Interactive Lecture
Mechanical Advantage, Velocity Ratio, and Load Sharing
by Dr Arnaldo Delli Carri
Pulleys are simple machines that use grooved wheels and ropes to change the direction of forces and provide mechanical advantage. They are fundamental components in lifting systems, reducing the effort force required to lift heavy loads.
Pulleys let you lift heavy loads with less force by trading force for distance. The more rope segments supporting the load, the greater the mechanical advantage.
There are three basic types of pulley systems:
A fixed pulley is attached to a stationary point. It changes the direction of the applied force but provides no mechanical advantage.
Fixed Pulley:
\[F_{\text{effort}} = F_{\text{load}}\]
\[\text{Mechanical Advantage (MA)} = 1\]
Where:
A movable pulley is attached to the load and moves with it. Two rope segments support the load, providing a mechanical advantage of 2 in the ideal case.
Movable Pulley (Ideal):
\[F_{\text{effort}} = \frac{F_{\text{load}}}{2}\]
\[\text{MA} = 2\]
A compound pulley combines fixed and movable pulleys to achieve greater mechanical advantage.
Mechanical advantage (MA) is the ratio of load force to effort force. It tells us how much a machine multiplies the input force.
Actual Mechanical Advantage (AMA):
\[\text{AMA} = \frac{F_{\text{load}}}{F_{\text{effort}}}\]
Ideal Mechanical Advantage (IMA):
\[\text{IMA} = \frac{d_{\text{effort}}}{d_{\text{load}}} = n\]
Where \(n\) is the number of rope segments supporting the load.
Important: The IMA equals the number of rope segments supporting the movable pulley. Each segment shares the load equally in an ideal system.
The velocity ratio (VR) is the ratio of the distance moved by the effort to the distance moved by the load.
Velocity Ratio:
\[\text{VR} = \frac{d_{\text{effort}}}{d_{\text{load}}}\]
For an ideal pulley system: \(\text{VR} = \text{IMA}\)
Real pulley systems have friction in the bearings and rope, making them less than 100% efficient.
Efficiency:
\[\eta = \frac{\text{AMA}}{\text{IMA}} \times 100\% = \frac{W_{\text{output}}}{W_{\text{input}}} \times 100\%\]
Work Relations:
\[W_{\text{input}} = F_{\text{effort}} \times d_{\text{effort}}\]
\[W_{\text{output}} = F_{\text{load}} \times d_{\text{load}}\]
\[W_{\text{lost}} = W_{\text{input}} - W_{\text{output}}\]
When multiple pulleys are combined, the overall mechanical advantage can be calculated by:
For pulleys in series:
\[\text{MA}_{\text{total}} = \text{MA}_1 \times \text{MA}_2 \times \ldots \times \text{MA}_n\]
For typical block and tackle:
\[\text{IMA} = n = \text{number of supporting rope segments}\]
Energy Conservation: Even though pulleys provide mechanical advantage in force, the total work done remains the same (neglecting friction). What you gain in force, you sacrifice in distance: \(F_{\text{effort}} d_{\text{effort}} = F_{\text{load}} d_{\text{load}}\)
Explore different pulley configurations and observe how mechanical advantage affects the effort force required to lift a load. Adjust the load weight and friction to see how they impact efficiency.
Observation: Notice how the compound system reduces the effort force significantly, but requires pulling much more rope. The work done remains constant (accounting for friction losses).
| Concept | Equation | Description |
|---|---|---|
| Ideal Mechanical Advantage | \(\text{IMA} = n\) | Number of rope segments supporting load |
| Actual Mechanical Advantage | \(\text{AMA} = \frac{F_{\text{load}}}{F_{\text{effort}}}\) | Ratio of load to effort force |
| Velocity Ratio | \(\text{VR} = \frac{d_{\text{effort}}}{d_{\text{load}}}\) | Ratio of distances moved |
| Efficiency | \(\eta = \frac{\text{AMA}}{\text{IMA}} \times 100\%\) | Effectiveness of the system |
| Work Input | \(W_{\text{input}} = F_{\text{effort}} \times d_{\text{effort}}\) | Energy supplied to the system |
| Work Output | \(W_{\text{output}} = F_{\text{load}} \times d_{\text{load}}\) | Useful work done by the system |
| Fixed Pulley MA | \(\text{MA} = 1\) | Changes direction only |
| Movable Pulley MA | \(\text{MA} = 2\) | Two rope segments support load |
A worker uses a movable pulley to lift a 300 N crate. Assuming an ideal system with no friction:
Show Solution
Given: \(F_{\text{load}} = 300\) N, movable pulley (IMA = 2), ideal system
a) Effort force:
For an ideal movable pulley: \(\text{IMA} = 2\)
\[F_{\text{effort}} = \frac{F_{\text{load}}}{\text{IMA}} = \frac{300}{2} = 150 \text{ N}\]
b) Rope distance:
Given: \(d_{\text{load}} = 2\) m
\[d_{\text{effort}} = \text{IMA} \times d_{\text{load}} = 2 \times 2 = 4 \text{ m}\]
c) Work calculations:
\[W_{\text{input}} = F_{\text{effort}} \times d_{\text{effort}} = 150 \times 4 = 600 \text{ J}\]
\[W_{\text{output}} = F_{\text{load}} \times d_{\text{load}} = 300 \times 2 = 600 \text{ J}\]
Since the system is ideal, \(W_{\text{input}} = W_{\text{output}}\) ✓
Answer: a) 150 N, b) 4 m, c) \(W_{\text{input}} = W_{\text{output}} = 600\) J
A compound pulley system with 4 supporting rope segments is used to lift a 800 N load. Due to friction, the actual effort force required is 250 N. The load is lifted 1.5 metres.
Show Solution
Given: \(n = 4\) segments, \(F_{\text{load}} = 800\) N, \(F_{\text{effort}} = 250\) N, \(d_{\text{load}} = 1.5\) m
a) Ideal mechanical advantage:
\[\text{IMA} = n = 4\]
b) Actual mechanical advantage:
\[\text{AMA} = \frac{F_{\text{load}}}{F_{\text{effort}}} = \frac{800}{250} = 3.2\]
c) Efficiency:
\[\eta = \frac{\text{AMA}}{\text{IMA}} \times 100\% = \frac{3.2}{4} \times 100\% = 80\%\]
d) Rope distance:
\[d_{\text{effort}} = \text{IMA} \times d_{\text{load}} = 4 \times 1.5 = 6 \text{ m}\]
e) Work lost to friction:
\[W_{\text{input}} = F_{\text{effort}} \times d_{\text{effort}} = 250 \times 6 = 1500 \text{ J}\]
\[W_{\text{output}} = F_{\text{load}} \times d_{\text{load}} = 800 \times 1.5 = 1200 \text{ J}\]
\[W_{\text{lost}} = W_{\text{input}} - W_{\text{output}} = 1500 - 1200 = 300 \text{ J}\]
Answer: a) IMA = 4, b) AMA = 3.2, c) \(\eta\) = 80%, d) 6 m, e) 300 J lost
An engineer needs to design a pulley system to lift a 1200 N engine block. The maximum force a worker can comfortably apply is 300 N. The system should be at least 75% efficient.
Show Solution
Given: \(F_{\text{load}} = 1200\) N, \(F_{\text{effort,max}} = 300\) N, \(\eta = 75\% = 0.75\)
a) Minimum IMA required:
We need: \(\text{AMA} \geq \frac{F_{\text{load}}}{F_{\text{effort,max}}} = \frac{1200}{300} = 4\)
Using: \(\eta = \frac{\text{AMA}}{\text{IMA}}\)
\[\text{IMA} = \frac{\text{AMA}}{\eta} = \frac{4}{0.75} = 5.33\]
Round up to nearest integer: \(\text{IMA} = 6\)
b) Supporting rope segments:
\[n = \text{IMA} = 6 \text{ segments}\]
c) Actual effort force at 75% efficiency:
With IMA = 6 and η = 0.75:
\[\text{AMA} = \eta \times \text{IMA} = 0.75 \times 6 = 4.5\]
\[F_{\text{effort}} = \frac{F_{\text{load}}}{\text{AMA}} = \frac{1200}{4.5} = 267 \text{ N}\]
This is within the worker's 300 N limit ✓
Answer: a) IMA \(\geq\) 5.33, rounded up to 6, b) \(n\) = 6 segments, c) \(F_{\text{effort}} \approx 267\) N