Question
A screw jack whose pitch is 2mm is used to raise a motor car of mass 900kg through a height of 20cm. The length of the tommy bar of the jack is 40cm. If the jack is 60% efficient, calculate:
Velocity ratio;
Mechanical advantage;
Effort required;
Work done by the effort.g=10ms^-2
Velocity ratio;
Mechanical advantage;
Effort required;
Work done by the effort.g=10ms^-2
Answers
GPT-4o mini
To solve the problem, we need to calculate various mechanical properties related to the screw jack used to lift the motor car.
### 1. **Velocity Ratio (VR)**
The velocity ratio for a screw jack can be calculated using the formula:
\[
\text{VR} = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} = \frac{\text{Circumference of the circle traced by the tommy bar}}{\text{Pitch of the screw}}
\]
#### Step 1: Calculate the distance moved by effort.
The circumference (C) of the circle traced by the tommy bar is given by:
\[
C = 2\pi r
\]
where \( r \) is the length of the tommy bar (0.4 m).
\[
C = 2\pi(0.4) \approx 2.513 \, \text{m}
\]
#### Step 2: Calculate the pitch of the screw (distance moved by load).
The pitch of the screw is given as \( 2 \, \text{mm} = 0.002 \, \text{m} \).
#### Step 3: Calculate the VR.
\[
\text{VR} = \frac{C}{\text{Pitch}} = \frac{2.513}{0.002} = 1256.5
\]
### 2. **Mechanical Advantage (MA)**
Mechanical advantage is given by:
\[
\text{MA} = \frac{\text{Efficiency}}{\text{VR}}
\]
Where efficiency is given as 60% (or 0.6 in decimal).
\[
\text{MA} = \text{Efficiency} \times \text{VR} = 0.6 \times 1256.5 \approx 753.9
\]
### 3. **Effort Required**
The effort required can be calculated using the relation between load and effort, which is given by:
\[
\text{Load} = \text{MA} \times \text{Effort}
\]
Rearranging gives:
\[
\text{Effort} = \frac{\text{Load}}{\text{MA}}
\]
#### Step 1: Calculate Load (Force due to weight of the car)
Given:
- Mass of the car (\( m \)) = 900 kg
- \( g = 10 \, \text{m/s}^2 \)
\[
\text{Load} = m \cdot g = 900 \cdot 10 = 9000 \, \text{N}
\]
#### Step 2: Calculate Effort.
\[
\text{Effort} = \frac{9000}{753.9} \approx 11.93 \, \text{N}
\]
### 4. **Work Done by the Effort**
Work done by the effort can be calculated using the formula:
\[
\text{Work} = \text{Effort} \times \text{Distance moved by effort}
\]
#### Step 1: Using the distance moved by effort (Circumference of the tommy bar):
\[
\text{Work} = \text{Effort} \times C = 11.93 \times 2.513 \approx 30.0 \, \text{J}
\]
### Summary of Results:
- **Velocity Ratio (VR):** \( 1256.5 \)
- **Mechanical Advantage (MA):** \( 753.9 \)
- **Effort Required:** \( 11.93 \, \text{N} \)
- **Work Done by the Effort:** \( 30.0 \, \text{J} \)
These calculations illustrate the basic principles of using a screw jack to lift heavy objects based on its efficiency, mechanical advantage, and other parameters.
### 1. **Velocity Ratio (VR)**
The velocity ratio for a screw jack can be calculated using the formula:
\[
\text{VR} = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}} = \frac{\text{Circumference of the circle traced by the tommy bar}}{\text{Pitch of the screw}}
\]
#### Step 1: Calculate the distance moved by effort.
The circumference (C) of the circle traced by the tommy bar is given by:
\[
C = 2\pi r
\]
where \( r \) is the length of the tommy bar (0.4 m).
\[
C = 2\pi(0.4) \approx 2.513 \, \text{m}
\]
#### Step 2: Calculate the pitch of the screw (distance moved by load).
The pitch of the screw is given as \( 2 \, \text{mm} = 0.002 \, \text{m} \).
#### Step 3: Calculate the VR.
\[
\text{VR} = \frac{C}{\text{Pitch}} = \frac{2.513}{0.002} = 1256.5
\]
### 2. **Mechanical Advantage (MA)**
Mechanical advantage is given by:
\[
\text{MA} = \frac{\text{Efficiency}}{\text{VR}}
\]
Where efficiency is given as 60% (or 0.6 in decimal).
\[
\text{MA} = \text{Efficiency} \times \text{VR} = 0.6 \times 1256.5 \approx 753.9
\]
### 3. **Effort Required**
The effort required can be calculated using the relation between load and effort, which is given by:
\[
\text{Load} = \text{MA} \times \text{Effort}
\]
Rearranging gives:
\[
\text{Effort} = \frac{\text{Load}}{\text{MA}}
\]
#### Step 1: Calculate Load (Force due to weight of the car)
Given:
- Mass of the car (\( m \)) = 900 kg
- \( g = 10 \, \text{m/s}^2 \)
\[
\text{Load} = m \cdot g = 900 \cdot 10 = 9000 \, \text{N}
\]
#### Step 2: Calculate Effort.
\[
\text{Effort} = \frac{9000}{753.9} \approx 11.93 \, \text{N}
\]
### 4. **Work Done by the Effort**
Work done by the effort can be calculated using the formula:
\[
\text{Work} = \text{Effort} \times \text{Distance moved by effort}
\]
#### Step 1: Using the distance moved by effort (Circumference of the tommy bar):
\[
\text{Work} = \text{Effort} \times C = 11.93 \times 2.513 \approx 30.0 \, \text{J}
\]
### Summary of Results:
- **Velocity Ratio (VR):** \( 1256.5 \)
- **Mechanical Advantage (MA):** \( 753.9 \)
- **Effort Required:** \( 11.93 \, \text{N} \)
- **Work Done by the Effort:** \( 30.0 \, \text{J} \)
These calculations illustrate the basic principles of using a screw jack to lift heavy objects based on its efficiency, mechanical advantage, and other parameters.