined plane.

Solution

Data:

We deduce our data by looking at the question above

Assuming the required Force = P

There is also a force resisting the movement of the load up to the inclined plane which is

Fictional force on the plane F = μR

We also have a Reactional Force R = wcosθ

and we also have a force acting down the plane, which is force = wsinθ

w = weight = mg

Therefore, the total force acting down the plane is

T = μR + wsinθ

T = μmg cos θ + mg sin θ

We should also know that limiting friction μ = tan θ

Now, T = tan θ mg cos θ + mg sin θ

Since θ = 60^{0} , m = 2kg , and g = 10 ms^{-2}

We have

T = tan 60^{0} x 2 x 10 x cos 60^{0} + 2 x 10 sin 60^{0}

T = 1.73 x 20 x 0.5 + 20 x 0.866

Therefore

T = 34.64 N

### Problem 4

An effort of 20N is applied to lift a load of 120N with the aid of a machine. The distances covered by the effort and the loads are 50 meters and 200 meters respectively. Calculate the velocity ratio of the machine.

**Solution:**

**Data**: You need to extract your data from the question to simplify your work.

-Underline the figures

-take note of the s.i units

-relate the figures with the corresponding parameters

-apply the v.r formula to solve the problem

Now,

V.R = ?

Distance moved by the effort = 50 meters

Distance moved by the load = 200 meters

The formula for Velocity ratio (V.R) = Distance moved by the effort (50 m) / Distance moved by the load (200 m)

V.R = 50 / 200 = 0.25

### Problem 5

A 20-kilogram mass is to be pulleyed up a slope inclined at 30^{0} to the horizontal. If the efficiency of the plane is 75%, calculate the force required to pull the load up the plane.

**Solution:**

**Data:**

Load, L = 20kg = 200N

The angle, θ = 30^{0}

Efficiency, η = 75% = 75/100 = 0.75

The force required to pull the load up an inclined plane = Effort = E = ?

Now we apply the formula that says

V.R = 1 / Sin θ

Which implies that the velocity ratio of an inclined plane is

V.R = 1 / Sin 30^{0}

Thus,

The velocity ratio of an inclined plane, V.R = 1/0.5 = 2

Additionally, the mechanical advantage, M.A = Load / Effort = L / E = 200 / E

But the formula for efficiency is η = ( M.A / V.R ) x 100%

By substituting our values into the above formula for efficiency, we have

75 = [ (200/E) / 2 ] x 100

this is equal to

75/100 = (200/E) / 2

0.75 x 2 = 200/E

1.5 = 200 / E

Thus, E = 200 / 1.5 = 133.3N

### Problem 6

A body of mass 7.5 kilograms is to be pulled up along a plane that is inclined at an angle of 30 degrees to the horizontal. If the efficiency of the plane is 75%, Calculate the velocity ratio of the plane.

**Solution:**

**Data:**

mass, M = 7.5kg

Angle of inclination, θ = 30^{0}

Efficiency, η = 75% = 75/100 = 0.75

The velocity ratio of an inclined angle is V.R = 1 / Sin θ

Thus

V.R = 1 / Sin 30^{0} = 1 / 0.5 = 2

### Problem 7

A man pulls up a box of mass 70 kg using an inclined plane of an effective length 5 meters onto a platform 2.5 meters high at a uniform speed. If the frictional force between the box and plane is 100N. Calculate the velocity ratio of the angle of inclination.

**Solution**

**Data:**

Since Sin θ = opposite / hypotenuse

and we can see that opposite length = 2.5m

Hypotenuse = 5m

Thus

Sin θ = 2.5 / 5 = 0.5

Now we can write the above expression as θ = Sin^{-1} 0.5

Thus θ = 30^{0}

Applying the formula for the velocity ratio of an inclined plane (V.R = 1 / Sin θ), we now have

V.R = 1 / Sin 30^{0} = 1 / 0.5 = 2

Thus the velocity ratio of an inclined plane is 2

### Problem 8

A box of weight W resting on a plank PQ that has it is lower end P hinged to a horizontal floor PR. The plank makes an angle θ to the floor. If the box is pulled up from P to Q, what is the expression for the velocity ratio of an inclined plane?

**Solution**

**Data:**

Distance covered by the effort = PQ

Distance covered by the load = QR

We know that the velocity ratio (V.R) of an inclined plane

V.R = distance covered by the effort/distance covered by the load

This implies that

V.R = PQ / QR

### Problem 9

An object is released from the top of an inclined plane about 250 meters long and 50 meters high. What is the velocity ratio of the inclined plane?

**Solution**

**Data**

After reading the above question, we can come up with our data as follows:

Length of the plane, L = 250 meters

Height of the plane, h = 50 meters

Now, apply the formula for the velocity ratio below to find our answer

We initially said that Velocity ratio (V.R) = length of the plane (L)/height of a plane (h) = L/h

Thus, we can easily insert our data in the above formula velocity ratio of an inclined plane to represent the length of the plane (L) and height of the plane (h).

Therefore,

V.R = L/h = 250/50 = 5

**Hence, the velocity ratio of the inclined plane is 5 [Note: Velocity ratio does not have a unit]**

### Problem 10

Five laborers allowed a drum to roll along a horizontally inclined tower about 2 kilometers long and 500 meters high. Calculate the velocity ratio of the inclined tower.

**Solution**

**Data**

The length of the tower, L = 2 km = 2 x 1000 meters = 2000 meter

Height of the tower, h = 500 meters

We can now calculate the velocity ratio of the inclined tower by applying the formula that says

**We initially said that Velocity ratio (V.R) = length of the plane (L)/height of a plane (h) = L/h**

Thus,

V.R = L/h = 2,000 meters / 500 meters = 4

**Therefore, the velocity ratio of the inclined tower is 4**

### Problem 11

A body is released from the top of an inclined plane 150 meters long and 25 meters high. Find the velocity ratio if the angle of inclination is thirty degrees (30^{0}).

**Solution**

**Data**

Start by extracting your data from the question

Length of the plane, L = 150 meters

Height of the plane, h = 25 meters

We have two scenarios in this question. We need to apply the formula that says V.R = L/h to solve the problem, and then apply the other formula that was affected by the angle of inclination (slope). You can then multiply the two to obtain the total velocity ratio of the plane.

**Applying V.R = L/h**

Now, apply the formula for the velocity ratio below to find our answer

We initially said that Velocity ratio (V.R) = length of the plane (L)/height of a plane (h) = L/h

Therefore, V.R = 150 meters / 25 meters = 6

**Applying V.R = 1 / Sin θ**

To calculate the velocity ratio of an inclined plane formed by an angle of 30^{0}, we apply the formula, V.R = 1 / Sin θ

Thus

V.R = 1 / Sin 30^{0} = 1 / 0.5 = 2

**We can now multiply the two velocity ratios to arrive at the total velocity ratio of the plane **

V.R = 2 x 6 = 12

**Therefore, the velocity ratio of the plane is 12**

### Problem 12: For an inclined plane derive efficiency=l/h

For an inclined plane derive efficiency=l/h

**Solution**

Certainly! To derive the efficiency (( \eta )) of an inclined plane, which is defined as the ratio of useful work output (W_{out}) to the total work input (W_{in}), you can follow these steps:

**Useful Work Output (W**When an object is moved along an inclined plane, the useful work done is the weight of the object (F_{out}):_{load}= m x g) multiplied by the distance it moves horizontally (l). Therefore,**W**= F_{out}_{load}x l.**Total Work Input (W**_{in}**):**The total work done on the object is the applied force F_{effort}multiplied by the distance it moves horizontally (l). Thus, W_{in}= F_{effort}x l.**Efficiency (η) Calculation:**Divide the useful work output by the total work input and express it as a fraction: η =**W**/ W_{out}_{in}= (F_{load}x l) / (F_{effort}x l) = F_{load}/ F_{effort}**Substitute Load and Effort Forces:**In terms of load F_{load}and F_{effort}forces along the inclined plane, you can use trigonometry: F_{load}= F_{effort}x Sinθ, where θ is the angle of inclination.**Final Expression:**Substituting the load-force relationship into the efficiency equation: η = F_{load}/ F_{effort}= (F_{effort}x Sinθ) / F_{effort}. Now, we can relate Sinθ to the ratio of load height (h) to the inclined plane length (l) using trigonometry: Sinθ = h/l. Thus, the derived efficiency of the inclined plane is: η = h/l

This demonstrates that the efficiency of an inclined plane is directly proportional to the ratio of the load height to the inclined plane length.

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