IB MYP 4-5 Physics- Levers and torque- Study Notes - New Syllabus
IB MYP 4-5 Physics-Levers and torque- Study Notes
Key Concepts
- Levers and torque
Levers & Torque
Levers
A lever is a simple machine consisting of a rigid bar that pivots around a fixed point called the fulcrum.
It is used to amplify force, change the direction of force, or make tasks easier. The working of a lever depends on the position of the fulcrum, load, and effort.
\(\text{Mechanical Advantage (MA)} = \dfrac{\text{Load}}{\text{Effort}}\)
\(\text{Velocity Ratio (VR)} = \dfrac{\text{Distance moved by effort}}{\text{Distance moved by load}}\)
\(\text{Efficiency} = \dfrac{\text{MA}}{\text{VR}} \times 100\%\)
- First-Class Lever: Fulcrum between effort and load (e.g., scissors, crowbar).
- Second-Class Lever: Load between fulcrum and effort (e.g., wheelbarrow, nutcracker).
- Third-Class Lever: Effort between fulcrum and load (e.g., broom, tweezers).
Example:
A wheelbarrow is used to lift a load of \(200 \ \text{N}\). The effort arm is \(1.5 \ \text{m}\) and the load arm is \(0.5 \ \text{m}\). Calculate the effort required.
▶️ Answer/Explanation
From the principle of levers: \(\text{Effort} \times \text{Effort Arm} = \text{Load} \times \text{Load Arm}\)
\(E \times 1.5 = 200 \times 0.5\)
\(1.5E = 100\)
\(E = \dfrac{100}{1.5} = 66.67 \ \text{N}\)
\(\boxed{66.67 \ \text{N}}\)
Example:
A gardener uses a wheelbarrow to carry soil. The wheel acts as the pivot, the handles are where the effort is applied, and the load is the soil in the bucket. If the distance from the wheel (pivot) to the handles is \( 1.5 \, \text{m} \), and the load is \( 200 \, \text{N} \) placed \( 0.5 \, \text{m} \) from the pivot, find the effort force required to lift the load in equilibrium.
▶️ Answer/Explanation
Using the principle of moments:
\(\text{Effort} \times 1.5 = 200 \times 0.5\)
\(\text{Effort} \times 1.5 = 100\)
\(\text{Effort} = \dfrac{100}{1.5} \approx 66.7 \, \text{N}\)
Final Answer: \(\boxed{66.7 \, \text{N}}\)
Torque
Torque (or moment of force) is the turning effect produced when a force is applied at a distance from a pivot point.
\(\text{Torque} = \text{Force} \times \text{Perpendicular distance from pivot}\)
- Unit: Newton meter (N·m)
- Vector quantity – direction given by the right-hand rule.
- Positive torque – anticlockwise rotation.
- Negative torque – clockwise rotation.
Principle of Moments: For a body in rotational equilibrium:
\(\text{Sum of clockwise moments} = \text{Sum of anticlockwise moments}\)
Example:
A seesaw is balanced. A child of mass \(30 \ \text{kg}\) sits \(2 \ \text{m}\) from the pivot. At what distance should another child of mass \(40 \ \text{kg}\) sit to balance it?
▶️ Answer/Explanation
Weight of first child: \(30 \times 9.8 = 294 \ \text{N}\)
Weight of second child: \(40 \times 9.8 = 392 \ \text{N}\)
From the principle of moments: \(294 \times 2 = 392 \times d\)
\(588 = 392d\)
\(d = \dfrac{588}{392} = 1.5 \ \text{m}\)
\(\boxed{1.5 \ \text{m}}\)
Example:
A spanner of length \( 0.25 \, \text{m} \) is used to loosen a nut. If a force of \( 150 \, \text{N} \) is applied perpendicular to the spanner, calculate the torque produced.
▶️ Answer/Explanation
Torque formula:
\(\tau = F \times d\)
\(\tau = 150 \times 0.25\)
\(\tau = 37.5 \, \text{N·m}\)
Final Answer: \(\boxed{37.5 \, \text{N·m}}\)