Question
A man uses a rope to haul a box of weight 600N up an inclined wooden plank of effective length 3m and onto a platform 1m high. If in doing so, he exerts a force of 400N calculate;
The fictional force between the case and the plank
The velocity ratio of the machine
The mechanical advantage
The useful workdone
The fictional force between the case and the plank
The velocity ratio of the machine
The mechanical advantage
The useful workdone
Answers
GPT-4o mini
To solve the problem, we will calculate the frictional force, velocity ratio, mechanical advantage, and useful work done.
### 1. Calculate the frictional force between the box and the plank.
**Given:**
- Weight of the box (W) = 600 N
- Applied force (F) = 400 N
The net force acting on the box along the incline can be represented as:
\[
F_{\text{net}} = W_{\text{incline}} - F - F_{\text{friction}}
\]
Where:
- \(W_{\text{incline}}\) is the component of the weight acting along the incline.
- \(F_{\text{friction}}\) is the frictional force.
**Weight component along the incline:**
To find the weight component acting down the incline, we need the angle of the incline (θ). Since the height (h) of the platform is 1 m and the length of the plank (l) is 3 m, we can calculate θ using the sine function:
\[
\sin(\theta) = \frac{h}{l} = \frac{1}{3}
\]
\[
\theta = \arcsin\left(\frac{1}{3}\right)
\]
Now we can find the gravitational force component acting down the plank. This can be calculated as:
\[
W_{\text{incline}} = W \cdot \sin(\theta) = 600 \cdot \frac{1}{3} = 200 \text{ N}
\]
The net force equation becomes:
\[
F_{\text{net}} = W_{\text{incline}} - F - F_{\text{friction}}
\]
\[
F_{\text{net}} = 200 \text{ N} - 400 \text{ N} - F_{\text{friction}}
\]
Since the box is moving upward, we have:
\[
F_{\text{friction}} = W_{\text{incline}} - F_{\text{net}} = 200 - 400 + F_{\text{friction}} \Rightarrow 0 = 200 - 400 + F_{\text{friction}}
\]
\[
F_{\text{friction}} = 200 \text{ N}
\]
### 2. Calculate the velocity ratio (VR) of the machine.
The velocity ratio is defined as the ratio of the distance moved by the effort to the distance moved by the load:
\[
VR = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}}
\]
- Distance moved by the effort (along the plank) = 3 m
- Distance moved by the load (height of platform) = 1 m
So, we have:
\[
VR = \frac{3 m}{1 m} = 3
\]
### 3. Calculate the mechanical advantage (MA).
Mechanical advantage is defined as the ratio of the load lifted to the effort applied:
\[
MA = \frac{\text{Load (W)}}{\text{Effort (F)}}
\]
Substituting the values:
\[
MA = \frac{600 \text{ N}}{400 \text{ N}} = 1.5
\]
### 4. Calculate the useful work done.
Useful work done (W) is calculated as:
\[
\text{Useful work done} = \text{Load} \times \text{Height lifted}
\]
\[
\text{Useful work done} = 600 \text{ N} \times 1 \text{ m} = 600 \text{ J}
\]
### Summary of Results:
- Frictional force between the box and the plank: **200 N**
- Velocity ratio of the machine: **3**
- Mechanical advantage: **1.5**
- Useful work done: **600 J**
### 1. Calculate the frictional force between the box and the plank.
**Given:**
- Weight of the box (W) = 600 N
- Applied force (F) = 400 N
The net force acting on the box along the incline can be represented as:
\[
F_{\text{net}} = W_{\text{incline}} - F - F_{\text{friction}}
\]
Where:
- \(W_{\text{incline}}\) is the component of the weight acting along the incline.
- \(F_{\text{friction}}\) is the frictional force.
**Weight component along the incline:**
To find the weight component acting down the incline, we need the angle of the incline (θ). Since the height (h) of the platform is 1 m and the length of the plank (l) is 3 m, we can calculate θ using the sine function:
\[
\sin(\theta) = \frac{h}{l} = \frac{1}{3}
\]
\[
\theta = \arcsin\left(\frac{1}{3}\right)
\]
Now we can find the gravitational force component acting down the plank. This can be calculated as:
\[
W_{\text{incline}} = W \cdot \sin(\theta) = 600 \cdot \frac{1}{3} = 200 \text{ N}
\]
The net force equation becomes:
\[
F_{\text{net}} = W_{\text{incline}} - F - F_{\text{friction}}
\]
\[
F_{\text{net}} = 200 \text{ N} - 400 \text{ N} - F_{\text{friction}}
\]
Since the box is moving upward, we have:
\[
F_{\text{friction}} = W_{\text{incline}} - F_{\text{net}} = 200 - 400 + F_{\text{friction}} \Rightarrow 0 = 200 - 400 + F_{\text{friction}}
\]
\[
F_{\text{friction}} = 200 \text{ N}
\]
### 2. Calculate the velocity ratio (VR) of the machine.
The velocity ratio is defined as the ratio of the distance moved by the effort to the distance moved by the load:
\[
VR = \frac{\text{Distance moved by effort}}{\text{Distance moved by load}}
\]
- Distance moved by the effort (along the plank) = 3 m
- Distance moved by the load (height of platform) = 1 m
So, we have:
\[
VR = \frac{3 m}{1 m} = 3
\]
### 3. Calculate the mechanical advantage (MA).
Mechanical advantage is defined as the ratio of the load lifted to the effort applied:
\[
MA = \frac{\text{Load (W)}}{\text{Effort (F)}}
\]
Substituting the values:
\[
MA = \frac{600 \text{ N}}{400 \text{ N}} = 1.5
\]
### 4. Calculate the useful work done.
Useful work done (W) is calculated as:
\[
\text{Useful work done} = \text{Load} \times \text{Height lifted}
\]
\[
\text{Useful work done} = 600 \text{ N} \times 1 \text{ m} = 600 \text{ J}
\]
### Summary of Results:
- Frictional force between the box and the plank: **200 N**
- Velocity ratio of the machine: **3**
- Mechanical advantage: **1.5**
- Useful work done: **600 J**