Answer :
Certainly! Let's go through each question step-by-step to find the required solutions.
### Question 1
At crowbar is used to lift a load of 1200 N by placing it at a distance of 25 cm from the fulcrum. Calculate the force required at 60 cm from the fulcrum to balance the load.
#### Solution:
The principle of moments states that for a lever in equilibrium, the clockwise moment about the fulcrum is equal to the anticlockwise moment about the fulcrum.
Let's denote:
- Load (L) = 1200 N
- Distance of load from fulcrum (d_L) = 25 cm
- Effort (E) = ?
- Distance of effort from fulcrum (d_E) = 60 cm
The moment due to the load is:
[tex]\[ \text{Moment}_L = L \times d_L = 1200 \, \text{N} \times 25 \, \text{cm} \][/tex]
The moment due to the effort is:
[tex]\[ \text{Moment}_E = E \times d_E = E \times 60 \, \text{cm} \][/tex]
For equilibrium:
[tex]\[ \text{Moment}_L = \text{Moment}_E \][/tex]
[tex]\[ 1200 \, \text{N} \times 25 \, \text{cm} = E \times 60 \, \text{cm} \][/tex]
Solving for [tex]\( E \)[/tex]:
[tex]\[ E = \frac{1200 \, \text{N} \times 25 \, \text{cm}}{60 \, \text{cm}} \][/tex]
[tex]\[ E = 500 \, \text{N} \][/tex]
So, the force required is 500 N.
### Question 2
If a 500 N load is to be lifted with a 100 N effort using a first class lever, at what distance from the fulcrum must the effort be applied when the load is 20 cm away from the fulcrum? Show your answer in diagram form.
#### Solution:
Using the principle of moments again:
- Load (L) = 500 N
- Effort (E) = 100 N
- Distance of load from fulcrum (d_L) = 20 cm
- Distance of effort from fulcrum (d_E) = ?
For equilibrium:
[tex]\[ \text{Moment}_L = \text{Moment}_E \][/tex]
[tex]\[ L \times d_L = E \times d_E \][/tex]
[tex]\[ 500 \, \text{N} \times 20 \, \text{cm} = 100 \, \text{N} \times d_E \][/tex]
Solving for [tex]\( d_E \)[/tex]:
[tex]\[ d_E = \frac{500 \, \text{N} \times 20 \, \text{cm}}{100 \, \text{N}} \][/tex]
[tex]\[ d_E = 100 \, \text{cm} \][/tex]
So, the effort must be applied at a distance of 100 cm from the fulcrum.
Diagram:
Here is a basic diagram to illustrate this
```
Fulcrum (F) Load (L)
| |
| |
V (20 cm)
------------------------
^
|
E
(100 cm)
```
### Question 3
A machine raises a load through a distance of 30 cm by an effort which moves through 90 cm. What will be the Velocity Ratio (V.R)?
#### Solution:
The Velocity Ratio (V.R) is given by:
[tex]\[ \text{V.R} = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} \][/tex]
[tex]\[ \text{V.R} = \frac{90 \, \text{cm}}{30 \, \text{cm}} \][/tex]
[tex]\[ \text{V.R} = 3 \][/tex]
So, the Velocity Ratio is 3.
### Question 4
A crowbar of length 2 m is pivoted about 40 cm from its tip and a load of 800 N is lifted by applying an effort at the other end. Calculate the effort required.
#### Solution:
Let's denote:
- Load (L) = 800 N
- Distance of pivot from load end (d_L) = 40 cm
- Total length of crowbar = 200 cm
- Distance of effort from pivot = 200 cm - 40 cm = 160 cm
- Effort (E) = ?
Using the principle of moments:
[tex]\[ L \times d_L = E \times d_E \][/tex]
[tex]\[ 800 \, \text{N} \times 40 \, \text{cm} = E \times 160 \, \text{cm} \][/tex]
[tex]\[ E = \frac{800 \, \text{N} \times 40 \, \text{cm}}{160 \, \text{cm}} \][/tex]
[tex]\[ E = 200 \, \text{N} \][/tex]
So, the effort required is 200 N.
### Question 5
A lever with a V.R of 30 overcomes a load of \_\_\_\_.
#### Solution:
The question seems incomplete as it does not provide complete data. Velocity Ratio (V.R) is the ratio of the distance moved by the effort to the distance moved by the load. Without additional information such as effort or exact load, we cannot provide a precise numerical value.
However, let's assume you want the force needed to sustain a load with a V.R of 30. In general, given V.R of a lever system, we can write:
[tex]\[ \text{Load} = \text{V.R} \times \text{Effort} \][/tex]
If more details are provided, we can apply them to determine the exact value.
Please clarify the question or provide the missing data, and I will be happy to complete the solution.
### Question 1
At crowbar is used to lift a load of 1200 N by placing it at a distance of 25 cm from the fulcrum. Calculate the force required at 60 cm from the fulcrum to balance the load.
#### Solution:
The principle of moments states that for a lever in equilibrium, the clockwise moment about the fulcrum is equal to the anticlockwise moment about the fulcrum.
Let's denote:
- Load (L) = 1200 N
- Distance of load from fulcrum (d_L) = 25 cm
- Effort (E) = ?
- Distance of effort from fulcrum (d_E) = 60 cm
The moment due to the load is:
[tex]\[ \text{Moment}_L = L \times d_L = 1200 \, \text{N} \times 25 \, \text{cm} \][/tex]
The moment due to the effort is:
[tex]\[ \text{Moment}_E = E \times d_E = E \times 60 \, \text{cm} \][/tex]
For equilibrium:
[tex]\[ \text{Moment}_L = \text{Moment}_E \][/tex]
[tex]\[ 1200 \, \text{N} \times 25 \, \text{cm} = E \times 60 \, \text{cm} \][/tex]
Solving for [tex]\( E \)[/tex]:
[tex]\[ E = \frac{1200 \, \text{N} \times 25 \, \text{cm}}{60 \, \text{cm}} \][/tex]
[tex]\[ E = 500 \, \text{N} \][/tex]
So, the force required is 500 N.
### Question 2
If a 500 N load is to be lifted with a 100 N effort using a first class lever, at what distance from the fulcrum must the effort be applied when the load is 20 cm away from the fulcrum? Show your answer in diagram form.
#### Solution:
Using the principle of moments again:
- Load (L) = 500 N
- Effort (E) = 100 N
- Distance of load from fulcrum (d_L) = 20 cm
- Distance of effort from fulcrum (d_E) = ?
For equilibrium:
[tex]\[ \text{Moment}_L = \text{Moment}_E \][/tex]
[tex]\[ L \times d_L = E \times d_E \][/tex]
[tex]\[ 500 \, \text{N} \times 20 \, \text{cm} = 100 \, \text{N} \times d_E \][/tex]
Solving for [tex]\( d_E \)[/tex]:
[tex]\[ d_E = \frac{500 \, \text{N} \times 20 \, \text{cm}}{100 \, \text{N}} \][/tex]
[tex]\[ d_E = 100 \, \text{cm} \][/tex]
So, the effort must be applied at a distance of 100 cm from the fulcrum.
Diagram:
Here is a basic diagram to illustrate this
```
Fulcrum (F) Load (L)
| |
| |
V (20 cm)
------------------------
^
|
E
(100 cm)
```
### Question 3
A machine raises a load through a distance of 30 cm by an effort which moves through 90 cm. What will be the Velocity Ratio (V.R)?
#### Solution:
The Velocity Ratio (V.R) is given by:
[tex]\[ \text{V.R} = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} \][/tex]
[tex]\[ \text{V.R} = \frac{90 \, \text{cm}}{30 \, \text{cm}} \][/tex]
[tex]\[ \text{V.R} = 3 \][/tex]
So, the Velocity Ratio is 3.
### Question 4
A crowbar of length 2 m is pivoted about 40 cm from its tip and a load of 800 N is lifted by applying an effort at the other end. Calculate the effort required.
#### Solution:
Let's denote:
- Load (L) = 800 N
- Distance of pivot from load end (d_L) = 40 cm
- Total length of crowbar = 200 cm
- Distance of effort from pivot = 200 cm - 40 cm = 160 cm
- Effort (E) = ?
Using the principle of moments:
[tex]\[ L \times d_L = E \times d_E \][/tex]
[tex]\[ 800 \, \text{N} \times 40 \, \text{cm} = E \times 160 \, \text{cm} \][/tex]
[tex]\[ E = \frac{800 \, \text{N} \times 40 \, \text{cm}}{160 \, \text{cm}} \][/tex]
[tex]\[ E = 200 \, \text{N} \][/tex]
So, the effort required is 200 N.
### Question 5
A lever with a V.R of 30 overcomes a load of \_\_\_\_.
#### Solution:
The question seems incomplete as it does not provide complete data. Velocity Ratio (V.R) is the ratio of the distance moved by the effort to the distance moved by the load. Without additional information such as effort or exact load, we cannot provide a precise numerical value.
However, let's assume you want the force needed to sustain a load with a V.R of 30. In general, given V.R of a lever system, we can write:
[tex]\[ \text{Load} = \text{V.R} \times \text{Effort} \][/tex]
If more details are provided, we can apply them to determine the exact value.
Please clarify the question or provide the missing data, and I will be happy to complete the solution.
