A block and tackle system has velocity ratio 3. draw a labelled diagram of the system indicating the points of application and the direction of load L and effort E. A man can exert a pull of 200 kgf. (a) What is the maximum load he can raise with this pulley system if its efficiency is 60% ? (b) if the effort end moves a distance 60 cm, what distance does the load move ?

Answer

### Step-by-Step Solution

**Step 1: Draw a labeled diagram of the block and tackle system.**

- A block and tackle system typically consists of multiple pulleys. Since the velocity ratio (VR) is given as 3, we can represent this with 3 pulleys.
- In the diagram, indicate:
- The load (L) acting downward.
- The effort (E) applied downward.
- The direction of tension in the strings acting upward.

**Diagram:**
```
E
↓
----
| |
| | (Pulley 1)
| |
----
|
|
----
| |
| | (Pulley 2)
| |
----
|
|
----
| |
| | (Pulley 3)
| |
----
|
↓
L
```

**Step 2: Calculate the maximum load that can be raised.**

Given:
- Effort (E) = 200 kgf
- Efficiency (η) = 60% = 60/100 = 0.6
- Velocity Ratio (VR) = 3

Using the formula for efficiency:
\[ \text{Efficiency} = \frac{\text{Mechanical Advantage (MA)}}{\text{Velocity Ratio (VR)}} \]

Rearranging gives:
\[ \text{Mechanical Advantage (MA)} = \text{Efficiency} \times \text{Velocity Ratio (VR)} \]
\[ \text{MA} = 0.6 \times 3 = 1.8 \]

Now, using the formula for mechanical advantage:
\[ \text{MA} = \frac{\text{Load (L)}}{\text{Effort (E)}} \]

Substituting the values:
\[ 1.8 = \frac{L}{200} \]

Solving for Load (L):
\[ L = 1.8 \times 200 = 360 \text{ kgf} \]

**Step 3: Calculate the distance moved by the load.**

Given:
- Distance moved by effort (D_E) = 60 cm
- Velocity Ratio (VR) = 3

Using the formula for velocity ratio:
\[ \text{VR} = \frac{\text{Distance moved by effort (D_E)}}{\text{Distance moved by load (D_L)}} \]

Rearranging gives:
\[ D_L = \frac{D_E}{VR} \]
\[ D_L = \frac{60 \text{ cm}}{3} = 20 \text{ cm} \]

### Final Answers:
(a) The maximum load that can be raised is **360 kgf**.

(b) The distance moved by the load is **20 cm**.

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