Maths topic 1 forces PDF

Preview

Summary

in depth notes on the SM123 module - all topics covered...

Description

Faculty of Science, Technology, Engineering and Mathematics SM123 Physics and Space

Topic 1 Forces around you: Mathematics Practice Introduction In Topic 1 you were introduced to a number of equations describing the relationship between physical quantities. For example, in your experiment in Subsection 1.3, you found that there is a linear relationship between the amount that an elastic band stretches and the amount of force that is applied. This was written in terms of a linear equation called Hooke’s Law: Fstr = k strx,

(M1)

where F str is the magnitude of the applied force, x the extension and kstr the spring constant. This is an example of mathematical modelling, which you were introduced to in the introduction to Unit 2. More specifically, Hooke’s Law is an example of a linear mathematical model. Mathematical modelling is an extremely important part of physics and applied mathematics, and you will meet it again at all levels of your study. Here we will further explore the mathematical models you met in Topic 1 in the context of linear equations and quadratics from Unit 2.

1

Hooke’s Law as a linear mathematical model

The MST124 handbook describes mathematical modelling as “a collection of assumptions and mathematical statements that is intended to describe how some phenomenon in the real world behaves, and to enable predictions to be made about its behaviour”. Activity 1 (a)

What phenomenon is described by Hooke’s law?

(b)

What mathematical statements make up this mathematical model?

(c)

What predictions does this model allow us to make about the behaviour of this real-world phenomenon?

(d)

When you plotted your graph of weight versus extension in Subsection 1.3, you probably didn’t quite get a straight line going through both your points and the origin. Can you think of any reasons why that might be?

Linear mathematical models consist of one or more linear equations, or equations that describe straight lines. As you learnt in Unit 2, Subsection 2.2, such equations take the form y = mx + c

1

Hooke’s Law as a linear mathematical model

where m is the gradient and c the y-intercept. You also learnt that you can obtain the gradient of the line joining two distinct points (x1, y1) and (x 2, y2) as m=

y2 − y1

.

(M2)

x 2 − x1 You can then obtain the equation of the line from the gradient and one of the points, say (x1, y1), by the following formula: y − y 1 = m(x − x 1).

(M3)

Instead of using software as you did in Subsection 1.3, you are now going to use the above mathematics to obtain a linear model of your elastic band. er and a unit, such as 5 kg, 7 N or 8 m s−1, for instance. (There are exceptions: using n to represent a certain number ather complicated, and you must take special care not to confuse symbols with units, such as mass, m, and the unit m

Activity 2 (a)

Use the two points from your experiment in Subsection 1.3 and Equation M2 to determine the gradient of the straight line that joins them.

(b)

Use Equation M3, your gradient from part (a) and one of your points to find the equation of the line.

(c)

Find the x- and y-intercepts of your line.

(d)

Use your x- and y-intercepts, and your two points, to sketch a graph of your linear model.

(e)

Comment on the validity of your model when the applied force is 0 N.

We used given formulae for the gradient and the equation of a line in the above activity, in order to find the equation for the line describing our linear model. We could instead have used simultaneous linear equations (this is actually one way of deriving the formulae we used). You met these in Unit 2, Section 3.

2

2

Newton’s second law and simultaneous linear

Activity 3 (a)

Write a pair of simultaneous linear equations by substituting each of the points ( x1, y1) and (x2, y 2) from your experiment in Subsection 1.3 into the equation y i = mxi + c, where i = 1, 2. Each point should give you one equation.

(b)

Solve the pair of simultaneous linear equations in part (a) and check that your answer matches your answer to Activity 2 part (b).

In the next section we will continue our study of simultaneous linear equations, and their application to describing real-world behaviour under the influence of forces.

2

Newton’s second law and simultaneous linear equations

In Activity 1 (c), we noted that with Hooke’s law applied to a spring with a certain spring constant, we can find the extension given an applied force, or we can find the applied force given an extension. This is an example of using a linear equation to find an unknown value. Many real-world systems can be modelled as several linear equations that must all hold simultaneously, and this allows us to find the value of several unknowns. For example, you may meet systems of springs, each modelled using Hooke’s law, at Level 2 and 3. For now, however, we will leave Hooke’s law and consider Newton’s second law of motion from Subsection 2.4. Newton’s second law is another example of a linear mathematical model, this time relating the size of the applied force, F , to the amount of acceleration, a, of an object of mass m. To see how simultaneous linear equations from Unit 2, Section 3 can arise from Newton’s second law, we are going to consider a system with two masses. Suppose two masses, m1 = 110 g and m2 = 150 g, are attached to either end of a massless, inextensible string, and that the string is resting on a smooth pulley. Initially, m1 is held 1.3 m below the pulley while m2 is hanging freely 0.30 m below the pulley, as shown in the diagram below. At time t = 0, m1 is released and the system moves under the influence of gravity only. We wish to know the acceleration of the masses (it must be the same size for both masses, but in opposite directions, since they are connected by an inextensible string) and the tension in the string between them. 0.30m m2 = 150g 1.3m

m 1 = 110g Note all the assumptions that are made in this mathematical model: we assume the mass of the string has no effect, that there is no friction as it passes over the pulley and that the string cannot extend. This allows us to apply our simple mathematical models to make predictions about a complex, real physical system. When applying such models to a real system, however, their validity should always be checked.

3

3

Coulomb’s Law and quadratics

Activity 4 (a)

Draw separate free-body diagrams for each of the two masses, labelling the unknown tension of the string T .

(b)

Once m1 is released at t = 0 the masses are not at rest: they are moving with acceleration a and −a. Thus the net force on each mass is not 0, but rather m1a and −m2a. Find the net force from each free-body diagram and set each equal to m1a and − m2a, thereby writing a pair of linear simultaneous equations in the unknowns T and a. Remember that g = 9.81 m s −2.

(c)

Solve the simultaneous equations from part (b) to find T and a.

We are now going to leave the world of linear equations and consider a slightly more complex mathematical model.

3

Coulomb’s Law and quadratics

In Topic 1, Part 3 you were introduced to Coulomb’s Law which relates the magnitude of the electric force Fel between two charges to the value of the charges, Q1 and Q2, and the distance between them, r:

Fel

= ke

Q1Q 2 r2

(M4)

Coulomb’s law is also a mathematical model, but it is not linear. The form of equation to solve will depend on which values are known, and which are not. If all values are known apart from r, then Coloumb’s law becomes a quadratic in r. In Unit 2, Section 4 you learned that quadratics are equations of the form 0 = ax2 + bx + c, where a, b and c are constants and a =/ 0. Note that we could call the independent variable r instead and write a quadratic equation in r as 0 = ar 2 + br + c. Activity 5 (a)

Write Coulomb’s law as a quadratic equation in r, by assuming all values other than r are known. Start by subtracting Fel from both sides of the equation and then rearrange the equation into quadratic form. Identify the values of the constants a, b and c. Is a non-zero as required?

(b)

In Subsection 3.1.3 you found that the electric force between the proton and electron at the average distance between them in a hydrogen atom is —8.24 × 10 −8 N. Suppose instead that the electric force were known, and that we wanted to find the average distance between the electron and the proton. Use the quadratic equation you found in part (a) to find this distance of separation, r. Remember that the charge of an electron is −1.602 × 10−19 C, the charge of a proton is 1.602 × 10−19 C and ke = 8.99 × 109 N m2 C−2.

(c)

Quadratic equations also arise from Coulomb’s law in other ways. Suppose two charges separated by a distance of r = 10 cm repel each other with a force of 10 N. We know that the sum of the charges is 910 × 10−5 C. What are the two charges? Use ke = 9 × 109 N m2 C−2 in this Activity. Hint: Denote the unknown charges by Q1 and Q2 and substitute the given values into Coulomb’s equation. This gives you one equation with two unknown values. Write the 4

4

Conclusion

relationship between the size of the charges as a second equation in terms of Q1 and Q2, and treat the two equations as simultaneous equations. Solve these using the substitution method. (d)

10 Sketch a graph of the quadratic equation y = x2 − 9 × 10 −5x + 1 ×9 10−10 and comment on the meaning of its x-intercepts. Hint: You may find it helpful to review Unit 2, Subsection 4.9.

4

Conclusion

We have brought together a number of ideas from Topic 1 and Unit 2. You have used your knowledge of equations of straight lines to find a mathematical model for your elastic band experiment; you have seen how simultaneous linear equations arise when we balance the forces on two linked masses; and you have solved quadratic equations that arose from Coulomb’s law. As you continue your study of physics and applied mathematics, you will discover many more mathematical models linking our understanding of physical phenomena with more complex mathematical techniques.

5

Solutions to activities

Solutions to activities Solution to Activity 1 (a)

Hooke’s law describes the phenomenon of an elastic object (such as a spring or elastic band) stretching when a force is applied to it.

(b)

There is only one mathematical statement in this model, and it is Hooke’s law as given by Equation M1.

(c)

We can predict how much an elastic object will stretch if we apply a force of a certain size. We can also predict the size of force required to stretch the elastic object a certain amount.

(d)

Real elastic bands are unlikely to display the exact linear behaviour of Hooke’s law, for example because of the age and shape of the elastic band, how close to its maximum stretched length it gets, the temperature of its surroundings and inaccuracies in measurements. However, a linear model probably describes the behaviour of your elastic band pretty well, and could be used to give useful predictions. In fact, it is much more useful and widely applicable than if you had found a model to describe your particular elastic band more accurately. The most useful mathematical model is the simplest model that is sufficient for our purposes, and Hooke’s law is widely applicable to numerous real-world applications with sufficient accuracy. To achieve such a model we have to assume that a number of properties of the real-world elastic band and its surroundings are negligible. Determining the important aspects to include in a mathematical model, and deciding which can be ignored, is an important skill in mathematical modelling. Hooke’s law is used in many more complex mathematical models. In these we may state as one of our assumptions that elastic parts of our model act as model springs. A model spring is assumed to have an exact linear relationship between force and extension, so this means that we assume that Hooke’s law applies.

Solution to Activity 2 (a)

My two points are (0.060 m, 5.1 N) and (0.029 m, 2.6 N). The gradient of the straight line that joins these two points is m = = 5 .1 N−2.6 N = 80.645...N m−1 ≈ 81 N m −1 (to y2−y1 x 2−x1

0.060 m−0.029 m

two sig. figs.). So the gradient is 81 N m−1 . (b)

I am going to use the larger point, (0.060 m, 5.1 N), and I am going to use the gradient to 5 significant figures within my calculations. Thus the equation of my line is y − 5.1 N = 80.645 N m−1(x − 0.060 m). Rearranging to make y the subject and rounding the final answer to 2 sig. figs. gives y = 81 N m−1x + 0.26 N.

(c)

The x-intercept is the value of x when y = 0. Setting y = 0 N gives 0 N = 81 N m−1 x + 0.26 N. Solving this equation gives x = − 0.0032 m (to two significant figures). The y-intercept is the value of y when x = 0. Setting x = 0 m gives y = 81 N m−1 × 0 m + 0.26 N. Solving this equation gives y = 0.26 N. Thus the x-intercept is −0.0032 m and the y-intercept is 0.26 N.

(d)

A sketch of my linear model is shown below. Your sketch should show the x- and y-intercepts, the two measured points, be a straight line and have labelled axes with some points marked.

6

force (/N) y = (81 N m−1)x + 0.26 N 6 Measured point (0.060, 5.1)

4

Measured point (0.029, 2.6)

2

(0, 0.26) 0.02

0.04

0.06

extension (/m)

(−0.0032, 0)

(e)

My model predicts an extension of − 0.0032 m when a force of 0 N is applied. This is clearly nonsense! However, it is not that far off the 0 m extension we would expect, either. In general, care must be taken when applying mathematical models to ensure they make physical sense, and the models are most reliable when used within (or near) the range of measurements that informed the original model. For example, a prediction of the extension for an applied force of 3 N is more likely to be reliable.

Solution to Activity 3 (a)

My two points are (0.060 m, 5.1 N) and (0.029 m, 2.6 N), so my two equations become: 5.1 N = (0.060 m)m + c 2.6 N = (0.029 m)m + c.

(b)

(S1) (S2)

Subtracting Equation S2 from Equation S1 gives: 5.1 N − 2.6 N = (0.060 m − 0.029 m)m 2.5 N = (0.031 m)m m = 80.645... N m−1 ≈ 81 N m −1 Now substituting this value for m (to 5 sig. figs. in this intermediate calculation) into Equation S1 gives: 5.1 N = 0.060 m × 80.645 N m −1 + c c = 5.1 N − 0.060 m × 80.645 N m−1 c = 0.2631 N ≈ 0.26 N This does match the equation we found in Activity 2 (b). Depending on when you did your rounding you may have got a slightly different answer to that in the previous Activity.

Solution to Activity 4 (a)

(b)

Note that all values have been converted to SI units. T T m1

m2

(0.11kg)g

(0.15kg)g

From the first free-body diagram, the net force on m1 is T − (0.11 kg)g = T − 1.0791 N. Setting this equal to m1a = (0.11 kg)a gives the equation T − 1.0791 N = (0.11 kg)a. From the second free-body diagram, the net force on m2 is T − (0.15 kg)g = T − 1.4715 N. Setting this equal to —m2a = ( −0.15 kg)a gives the equation T − 1.4715 N= ( −0.15 kg)a. We thus have the following pair of simultaneous equations: (S3) (S4)

T − 1.0791 N = (0.11 kg)a T − 1.4715 N = (−0.15 kg)a (c)

Subtracting Equation S4 from Equation S3 gives −1.0791 N + 1.4715 N = (0.11 kg + 0.15 kg)a. Rearranging to make a the subject gives a = 1.5092 m s−2 (to 5 sig. figs. for intermediate calculations). Substituting this value for a back into Equation S3 gives T − 1.0791 N = 0.1660 N. Rearranging to make T the subject gives T = 1.2451 N. So the tension in the string is 1.2 N, and both masses accelerate at 1.5 m s−2, with m1 accelerating upwards and m2 accelerating downwards.

Solution to Activity 5 Q

(a)

Rearranging as suggested gives ke Q12 2 − F el = 0. Multiplying through by −r2 gives r Felr 2 − k eQ 1Q 2 = 0. This is of the required form for a quadratic equation with a = Fel, b = 0 and c = −keQ 1Q 2. As long as Fel /= 0 (if the electric force were equal to zero we would have a rather uninteresting problem!), a /= 0 and we have a quadratic.

(b)

Substituting in the given values gives 0 = (−8.24 × 10 −8 N)r2 − (8.99 × 10 9 N m2 C−2) × (−1.602 × 10−19 C) × (1.602 × 10−19 C) 0 = (−8.24 × 10 −8 N)r2 + (2.3072 × 10−28 N m2) r 2 = 0.280 × 10 −20 m2 r = ±5.29 × 10−11 m Since r is a distance, only the positive value makes sense in this context, so we have, of course, obtained the distance given in the original exercise of r = 5.29 × 10−11 m. This also illustrates how the difficulty of finding an unknown value depends on the type of equation that results after substituting in the known values.

(c)

Substituting the given values into Coulomb’s law, and writing the relationship between the

two charges as a second equation, gives the pair of simultaneous equations Q1Q2 10 N = (9 × 109 N m2 C −2 ) (0.1 m)2 Q 1 + Q2 =

10 9

−5 × 10 C.

(S5)

(S6)

Rearranging Equation S6 to make Q2 the subject gives Q 2 = ( 10 × 10−5 C) − Q1. 9 Substituting this in for Q2 in Equation S5 gives −5 10 10 N = (9 × 10 9 N m2 C−2 ) Q1 ( 9 × 10 C − Q1) . (0.1 m)2 Multiplying out the brackets and writing this as a quadratic equation gives

Q2 −

10

× 10 −5 C Q +

1

1

9 Q − 10−5 C Q

1

× 10 −10 C 2 = 0

(S7)

9 1

10−5 C = 0 1 × = × 10 −5 C. 1 − 9 9 Q1 = 10−5 C or Q 1

1

(You may have decided to use the quadratic formula rather than factorise here.) Substituting these values into Equation S6 gives the two solutions Q 1 = 10 −5 C, Q = 1

1

× 10

Q2 −5

C,

1 =

−5 × 10 C

or

9 Q = 10−5 C. 2

9 As it doesn’t matter which of the two charges we label Q1 and which we label Q 2, we can see that a pair of charges of sizes 19 × 10−5 C and 10 −5 C meet the requirements. These repel each other as they have the same sign. Aside: You may have noticed that we are not treating Fel as a vector here, but rather we are finding its magnitude and using the sign to determine whether the charges attract or repel. We treated Fstr and F = ma similarly. As forces, F el (and Fstr and F = ma) is, of course, a vector. However, a full vector treatment is beyond the scope of SM123. (d)

A sketch of the quadratic equation y = x2 − 109 × 10 −5 x + 1 ×9 10 −10 is shown below. Your quadratic should be u-shaped since a = 1 is positive. The y-intercept should be 19 × 10−10, and the x-intercepts were found in part (c): 91 × 10−5 and 10−5. The axis of symmetry is at 10−5 and the vertex is at ( 5 9× 10 −5, − 1681× 10−10) (the vertex x = ( 19× 10 −5 + 10−5)/2 = 5 × 9 passes through the axis of symmetry, giving the x-coordinate, and the y-coordinate is found by substituting the value of the x-coordinate into the original equation).

y 10 y =2x −× 9

0.5 × 10−10

1 (0,× 9

−10 10)

...