Physics
Key Stage 4
Module 4 Force
Tutorial 4.4 Force and
elasticity Tutorial 4.3
Resolution of forces Tutorial 4.2
Resultant forces Tutorial 4.1
Forces
Knowledge Check
In this module we will look at:
1. Drawing and understanding free body diagrams 2. Resolving forces
3. Understanding Hooke’s law
TO BE ADDED – info about completing knowledge check for this module…
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Tutorial 4.1 Forces
In this tutorial we will look at:
1. Free body diagrams
2. Examples of forces
3. Weight
Free body diagrams
In physics, there are scalar quantities and vector quantities:
o Scalar quantities have magnitude only
o Vector quantities have magnitude and an associated direction For example, energy is a scalar quantity. It has a magnitude only.
Force is a vector quantity. It has a magnitude and an associated direction. Stating the direction of the force is important, because it matters in what direction the force is acting.
We can represent the magnitude and direction of the forces acting on an object using a free body diagram.
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Free body diagrams
The free body diagram below shows a 5 N force acting leftward on a body, and a 10 N force acting rightward on the body. The free body diagram is drawn to scale using the scale of 1cm = 1N.
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Free body diagrams
The free body diagram shows a 5 N force acting rightward on a body at 45° to the horizontal, and a 2 N force acting vertically downward.
The free body diagram is drawn to scale using the scale of 1 cm = 1 N and a protractor is used to set the angle.
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Free body diagrams
Which row describes the forces shown in the free body diagram?
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Vertical Horizontal
1 5 N 4 N
2 4 N 5 N
3 5 N 2 N
4 2 N 5 N
Examples of forces
A force is a push, pull or twist that acts on an object due to an interaction with another object.
All forces acting between objects are either:
o contact forces: the objects are physically touching. Examples of contact forces include friction, air resistance, tension, upthrust, lift and normal contact force.
o non-contact forces: the objects are physically separated.
Examples of non-contact forces are gravitational force (weight), electrostatic force and magnetic force.
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Examples of forces
A child pulls a block by a string across the floor. The force on the block due to the tension in the string is 15 N. The friction exerted on the block due to the floor is 15 N. The weight of the block is 20 N. The normal contact force exerted on the block due to the floor is 20 N.
Draw a free body diagram to scale representing this situation. Label the forces.
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Examples of forces
An aircraft ascends to cruising altitude. The thrust (forward force) on the plane is 2.5 kN. The air resistance acting on the aircraft is 2.5 kN.
The lift on the aircraft is 7.5 kN. The weight of the aircraft is 5 kN. Draw a free body diagram to scale representing this situation. Label the forces.
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Examples of forces
A book sits on a table. The free body diagram for the book is shown below. Which row is correct?
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Force A Force B
1 Weight Normal reaction force
2 Weight Upthrust
3 Normal reaction force Weight
4 Upthrust Weight
Weight
Weight is the force acting on an object due to gravity. The weight of an object depends on the gravitational field strength at the point
where the object is. The weight of an object can be calculated using the equation:
𝑊 = 𝑚𝑔
where:
o W is weight (in units of newtons, N) o m is mass (in units of kilograms, kg)
o g is gravitational field strength (in units of newtons per kilogram, N/kg) The weight of an object may be considered to act at a single point referred to as the object’s ‘centre of mass’. The centre of mass is the dot shown in the free body diagrams above.
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Weight
Calculate the weight of a 5 kg object on (a) the Earth, where the gravitational field strength is 9.8 N/kg (b) the Moon, where the gravitational field strength is 1.6 N/kg.
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m = 5 kg g = 9.8 N/kg
W = mg = 5 x 9.8 = 49 N
On Moon:
m = 5 kg g = 1.6 N/kg
W = mg = 5 x 1.6 = 8 N
The object is about 6 times lighter on the Moon compared to on the Earth.
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Physics
Key Stage 4
Tutorial 4.2
Resultant forces
In this tutorial we will look at:
Finding resultant force for forces acting along a line
Finding resultant force for forces acting at right angles
Finding resultant force for forces at an angle to each other
Finding resultant force for forces acting along a line
A number of forces acting on an object may be replaced by a single force that has the same effect as all the original forces acting
together. This single force is called the resultant force.
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Finding resultant force for forces acting along a line
Question
The free body diagram for a book sitting on a table is shown. What is the resultant force on the book?
Answer
If we take upward as positive and
downward as negative, then the resultant force F = + 15 – 15 = 0 N. There is no resultant force on the book – the forces balanced and cancel out.
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Finding resultant force for forces acting along a line
Question
The free body diagram for a book being pushed across a table is shown. Draw the free body diagram for the book showing the resultant force.
Answer
We established in the previous example that the normal contact force and weight cancel out. We are thus left with the friction and push. The resultant force will be 10 N horizontally to the right:
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Finding resultant force for forces acting along a line
In which of these situations is the magnitude of the resultant force largest?
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Finding resultant force for forces acting along at right angles
How do we determine the resultant force on the body in the vector diagram? We need to add the forces to determine the resultant force. We can use the ‘parallelogram of forces’ in the vector diagram.
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Finding resultant force for forces acting along at right angles
Step 1: Draw a parallelogram in which the two forces are adjacent sides of the parallelogram. In this case, the parallelogram is simply a rectangle (dashed line)
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Finding resultant force for forces acting along at right angles
Step 2: Draw an arrow across the diagonal of the parallelogram (rectangle), from the body to the opposite corner. This arrow
represents the addition of the two forces, and the resultant force on the body.
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Finding resultant force for forces acting along at right angles
Step 3: Use the scale to determine the magnitude of the resultant force and use a protractor to measure an angle so you can state its direction. In this case, the resultant force is 5.4 N and directed 22° to the horizontal.
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Finding resultant force for forces acting along at right angles
Step 3: Use the scale to determine the magnitude of the resultant force and use a protractor to measure an angle so you can state its direction. In this case, the resultant force is 5.4 N and directed 22° to the horizontal.
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Finding resultant force for forces acting along at right angles
Two forces (solid arrows) act at a right angle on a body. Which dashed arrow represents the resultant force on the body?
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Finding resultant force for forces at an angle to each other
How do we determine the resultant force on the body in the vector diagram? We need to add the forces to determine the resultant force. We can use the ‘parallelogram of forces’ in the vector diagram.
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Finding resultant force for forces at an angle to each other
Step 1: Draw a parallelogram in which the two forces are adjacent sides of the parallelogram.
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Finding resultant force for forces at an angle to each other
Step 2: Draw an arrow across the diagonal of the parallelogram, from the body to the opposite corner. This arrow represents the addition of the two forces, and the resultant force on the body.
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Finding resultant force for forces at an angle to each other
Step 3: Use the scale to determine the magnitude of the resultant force and use a protractor to measure an angle so you can state its direction. In this case, the resultant force is 7.2 N and the directed 33°
to the horizontal.
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Finding resultant force for forces at an angle to each other
Two forces (solid arrows) act on a body. Which dashed arrow represents the resultant force on the body?
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Physics
Key Stage 4
Tutorial 4.3
Resolution of forces
In this tutorial we will look at:
Resolving a force into components
Forces on an inclined plane
Resolution of weight on an inclined plane
Resolving a force into components
A single force can be resolved into two components acting at right angles to each other. The two component forces together have the same effect as the single force.
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The vector diagram shows a force acting on an object. Resolve the force into its horizontal and vertical components.
Answer
To resolve the force into its
components, we can apply the steps that were used for finding the
resultant force – but in reverse.
Resolving a force into components
A single force can be resolved into two components acting at right angles to each other. The two component forces together have the same effect as the single force.
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Draw a rectangle around the force vector, such that the vector is the diagonal of the rectangle.
Resolving a force into components
A single force can be resolved into two components acting at right angles to each other. The two component forces together have the same effect as the single force.
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Draw horizontal and vertical force vectors from the body, along the sides of the rectangle. Use the scale to determine the magnitude of these horizontal and vertical components.
Resolving a force into components
The vector diagram shows a force acting on a body. Which statement is correct?
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1. The horizontal and vertical components of the force are equal in magnitude
2. The horizontal component of the force is greater in magnitude than the vertical component
3. The horizontal component of the force is lesser in magnitude than the vertical component
Forces on an inclined plane
‘Inclined plane’ or ‘ramp’ problems are common in physics.
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The vector diagram shows a body on an inclined plane. Three forces act on the body:
o Weight due to gravity. This is vertically downwards.
o Normal contact force. This is the
reaction force on the body due to the ramp, which acts normal
(perpendicular) to the ramp.
o Friction on the body due to the surface of the ramp. This acts along the incline, and in a direction opposite to any
motion that the body might have.
Forces on an inclined plane
The free-body diagram is for a mass on an inclined plane. Which statement is incorrect?
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1. The forces are weight, normal contact force and friction.
2. The normal contact force is perpendicular to the plane.
3. The weight of the mass is perpendicular to the plane.
4. The force of friction is parallel to the plane.
Resolution of weight on an inclined plane
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The vector diagram shows the weight of a body on an inclined plane (other forces, such as normal contact force, have not been drawn). Resolve the weight into components that parallel and perpendicular to the plane, and determine the magnitude of these components.
Resolution of weight on an inclined plane
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Step 1
Draw a rectangle - in this case, a
square - around the weight vector, so that the weight vector is the diagonal of the square.
Resolution of weight on an inclined plane
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Step 2
Draw force vectors from the body along the sides of the square that are parallel and perpendicular to the inclined plane. These are the parallel and perpendicular components of the weight.
Resolution of weight on an inclined plane
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Step 3
These components represent the weight (hence, the weight vector is no longer shown in the vector
diagram). Using the scale, we can see that the component of the
weight that is parallel to the ramp is 1.4 N, and the component of the weight that is perpendicular to the ramp is 1.4 N.
Resolution of weight on an inclined plane
The diagram shows a body on a ramp. The components of the body’s weight parallel and perpendicular to the ramp have been drawn. What is the weight of the body?
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2. 40 N 3. 50 N 4. 60N
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Physics
Key Stage 4
Tutorial 4.4
Force and elasticity
In this tutorial we will look at:
Elasticity
Hooke’s Law
Equation for Hooke’s Law
Elasticity
The shape of an object can be changed by applying forces to the object. The change in shape is called a deformation. Elastic objects return to their original shape when the forces are removed. Inelastic objects do not return to original shape when the forces are removed.
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Elasticity
An example of an elastic object is a spring. The image shows springs undergoing deformations
(compression or extension) when a force is applied. In this case, the force is the weight of the mass. The deformations will be reversed when the force is
removed.
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Elasticity
A student investigates an extended spring. Which calculation is correct?
1. new length = natural length – extension 2. extension = natural length + new length 3. natural length = extension + new length 4. new length = natural length + extension
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Hooke’s Law
This graph shows the extension of a spring against force applied, which illustrated Hooke’s Law.
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The extension of an elastic object, such as a spring, is directly proportional to the force applied, provided that the limit of proportionality is not exceeded.
Hooke’s Law
A student plotted extension vs force data for two elastic bands.
Which statement is incorrect?
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1. X extends more than Y under 5 N of force.
2. Y exceeds the limit of proportionality
3. More force is required to extend X by 0.05 m than to extend Y by 0.05 m.
4. Between 0 and 5 N, Y is stiffer than X.
Equation for Hooke’s Law
For the range of forces over which there is a direct proportionality between extension and force, Hooke’s law can be stated as an equation:
F=ke
where,
o F is force (in units of newtons, N) o e is extension (in units of metres, m)
o k is the stiffness or spring constant (in units of newtons per metre, N/m)
By comparing the equation to the equation for a straight-line (y=mx), we can see the k is the gradient of the graph F against e. Alternatively, we can see that 1/k is the gradient of the graph e against F.
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Equation for Hooke’s Law
A rubber band with stiffness 2000 N/m is extended by 2 cm.
Determine the force applied to the rubber band.
Answer
e = 2 cm = 0.02 m ← convert units straight away k = 2000 N/m
F = ke = 2000 x 0.02 = 40 N
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F=ke
Equation for Hooke’s Law
A spring is compressed by 0.1 m when a force of 10 N is applied to it. The spring constant for the spring is:
1. 0.01 N/m 2. 1 N/m 3. 10 N/m 4. 100 N/m
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Knowledge Check
In this module we will looked at:
1. Drawing and understanding free body diagrams 2. Resolving forces
3. Understanding Hooke’s law
TO BE ADDED – info about completing knowledge check for this module…
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Reflection
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