IB MYP 4-5 Physics- Types of forces and their effects on motion- Study Notes - New Syllabus
IB MYP 4-5 Physics-Types of forces and their effects on motion- Study Notes
Key Concepts
- Types of forces and their effects on motion
Types of forces and their effects on motion
Force
A force is a push or pull that can change the motion or shape of an object.
- It is a vector quantity, meaning it has both magnitude and direction.
- SI unit: Newton (N).
- Forces can cause an object to start moving, stop moving, change speed, change direction, or change shape.
Mathematically: \( F = m \times a \) (Newton’s Second Law)
Balanced and Unbalanced Forces
- Balanced Forces: Occur when the net force acting on an object is zero. The object remains at rest or moves with constant velocity (no acceleration).
- Unbalanced Forces: Occur when the net force is not zero. The object will accelerate in the direction of the net force.
Mathematically: \(\Sigma F = 0 \quad \text{(Balanced)} \quad \Rightarrow \quad a = 0\) \(\Sigma F \neq 0 \quad \text{(Unbalanced)} \quad \Rightarrow \quad a \neq 0\)
Effects of Balanced Forces
- No change in speed or direction of motion.
- If at rest, stays at rest; if moving, continues with the same velocity.
Effects of Unbalanced Forces
- Changes the speed of an object (acceleration or deceleration).
- Changes the direction of motion.
- Can change the shape of an object.
Adding Forces (Resultant Force)
- When multiple forces act on an object, they can be combined into a single force called the resultant force.
- If forces are in the same direction: \( F_{\text{net}} = F_1 + F_2 \)
- If forces are in opposite directions: \( F_{\text{net}} = |F_1 – F_2| \)
- If forces act at an angle, use the vector addition formula: \( F_{\text{net}} = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \)
Example:
Two forces of \( 8\,\text{N} \) and \( 6\,\text{N} \) act on an object in the same direction. What is the resultant force? What happens if they act in opposite directions?
▶️ Answer/Explanation
Case 1 (Same Direction): \( F_{\text{net}} = 8 + 6 = \boxed{14\,\text{N}} \) in the same direction.
Case 2 (Opposite Directions): \( F_{\text{net}} = |8 – 6| = \boxed{2\,\text{N}} \) in the direction of the larger force.
Example:
Two people push a box in the same direction with forces of \(40 \, \text{N}\) and \(60 \, \text{N}\). Find the net force.
▶️ Answer/Explanation
Since the forces are in the same direction: \(F_{\text{net}} = 40 + 60 = \boxed{100 \, \text{N}}\)
Example:
A tug-of-war team pulls with \(500 \, \text{N}\) to the left and the other team pulls with \(480 \, \text{N}\) to the right. Determine the net force and direction.
▶️ Answer/Explanation
Since forces are in opposite directions: \(F_{\text{net}} = 500 – 480 = 20 \, \text{N}\) to the left.
\(\boxed{20 \, \text{N to the left}}\)
Example:
Two forces of \(6 \, \text{N}\) and \(8 \, \text{N}\) act at right angles. Find the resultant force.
▶️ Answer/Explanation
Using Pythagoras: \(F_{\text{net}} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = \boxed{10 \, \text{N}}\)
Example:
A box is pulled by \(50 \, \text{N}\) to the east and \(30 \, \text{N}\) to the north. Find the resultant force and its direction.
▶️ Answer/Explanation
Resultant: \(F_{\text{net}} = \sqrt{50^2 + 30^2} = \sqrt{2500 + 900} = \sqrt{3400} \approx 58.31 \, \text{N}\)
Direction: \(\theta = \tan^{-1} \left( \dfrac{30}{50} \right) \approx 30.96^\circ\) north of east.
\(\boxed{58.31 \, \text{N at } 30.96^\circ \, \text{N of E}}\)
Tension Force
The tension force is the pulling force transmitted through a rope, string, cable, or chain when it is pulled tight by forces acting at both ends. It always acts along the length of the rope and away from the object it is attached to.
Symbol: \( T \)
SI Unit: Newton (\( N \))
- Tension is the same throughout an ideal (massless, frictionless) rope in equilibrium.
- If the rope has mass or acceleration is involved, tension can vary along its length.
- Tension always pulls, never pushes.
Example:
A 20 kg mass hangs vertically from a rope. Find the tension in the rope.
▶️ Answer/Explanation
Given: \( m = 20\ \text{kg},\ g = 9.8\ \text{m/s}^2 \)
Since the object is in equilibrium: \( T = mg = (20)(9.8) = 196\ \text{N} \)
\(\boxed{T = 196\ \text{N}}\)
Example:
A 5 kg block is pulled along a frictionless horizontal surface by a rope making a \( 40^\circ \) angle with the horizontal. The acceleration is \( 2\ \text{m/s}^2 \). Find the tension in the rope.
▶️ Answer/Explanation
Given: \( m = 5\ \text{kg},\ a = 2\ \text{m/s}^2,\ \theta = 40^\circ \)
Horizontal direction: \( T\cos\theta = ma \)
\( T = \dfrac{ma}{\cos\theta} = \dfrac{(5)(2)}{\cos 40^\circ} \)
\( T = \dfrac{10}{0.766} \approx 13.05\ \text{N} \)
\(\boxed{T \approx 13.05\ \text{N}}\)