3.1.1 Force Diagrams

Force diagrams, also known as free-body diagrams (FBDs), are essential tools in mechanics. They help visualize the forces acting on an object, making it easier to analyse the object's motion or equilibrium. Here's a guide to understanding and creating force diagrams:

1. What is a Force Diagram?

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A force diagram is a simplified representation of an object (or system) with all the forces acting on it illustrated as vectors.
The object is usually represented by a simple shape, like a dot or a box, and the forces are shown as arrows pointing in the direction in which the forces act.

2. Steps to Draw a Force Diagram

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Identify the object: Determine which object you are analysing and isolate it from the surroundings.
Draw the object: Represent the object by a simple shape, often a point (for simplicity) or a box if more context is needed.
Identify all the forces acting on the object: Consider forces such as gravity, normal force, tension, friction, applied forces, and air resistance.
Draw the forces as arrows:

The length of each arrow represents the magnitude of the force.
The direction of the arrow shows the direction in which the force acts.
Label each force clearly (e.g., $F_{\text{gravity}}$ , $F_{\text{normal}}$ , $F_{\text{friction}}$ , etc.).

Check for equilibrium or net force:

If the object is in equilibrium (stationary or moving at constant velocity), the forces should balance, meaning the arrows should add up to zero.
If the object is accelerating, there will be a net force in the direction of acceleration.

3. Common Forces to Include

Weight $( F_{\text{gravity}} )$ : Acts downward, equal to the mass of the object times the acceleration due to gravity $( mg )$ .
Normal Force $( F_{\text{normal}} )$ : Acts perpendicular to the surface on which the object rests, opposing the weight if on a horizontal surface.
Tension $( F_{\text{tension}} )$ : If the object is connected to a rope or string, tension acts along the string.
Friction $( F_{\text{friction}} )$ : Opposes the motion or attempted motion between two surfaces in contact.
Applied Force $( F_{\text{applied}} )$ : Any external force applied to the object.
Air Resistance $( F_{\text{air}} )$ : Acts opposite to the direction of motion, more significant at higher speeds.

4. Worked Example

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Problem: A box of mass $5$ $kg$ is on a flat surface. A force of $20$ $N$ is applied horizontally to the right. The coefficient of friction between the box and the surface is $0.3$ . Draw the force diagram.

Step 1: Identify forces acting on the box.

Weight $( F_{\text{gravity}} )$ : Downward, $F_{\text{gravity}} = mg = 5 \times 9.81 = :highlight[49.05 \, \text{N}]$ .
Normal Force $( F_{\text{normal}} )$ : Upward, balancing the weight, so $F_{\text{normal}} = :highlight[49.05 \, \text{N}]$ .
Applied Force $( F_{\text{applied}} )$ : $20$ $N$ to the right.
Frictional Force $( F_{\text{friction}} )$ : Opposes motion, calculated using $F_{\text{friction}} = \mu F_{\text{normal}} = 0.3 \times 49.05 = :highlight[14.715 \, \text{N}]$ to the left.

Step 2: Draw the diagram.

Draw a box or point to represent the object.
Draw arrows representing each force:
Weight pointing downwards.
Normal force pointing upwards.
Applied force pointing to the right.
Frictional force pointing to the left.

Final Diagram: The box will have:

A downward arrow labelled $49.05 \, \text{N}$ for gravity.
An upward arrow labelled $49.05 \, \text{N}$ for the normal force.
A rightward arrow labelled $20 \, \text{N}$ for the applied force.
A leftward arrow labelled $14.715 \, \text{N}$ for friction.

5. Interpreting the Diagram

Net Force: You can see that the applied force and friction are not equal, so the box will accelerate to the right. The net force in the horizontal direction is $20 \, \text{N} - 14.715 \, \text{N} = 5.285 \, \text{N}$ .
Equilibrium: If all forces balanced out (e.g., if friction equalled the applied force), the object would be in equilibrium, and there would be no acceleration.

Resolving Non-Perpendicular Forces

Problem Statement:

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Question : Find the magnitude and direction of the resultant of the given forces.

Solution:

Step 1: Resolve the Forces Horizontally and Vertically

Horizontal Component:

15 \cos 40^\circ - 12 \cos 35^\circ \approx :highlight[1.6608 \, \text{N}]

(Any forces with a component acting horizontally should be considered.)

Vertical Component:

10 + 15 \sin 40^\circ - 12 \sin 35^\circ \approx :highlight[12.7589 \, \text{N}]

(Any forces with a component acting vertically should be considered.)

Step 2: Calculate the Magnitude and Direction

Magnitude of the Resultant Force $|F|$ :

|F| = \sqrt{(1.6608)^2 + (12.7589)^2} \approx :highlight[12.87 \, \text{N}]

Direction $\alpha$ relative to the positive horizontal:

\alpha = \tan^{-1}\left(\frac{12.7589}{1.6608}\right) \approx :highlight[82.58^\circ]

Final Answer:

Magnitude: $|F| \approx 12.87 \, \text{N}$
Direction: $82.58^\circ$ above the positive horizontal (Note: Always provide a reference line for the direction.)

Force Diagrams Simplified Revision Notes for A-Level AQA Maths Mechanics

3.1.1 Force Diagrams

1. What is a Force Diagram?

2. Steps to Draw a Force Diagram

3. Common Forces to Include

4. Worked Example

Problem: A box of mass $5$ $kg$ is on a flat surface. A force of $20$ $N$ is applied horizontally to the right. The coefficient of friction between the box and the surface is $0.3$ . Draw the force diagram.

5. Interpreting the Diagram

Resolving Non-Perpendicular Forces

Problem Statement:

Solution:

Step 1: Resolve the Forces Horizontally and Vertically

Step 2: Calculate the Magnitude and Direction

500K+ Students Use These Powerful Tools to Master Force Diagrams For their A-Level Exams.

Other Revision Notes related to Force Diagrams you should explore

Force Diagrams Simplified Revision Notes for A-Level AQA Maths Mechanics

3.1.1 Force Diagrams

1. What is a Force Diagram?

2. Steps to Draw a Force Diagram

3. Common Forces to Include

4. Worked Example

Problem: A box of mass 555 kgkgkg is on a flat surface. A force of 202020 NNN is applied horizontally to the right. The coefficient of friction between the box and the surface is 0.30.30.3. Draw the force diagram.

5. Interpreting the Diagram

Resolving Non-Perpendicular Forces

Problem Statement:

Solution:

Step 1: Resolve the Forces Horizontally and Vertically

Step 2: Calculate the Magnitude and Direction

500K+ Students Use These Powerful Tools to Master Force Diagrams For their A-Level Exams.

Other Revision Notes related to Force Diagrams you should explore

Problem: A box of mass $5$ $kg$ is on a flat surface. A force of $20$ $N$ is applied horizontally to the right. The coefficient of friction between the box and the surface is $0.3$ . Draw the force diagram.