Section B Algebra
3.3 Manipulation of formulae
Rt = R0(1 + αt)
where Rt is the resistance at temperature t, R0 the resistance at 0°C and α a constant called the temperature coefficient of resistance. We might need to rearrange the equation so that we express α in terms of the other variables.
As a first step we can multiply out the brackets to give:
Rt = R0 + R0αt
Transposing the R0 from the left-hand to the right-hand side of the equation gives:
Rt — R0 = R0 αt
Reversing the sides of the equation so that we have the term involving a on the left-hand side, we then have
R0αt = Rt — R0
Transposing the R0t from the left-hand to right-hand sides gives α = Rt - R0
R0t Example
Rearrange the following equation for resistances in parallel to obtain R1
in terms of the other variables:
1 R1 + 1
R2 = 1 R
Transposing 1/R2 from the left-hand side to equation gives
1 R1 = 1
R -1 R2
Arranging the fractions on the right-hand side of the equation with a common denominator, then
1
R1 = R2 - R RR2
We can invert the fractions provided we do the same to both sides of the equation. Effectively what we are doing here is transposing the numerators and denominators on both sides of the equation. Thus
R1 = RR2 R2 - R
the right-hand side of the
Example
The surface area A of a closed cylinder is given by the equation A = 2πr2 + 2πrh
where r is the radius of the cylinder and h its height. Rearrange the equation to give h in terms of the other variables.
Transposing the 2πr2 term gives:
A - 2πr2 = 2πrh
Reversing the sides the equation gives:
2πrh = A - 2πr2
Transposing the 2πr from the left-hand to right-hand sides gives:
h = A - 2πr2
2πr Example
The power P dissipated by passing a current I through a resistance R is given by the equation P = I2R. Rearrange the equation to give I in terms of the other variables.
Writing the equation as I2R = P and then transposing the R gives:
I2 = P R
Taking the square roots of both sides of the equation then gives I =
√
PR Revision
5 Rearrange the following equations to give the indicated variable in terms of the remaining variables:
(a) v = u + αt, for α, (b) F1r1 = F2r2, for r1, (c) F
A = E x x L , for x, (d) T2 = 4π2L
g , for L, (e) Z2= R2 + X2, for X, (f) pV = mRT, for m, (g) E = ½mv2, for v, (h) s = ut + ½αt2, for α, (i) Z = bd2
6 , for d.
6 The length L of a bar of metal at t°C is related to the length L0 at 0°C by the equation L = L0(l + at). If the length at 100°C is 1.004 m for a bar of length 1 m at 0°C, what is the value of a?
7 The stress σ acting on a material is related to the force F acting over a cross-sectional area A of the material by the equation
σ = F A
If the stress is 1000 N/mm2 and the area 4 mm2, what is the force?
8 The relationship between the pressure p, the volume V and the temperature T on the kelvin scale can be written as:
pV T = mR
where m is the mass of the gas and R the characteristic gas constant.
What will be the volume of a mass of 2.0 kg of air at a temperature of 293 K and a pressure of 1 100 000 N/m2 (Pa) if the characteristic gas constant is 287 J/kg K?
It is not only numerical values that must balance when there is an equation.
The units of the quantities must also balance. For example, the area A of a rectangle is the product of the lengths b and w of the sides, i.e. A = bw. The units of the area must therefore be the product of the units of b and w if the units on both sides of the equation are to balance. Thus if b and w are both in metres then the unit of area is metre x metre, or square metres (m2). The equation for the stress a acting on a body when it is subject to a force F acting on a length of the material with a cross-sectional area A is
σ = F A
Thus if the units of F are newtons (N) and the area square metres (m2) then, for the units to be the same on both sides of the equation:
or, using the unit symbols:
unit of stress = N
m2 = N/m2
This unit of N/m2 is given a special name of the pascal (Pa).
Velocity is distance/time and so the unit of velocity can be written as the unit of distance divided by the unit of time. For the distance in metres and the time in seconds, then the unit of velocity is metres per second, i.e. m/s or m s-1. For the equation describing straight line motion of v = at, if the 3.4 Manipulation of
units
unit of stress = newtons
square metres = newtons/square metre
unit of v is metres/second then the unit of the at must be metres per second.
Thus, if the unit of t is seconds, we must have metre
second = metre
second2 x second or, in symbols,
m s =
m s2
X S
Hence the unit of the acceleration a must be m/s2 for the units on both sides of the equation to balance.
Example
When a force F acts on a body of mass m it accelerates with an accel
eration a given by the equation F = ma. If m has the unit of kg and a the unit m/s2, what is the unit of F?
For equality of units to occur on both sides of the equation, unit of F = kg x m
s2 = kg m/s2
This unit is given a special name of the newton (N).
Example
The pressure p due to a column of liquid of height h and density ρ is given by p = hρg, where g is the acceleration due to gravity. If h has the unit of m, p the unit kg/m3 and g the unit m/s2, what is the unit of the pressure?
For equality of units to occur on both sides of the equation, unit of pressure = m x kg
m3 x m
s2 = kg m x s2
Note that in the previous example we obtained the unit of newton (N) as being the special name for kg m/s2. Thus we can write the unit of pressure as
unit of pressure = kg x m s2 X
1
m2 = N/m2
This unit of N/m2 is given a special name, the pascal (Pa).
Example
For uniformly accelerated motion in a straight line, the velocity v after a time t is given by v = u + at, where u is the initial velocity and a the
acceleration. If v has the unit m/s, u the unit m/s and t the unit s, what must be the unit of a?
For the units to balance we must have the unit of each term on the right-hand side of the equation the same as the unit on the left-hand side. Thus the unit of at must be m/s. Hence, since t has the unit of s:
9 The pressure p acting on a surface of area A when it is subject to a force F is given by p = F/A. If A has the unit of m2 and F the unit of N, what will be the unit of the pressure?
10 The torque T resulting from a force F being applied at the end of a lever arm of radius r is given by T = Fr. What is the unit of the torque when F has the unit of N and r the unit of m?
11 The specific heat capacity c of a material is defined by the equation c = Qlmt, where Q is the heat transfer needed for a mass m of a material to change its temperature by t. If Q has the unit of joule (J), m the unit of kg and t the unit of kelvin (K), what is the unit of c?
12 The electrical resistivity ρ of a conductor is given by the equation ρ = RAIL, where R is the resistance of a strip of length L and cross-sectional area A. What is the unit of resistivity if R has the unit Ω, L the unit m and A the unit m2?
13 Kinetic energy E is given by the equation E = ½mv2, where m is the mass in kg, v the velocity in m/s. What will be the unit of E?
1 Solve the following equations:
(a) 2x + 4 = 5 - x, (b) x - 4 = 12, (c ) 2 x = 3 4 , (d) 3
2 Simplify the following by removing the brackets:
(a) (x + 5) + 3x, (b) 4 - 2(x + 3), (c) 2(x + l) + 5(x - 2), Problems
(d) 3(x - 2) + 5(x + 1) - 2(2x - 1), (e) x(2x + 3) + 5(x + 1), (f) (x + 3)(x - 2), (g) (x2 + l)(x + 2), (h) (x - 3)2, (i) (2x - 1)2, (j) x2 + (2x - 3)2, (k) (x + 3)(x - 3), (1) (2x - l)(2x + 1).
3 Express each of the following as single fractions:
(a) x = 5 2
y , (b) x + 1 -4 1
2x + 3 , (c) x
x + 5 + x - 1 2x , (d) 3 x + 1
-x x - l , 4 The area A of a circle is given by the equation A = 1/4πd2. Solve the
equation for d given that the area is 5000 mm2.
5 The velocity v of an object is given by the equation v = 10 + 2t. Solve the equation for the time t when v = 20 m/s.
6 The length L of a bar of metal at t°C is related to the length L0 at 0°C by the equation L = L0(1 + 0.0002t). What is the length at 100°C for a bar of length 1 m at 0°C?
7 The specific latent heat L of a material is defined by the equation L = Q/m, where Q is the heat transfer needed to change the state of a mass m of the material. If Q has the unit of joule (J) and m the unit of kg, what is the unit of L?
8 The density ρ of a material is defined by the equation ρ = m/V. What is the unit of density if m has the unit kg and V the unit m3?
9 For the bending of a beam, the bending moment M is given by the equation M = EI/R. If E, the tensile modulus, has the unit N/m2, I, the second moment of area, the unit m4, and R the radius of a bent beam the unit of m, what must be the unit of M?
10 Rearrange the following equations to give the indicated variable in terms of the remaining variables:
(a) p1 - p2 = hρg, for h, (b) ρ = RA
L , for R, (c) E = V - Ir, for r, (d) Ia = T - mr(a + g), for T, (e) s = ut + Ft2
2m , for F, (f) E = mgL
Ax , for x, (g) M I = E
R , for R, (e) E = ½mv2, for m, (i) V = V= πr2h, for r, (j) P = I2R, for R,
(k) m1a1 = - m2a2, for a2, (1) F = mv2
R J for v,
(m) C = ε0εr A d , for d.
11 For a gas that obeys Charles' law:
V1
T1 = V2
T2
where V1 is the initial volume and T1 the initial temperature, V2 the final volume and T2 the final temperature. Rearrange the equation to express
T2 in terms of the other variables. Hence determine the final temperature when the volume changes from 1 m3 at 300 K to 2 m3.
12 For a beam simply supported over a span of L and carrying a central load of F plus a uniformly distributed load of w per metre of beam, then the reaction R at the beam supports is given by the equation:
R = ½(W + wL)
Rearrange the equation so that w is expressed in terms of the other variables. Hence determine the value of w when L = 3 m, W = 1 0 kN and R = 11 kN.
4.1 Introduction The term linear equation is used for an equation in which the unknown quantity is only raised to the power 1. For example, 2x + 5 = 0 is a linear equation since the unknown, the x, is only to the power 1. An equation which is not linear is 2x2 + 5 = 0. In this equation the x is raised to the power 2. Such an equation is termed a quadratic equation. Quadratic equations are discussed in chapter 5.
Many equations in engineering are linear equations. For example, for motion in a straight line with a constant acceleration we have v = u + at, with perhaps the velocity v after a time t being the unknown, u a given initial velocity and a a given acceleration. Putting given values of, say, u = 10 m/s, a = 4 m/s2 and t = 2 s in the equation we obtain as the solution v = 10 + 4 x 2 =18 m/s. There is just one value of v which fits the equation.
This is always the case with a linear equation (but not with a quadratic equation, see chapter 5).
The above is a linear equation with just one unknown. In analysing situations in engineering we often end up with linear equations which have two unknowns. For example, in analysing an electrical circuit which has two meshes with a current I1 in one and I2 in the other we might obtain the equations for each mesh as 2I1 - I2 = 3 and I1 + 3I2 = 4. Both these are linear equations since the unknowns are only to the power 1. We have thus two unknowns and two equations. Because both equations must give the same values for both the unknowns, they are referred to as simultaneous equations. Because they are linear equations we will still only obtain just one value of each unknown which fits the equations.
This chapter is about solving linear equations when they contain just one unknown and pairs of simultaneous linear equations when they contain two unknowns.
4.2 Solving linear The basic rules for solving linear equations in one unknown can be