BTEC Level 3 in Eng Unit 1: Mechanical Principles Assignment Answers

Pearson BTEC International Level 3 in Engineering 

Level: 3
Unit type: Internal set assignment
Guided learning hours: 60 

Unit 1 Introductin

Modern life depends on engineers to develop, support and control the mechanical products and systems that are all around us, for example cars, machinery and manufacturing and transport systems. To make a contribution as an engineer, you must be able to draw on an important range of principles developed by early engineering scientists, such as Archimedes, Isaac Newton and James Watt.

There is an increasing demand for ‘multi-skilled’ engineers who can apply principles from several engineering disciplines to develop solutions to engineering problems. This unit will develop your mathematical and physical science knowledge, and understanding to enable you to solve problems set in an engineering context. You will explore and apply the algebraic and trigonometric mathematical methods required to solve engineering problems. The mathematical and physical science principles covered in this unit’ or ‘the engineering principles covered in this unit.

This sits at the heart of the qualification and gives you a foundation to support you in any engineering technician role, a trainee job role with an employer, or to help with your progression to higher education.

Learning Aim A: Examine how algebraic and trigonometric mathematical methods can be used to solve engineering problems.

A.D1 Solve routine and non-routine problems accurately, using algebraic and trigonometric methods.

Answer:

To meet this criterion, you must be able to solve a variety of basic and non-routine mathematical problems algebraically and trigonometrically. Solutions should contain clear working, correct formula, accurate working and appropriate checking working. 

Algebraic Problem – Routine
Solutions:
 3x + 7 = 25
Subtracting 7 from both sides:
 3x=18
Divide both sides by 3:
 X=6
Therefore, the solution is: 
 X=6

This is a simple algebraic problem since it is following a clear sequence of operations.

Algebraic Problem – Non-routine

Solution:
The length of a rectangle is 5cm more than its width. Its area is 84cm284cm^2. Find its dimensions.

Let the width be x cm. 
Therefore, the length is: 
 X+5

Using the area formula: 
 X(x+5) = 84
Expanding:
 X2 + 5x – 84 = 0
Factorizing:
 (x+12) (x-7) = 0
Therefore:
 x=-12 or x=7

A negative length is not possible, so: 
 X=7
 Width = 7cm
Length: 
 7+5 = 12cm
Check
 7x12=84cm2

Therefore, the dimensions of the rectangle are:
 7 cm x 12 cm

Trigonometric Problem – routine

Solution:
Find the size of the side of the right-angled triangle opposite the angle of 35∘35^\circ, given that the hypotenuse is 10 cm. 
Using: 
 Sinθ = Opposite ÷ hypotenuse

Therefore: 
 Sin35°= x/10
Rearranging: 
 x=10sin 35°
 x≈5.74cm

Problem 4: Algebraic Word Problem

Solution:

A person is 25m away from a building. The angle of elevation to the top of the building is 62°. The person's height is 1.6m above the ground. Find the height of the building.

Let h be the vertical distance between the person’s eye level and the roof of the building.

Using: 
 Tanθ = opposite/adjacent
 Tan62° = h/25
Therefore:
 H=25 tan 62°
 H=47.02 m

The height of the person’s eye, above the ground, is 1.6m, so;
Building height = 47.02 =1.6
 48.62m
So, the height of the building is about 48.62 meters.

Accuracy and Checking

Algebraic and trigonometric techniques are illustrated in these examples in their use of simple and more complex problems. To increase accuracy:

• Finding the right mathematical techniques before performing the calculation;
• Demonstrating each step in the calculations:
• Using appropriate formulas;
• Keeping suitable decimal places;
• Substituting answers in the original equation or using original geometric relationship to check answers;
• Not accepting solutions that are not feasible, e.g a negative physical length.

Such examples are particularly important as they are not always easy to solve, especially since the method is not always obvious. The problem must first be interpreted, converted into a mathematical model, solved and then checked against the original situation. 

A.P1 Solve given routine problems using algebraic methods.

Answer:

Algebra provides the use of letters to represent unknown values and the use of mathematical rules to solve problems. Typical algebra problems typically have a very clear series of steps. The examples below illustrate how algebra can be used to determine unknown values accurately.

Problem 1: Solving a linear equation

Solution: 

• 4x+8 = 28
  First, subtract 8 from both sides:
• 4x=20
  Now divide both sides by 4:
•  x=5
  Therefore:
• x=5

Check:

• 4(5) + 8 = 28
• 20+8 = 28

The answer is correct

Problem 2: Equation involving Brackets

Solution

• 3(x+4) = 24
  Expand the brackets:
• 3x+12 = 24
  Subtract 12 from both sides:
• 3x=12
  Divide by 3:
• x=4
  Therefore: x=4

Problem 3: Solving a quadratic equation

Solve:

• x2 – 5x+6 = 0

Factorise the equation:

•  (x-2) (x-3) = 0

Therefore:

• x-2 = 0 or x-3 = 0
  So:
• x=2 or x=3

Therefore: the solutions are

• x= 2,3

Problem 4: Algebraic word Problem

Solution:

There is a number that is 30 more than 12. Find the number

Let the number the unknown represents be x.

The equation is:

• x+12 = 30

Subtract 12 from both sides:

•  x=18
  Therefore, the number is:
• 18 

Check:

• 18+12 = 30

Hence, the answer is right.

The following examples demonstrate how simple problems can be solved using simple algebraic skills such as rearranging equations, expanding brackets, factorising and substituting values. Include working at every step to make sure that answers are correct and can be verified easily.

A.P2 Solve given routine problems using trigonometric methods.

Answer:

The unknown sides and angles of a triangle can be found using trigonometry. The three primary trigonometric ratios are sine, cosine and tangent (SOH-CAH-TOA) for right-angled triangles. The following is a list of routine problems that illustrate these methods.

Problem 1: Solving for an unknown side using sine.

Solution:

The hypotenuse of a right-angled triangle is 12cm and one of the angles is 40°. In a right-angled triangle, the hypotenuse is 12cm and one angle is 40^\circ40. Determine the length of the side opposite the angle.

If you are using the sine ratio:

•  sinθ= Opposite/Hypotenuse

Replace the values that you know:

• sin40° =x/12

Rearranging:

• x=12sin40°
• x≈7.71

Therefore, the length of the opposite side is:

Problem 2: Find the length of the unknown side with cosine.

Solution:

The hypotenuse of a right-angled triangle is 15 m and an angle is 35∘35^\circ. Find the length of the side adjacent to the angle.

Using cosine:

• cosθ=Adjacent/Hypotenuse
  Therefore:
• cos35°=x/15
  Rearranging:
•  x=15cos35°
  Therefore:
  x≈12.29m

Problem 3: Using Tangent to find an unknown Side.
Solution:

In a right-angled triangle, one of the angles measures 50∘ 50^\circ. If the side opposite this angle is 8cm, what is the length of the opposite side?
Using tangent:

• tanθ=Opposite/Adjacent
  
  Substituting the values:
  
  
• tan50°=x/8
  
  Rearranging:
  
  
• x=8tan50°
  Thus, the opposite side is:
  9.53cm

Problem 4: Finding an unknown angle
Solution:

Find the angle θ\theta of a right angled triangle with an opposite side of length 6cm and the adjacent side of length 10cm.

Using tangent:

• tanθ=Opposite/Adjacent
• tanθ=6/10
• tanθ=0.6

Computing the arctan:

• θ=tan-1(0.6)
• θ≈30.96°

Therefore:

• θ≈30.96°

Checking the answers

To verify the answer, substitute the side or angle that is found into the trigonometric ratio. When the angles are in degrees, it is also important to ensure that the calculator is in degree mode. These examples illustrate how to solve common right triangle problems with the sine, cosine and tangent functions correctly.

 A.M1 Solve routine problems accurately and non-routine problems using both algebraic and trigonometric methods. 

Answer:

Straightforward routine problems and more complex non-routine problems can be solved using algebraic and trigonometric methods. Routine problems usually have a clear mathematical structure, while non-routine problems will need to interpret the situation before determining the appropriate method.

Routine algebraic problem

Solve:
•    5x−9=31
Add 9 to each side:
•    5x=40
Divide by 5:
•    x=8
Therefore:
•    5(8)-9=40-9=31
Check:
•    5(8) - 9=40−9=31
Therefore, the answer is right.

Routine trigonometric problem

A ladder is 6 m long and makes an angle of 65∘65^\circ with the ground. Find the height of the ladder.
The hypotenuse is the ladder, and the length that is required is the opposite leg. So, it makes sense to use sine:
•    sin65°=h/6
Rearranging:
•    h=6sin65°
•    h≈ 5.44m
Thus, the length of a ladder is about:
•    5.44m

Non-routine algebraic problem

Solution:

The area of a rectangular garden is 96m296m2. It is 4 m longer than wide. Determine the width and length of the garden.
Let the width be x metres.

Thus, the length is:
•    x+4
Based on the area formula:
•    x(x+4)=96
Expanding:
•    X2+4x−96=0
Factorising:
•    (x+12)(x−8)=0
Therefore:
•    x=−12 or x=8 
There is no negative measurement, so:
•    x=8
The width is 8m, and its length is:
•    8+4=12m
Check:
•    8×12=96m2
Hence, the dimensions are:
•    8m×12m

Non-routine trigonometric problem

A surveyor wishes to measure the height of a building. He is 30 m from the base and his angle of elevation to the top of the building is 58^\circ58°. His measuring instrument is at a height of 1.5 m above ground level. Determine the height of a building.

Firstly take into account the right-angled triangle between the instrument and the top of the building.

The adjacent side is 30 m long and the opposite side is the height from the instrument to the top. Therefore, the tangent can be used:

•    tan58°=h/30
Rearranging:
•    h=30tan58°
•    h≈ 47.99m
This is the distance above the instrument level. As the instrument is 1.5 m above ground:
•    Building height=47.99+1.5
•    47.99m
So, the height of the building is about 49.49 metres.

Accuracy and Method Selection

As these problems demonstrate, method selection will vary with the information provided. Algebraic methods can be used if the unknown quantity is included in an equation; trigonometric methods can be used if angles and sides of triangles are involved. In non-routine problems, it is important first to establish what is known, what is needed, and then to set up the mathematical relationship to perform the calculations.

Simplify each step when showing the calculation, ensure the correct formula used, ensure the calculator is in Degree mode for any trigonometric calculations, use suitable rounding and verify the final answer by comparing with the original problem. This will enable the solving of routine as well as non-routine problems in a logical and accurate manner.

Learning aim C: Examine how dynamic engineering systems can be used to solve engineering problems.

C.P5 Solve routine problems that involve kinetic and dynamic parameters.

Answer:

Kinematics is the study of the motion of objects, and dynamics refers to the study of forces that cause or affect the motion of objects. Parameters are distance, displacement, speed, velocity, acceleration, mass, force and time. If a problem comes with a set formula, then the given numbers can be substituted to find the answer to the problem.

Problem 1: Calculating speed

Solution

A car travels a distance of 180 m in 12 seconds. Find the average speed at which it travelled.

Speed is defined as:

Speed=Distance/Time
Substituting the values:
•    Speed=180/12
•    Speed=15m/s
The average speed of the car is then:
•    15m/s

Problem 2: Calculating acceleration

Solution

A motorcycle increases its velocity from 10 m/s to 25 m/s in 5 seconds. Calculate its acceleration.

The formula is:
•    a=v−u/t
Where:
•    u=10m/s (initial velocity)
•    v=25m/s (final velocity)
•    t=5s
Therefore:
•    a = (25−10)/5 
•    a= 3m/s2
Hence, the acceleration is:
•    3m/s2

Problem 3: Calculating force

Solution 
A trolley with a mass of 50kg accelerates at 2m/s22m/sec^2. Work out the resultant force on it.

Newton's 2nd Law:
•    F=ma
Substituting the values:
•    F=50×2
•    F=100N
Hence, the resultant force is:
•    100N

Problem 4: Calculating weight

Solution:
Given mass of an object = 20 kg, what is its weight with g = 9.81m/s2g = 9.81m/s^2?
The formula is:
•    W=mg
Therefore:
•    W=20×9.81
•    W=196.2N
So, the mass of the object is:
•    196.2N

Problem 5: Calculating distance travelled
Solution:

A car is initially still and accelerates at 3m/s2 for 8 seconds. Work out the distance travelled.

The initial velocity is:
•    u=0m/s
Using the equation:
•    s=ut+1/2at2
Substituting the values:
•    s= (0)(8) +1/2(3)(82)
•    s=1.5×64
•    s=96m
Therefore, the vehicle travels:
•    96m

The following examples show how to solve simple problems with kinetic and dynamic parameters by identifying what is given, choosing the right formula, and substituting the values correctly. Check these steps during the work to ensure the final answer is in the correct form.

C.P6 Solve routine problems that involve angular parameters.

Answer:

The rotation and motion of objects are described using angular parameters. Some commonly used angular quantities are angular displacement, angular velocity, angular acceleration, frequency and time period. The following are significant parameters in rotary engineering systems such as shafts, wheels, gears and motors.

Problem 1: Compute angular displacement.

Solution:

A wheel makes 5 complete revolutions. Find the angle it has turned through in radians.

There is one complete revolution every:
•    2pi radians
Therefore:
•    θ= 5(2π)
•    θ=10π
•    θ≈31.42 radians
So, the angle of displacement is:
•    31.42rad

Problem 2: How to calculate the angular velocity?
Solution:
A shaft makes a rotation of 20 radians in 4 seconds. Determine its average angular velocity.
The formula is:
•    ω= θ/t
Where:
•    ω= angular velocity
•    θ = angular displacement
•    t = Time
Substituting the values:
•    ω = 20/4
•    ω = 5 rad/s
Therefore:
•    5 rad/s

Problem 3: To convert revolutions per minute to angular velocity.
Solutions:
A motor is found with a rotation speed of 120 rpm. Determine its angular velocity in rad/s.
Convert revolutions per minute to revolutions per second:
•    120/60 =2rev/s
One revolution is the same as 2π radians:
•     ω = 2(2π)
•    ω= 4π
•    ω≈ 12.57 rad/s
Therefore:
•    12.57 rad/s
Problem 4: Angular acceleration is calculated
Solution:
This shaft is made to rotate from \(4\text{ rad/s}\) to \(16\text{ rad/s}\) in a time of 6 seconds. Determine its angular acceleration.
The formula is:
•    α= (ω-u)/t
Whereas u is the initial angular velocity.
Therefore:
•    α= (16-4)/6
•    α=12/6
•    α=2 rad/s2
So, the angular acceleration is:
•    2rad/s2

Problem 5: When the angular velocity is known, how do you find the linear?
Solution:
A wheel has a radius of 0.25 m and rotates at 8rad/s. Find the linear velocity at its circumference.
The ratio of linear velocity to angular velocity is:
•    v=rω
Substituting the values:
•    v= 0.25x8
•    v=2m/s
Therefore:
•    2m/s
These examples demonstrate that small and everyday problems involving angles can be solved by recognizing the appropriate angle parameter, and using the proper formula. Care must be taken with units, particularly when converting between revolutions, degrees and radians or between rpm and rad/s. Unit checking and the plugging in of the answer back into the appropriate relationship can help confirm accuracy of calculations.

C.M3 Solve routine and non-routine problems accurately that involve dynamic systems.

Answer:

Objects in motion or under the influence of forces are part of dynamic systems. Newton's laws of motion, equations of motion, momentum, work, energy and power may be necessary to solve problems of dynamic systems. Routine problems are usually problems where a formula is directly applied and non-routine problems are problems that involve multiple steps or the integration of multiple principles.
A problem dealing with the resultant force that is asked on a regular basis.

A vehicle has a mass of \(1,200\,kg\) and accelerates at \(2.5\,m/s^2\). Find the resultant force on the car.

From Newton's 2nd Law:

• F=ma
  Substituting the values:
• F=1200 x 2.5 
•  F=3000N
  So, the resultant force is:
• 3000N

Routine problem 2: Momentum

Solution:

A ball with a mass of \(0.5\,kg\) is travelling at \(20\,m/s\). Calculate its momentum.

The formula for momentum is:

• p=mv
  Therefore:
• p=0.5x20 
• p=10kgm/s
  Hence:
• 10kg m/s

Non-routine problem 1: Car speed and braking distance

A car with a mass of 1,000kg is travelling at 20 m/s. It comes to rest in 5 seconds.

Calculate:

1.    its acceleration;
2.    the braking force.

The initial velocity is:

• u=20 m/s
  The final velocity is:
• v=0 m/s
  and:
• t=5s
  Calculate the acceleration:
• a= (v-u)/t 
• a= (0-20)/5
• a= -4m/s2 
  The negative sign indicates that the car is slowing down.
  Use Newton's second law:
•  F=ma
•  F=1000(-4) 
• F=-4000N
  Thus, the braking force is equal to:
•   4000N
  and works in the opposite direction to the car.

Non-routine problem 2: Force and distance

Solution:

A trolley, of mass 20 kg, is given a constant resultant force of 60N for 10s, starting from rest. Find its acceleration, final velocity and distance.
First, calculate acceleration:

•  F=ma
  Rearranging:
•  a=F/m
•  a=60/20 =3m/s2
  The trolley is at rest, so:
  u=0m/s
  Using:
•  v=u+at
• v=0+(3)(10) 
• v=30m/s 
  Now find out how far he went:
•  s= ut+1/2at2 
•  s= (0)(10)+ 1/2(3)(102) 
• s= 150m 
  Therefore:
• a=3m/s2 
• v=30m/s 
• s=150m

Checking the results
Other equations can be used to validate the answers. For example:

• V2=u2+2as
  Substituting the values:
• 302=02+2(3) (150)
• 900=900

This verifies the values calculated are consistent.

The following examples illustrate solving routine, as well as non-routine, dynamic-system problems. Direct substitution into the standard equations is needed for the routine examples, while multiple equations need to be used in conjunction for the non-routine examples. The key to obtaining accurate solutions is to know what quantities are known, what quantities are unknown, which equation(s) to use, what SI units are involved, and whether the final solution values are reasonable.

Learning aim D: Examine how fluid engineering systems can be used to solve engineering problems.

D.P7 Solve routine problems that involve fluid systems.

Answer:

Fluid systems are those which contain fluids and demonstrate their motion and action. Pressure, density, volume, flow rate and velocity are common parameters. For common or everyday problems, the right formula for fluid mechanics can be picked and the known quantities can be replaced.

Problem 1: Calculating pressure
Solution:
A force of \(500N\) acts uniformly over an area of \(0.25m^2\). Determine the pressure of the product.
Pressure is given by the following formula:
•    P=F/A
Substituting the values:
•    P=500/0.25
•    P=2000Pa 
Therefore:
•    P=2000Pa

Problem 2: Pressure in a liquid
Solution:
Find the pressure exerted by a column of water 5m high. Assume that the density of water is w=1000kg/m3 and the acceleration of gravity is 9.81m/s2.
The equation for the hydrostatic pressure is:
•    P= gh
Therefore:
•    P=1000 x 9.81 x 5 
•    P=49,050Pa
Therefore:
•    P=49.05kPa

Problem 3: To determine the flow rate of water.
Solution:
Water is flowing in a pipe of cross section area \(0.02 m^2\) at a speed of \(3 m/s\). Find volume flow rate.
The formula is:
•    Q=Av 
Therefore:
•    Q=0.02 x 3 
•    Q=0.06m3/s
Hence:
•    Q=0.06m3/s

D.P8 Solve routine problems that involve immersed bodies.

Answer:

An immersed body is an object that is completely or partially surrounded by a fluid. One of the main principles applied in these problems is Archimedes' Principle – the upthrust on an immersed object is equal to the weight of the fluid dislodged.

Problem 1: Calculating upthrust
Solution:
A block displaces 0.015m3 of water. Find the upthrust of the block. Take the density of water as 1000kg/m3 and g=9.81m/s2.
The upthrust is:
 F_B=ρgV
Substituting:
 F_B=1000 x 9.81 x 0.015 
 F_B=147.15N 
Therefore:
 F_B=147.15N

Problem 2: This is a Question that asks whether or not an object is floating.
Solution:
An object has a mass of 20kg and displaces 0.025m3 of water. Predict whether an object will sink or float.
Measure the mass of the object (M):
 W=mg
 W=20 x 9.81 
 W=196.2
Now work out the maximum upthrust:
 F_B=ρgV
 F_B=1000 X 9.81 X 0.025
 F_B=245.25N
As the upthrust is greater than the weight of the object:
 245.25N>196.2N 
If the given displaced volume is the appropriate submerged volume, then the object will float.

D.M4 Solve routine and non-routine problems that involve fluid systems accurately.

Answer:

For more difficult problems of fluid systems, several calculations might be needed, and more than one principle of fluid mechanics may have to be used. The problem has to be translated to determine which equations are needed.

Non-routine problem: Flow in a reducing pipe

Water flows through a pipe that reduces from an area of \(0.08m^2\) to \(0.02m^2\). In the larger part the velocity is 2 m/s. Work out the velocity for the smaller part.

The continuity equation for an incompressible fluid is:

A_1 v_1= A_2 v_2
Therefore:
 0.08(2) = 0.02v_2 
 0.16=0.02v_2
Rearranging:
 v_2=0.16/0.02
 v_2=8m/s
So, the velocity in the smaller part is:
 8m/s

Also the result is physically correct since the pipe area has reduced. The velocity of the fluid needs to increase if the volume flow rate is to be kept the same.

Non-routine problem: Pressure at various depths
Solution:
Water is stored in a water tank that is 8 metres deep. Determine the gauge pressure at the bottom of the tank and the force on a cross-section of the bottom of an area of 0.5 m². Take:
 ρ= 1000kg/m3
and:
 g=9.81m/s2
The pressure is first to be calculated:
 P=ρgh
 P= 1000 x 9.81 x 8 
 P=78,480Pa

Therefore:
 P=78.48kPa
Now work out the force:
 P= F/A
Rearranging:
 F=PA 
 F=78,480 X 0.5 
 F=39,240N 
Therefore:
 F=39.24Kn

Accuracy and checking
The examples show how the principles of fluid systems are applied in routine and non-routine problems. The main equations used include:
 P= F/A
 P=ρgh 
 Q=Av 
 A_1 v_1=A_2 v_2
and:
 F_B=ρgV

Accurate determination of the fluid properties, SI units, correct substitution and sensible rounding. Another way of checking answers is to see whether they are physically reasonable – for example, if a fluid flows at a higher speed through one pipe than another, but the flow rate is the same, then the cross-sectional area of the first pipe must be smaller than the second pipe.