1. The magnitude of the force in the direction parallel to the surface
2. The magnitude of the force acting on the object perpendicular to the surface (the normal force N)
3. The coefficient of friction acting between the object and the surface (this  is a measure of the “roughness” of the surface between 0 and 1 where 0 is a completely smooth surface and with coefficient 1 the object couldn’t be moved however large the force μ.

Firction acts parallel to the surface opposing motion and the normal is perpendicular to the surface

We know that if the object is in equlibrium then the resultant force in the direction parallel to the surface must equal zero and hence the firctional force must be equal to the resultant force acting on the object parallel to the plane. However, what we are interested in finding is the minimum force that must be applied in order that the object begins to move, that is the maximum value that the firction can take before it moves.

To find this maximum we simply multiply the normal force by the coefficient of friction. That is the firctional force R on a static object on a surface being acted upon by a normal force N and a force parallel to the surface F with coefficent of friction between the object and the surface μ is

F ≤ Nμ

and if the object is on the point of moving (that is the we applying the minimum force required to make the object move)

F = Nμ

Example

An object of mass 5kg is on a rough slope at 30 degrees to the horizontal. The acceleration due to gravity is g and the object is on the point of slipping down the slope. Find the coefficient of friction between the slope and the object.

We know from Newtons laws that the force of gravity acting on the object will have a equal force opposing it since the object is at rest. To use the formulea above we want to break this into two components parallel and perpendicular to the slope. The component perpendicular to the slope is the normal force N and the component parallel to the slope is the force R which is trying to move the object down the slope.

Firstly it is always useful to draw a diagram of what we are interested in.

The forces acting upon the object on a slope

Since we know that the object is on the point of slipping down the slope we can the use the eqaulity from the equations above

F = Nμ

So to find μ we need to first calulate F and N. Because the object is in equilibrium F = R .

So we can resolve the weight into directions parallel and perpendicular to the slope to get

N = Wcos(30) = (5g√3)/2

and

F = Wsin(30) = 5g/2

so we can re arrange to get

Also see

• Will an object ever stop because of friction (11)

Example 1: simplifying single fractions

Consider

We want to find things that divide both the numerator (top) and denominator (bottom). In this case 2x does. We can then divide both the top and bottom by 2x (known as cancelling out the 2x) to gain the fraction

Note that you can only cancel things that divide the whole of the top and bottom not part of the top and bottom or things that are added to the top and bottom. For example

Sometimes the common factor that we are cancelling isn’t immediately obvious, for example if we have quadratic expressions. In these cases you should try to factorise the quadratic to see if there is anything you should cancel.

Note on quadratics: If a fraction has quadratics that can be factorised you should always leave the denominator factorised whilst it is common to multiply out the numerator (unless there is a need to leave it factorised or it is very complex)

Example 2: adding fractions

Consider the algebraic expression

Before we consider adding them we should make each of the fractions in the sum into its simplest form.

Firstly we want to have one denominator. To do this we need to find the simplest common denominator and then make the denominator of both these fractions this. This is done in a similar way in which we find the lowest common denominator when we are adding normal fractions. The easiest way to do this is to multiply numerator and the denominator of both factions by the denominator of the other so the expression becomes:

Now, as when adding normal fractions, because the denominator of both fractions is the same we can add their numerators together and put them over a single denominator to get

Example 3: Solving Equations

To solve an equation involving an algebraic fraction you should first multiply out the fractions as explained above.

Consider the equation

which is already simplified. To solve this we need to cross multiply, this involves multiplying both sides of the equation by the both the denominators and then simplifying both sides. However, because after the multiplication the denominator of each fraction will be multiplied by the numerator they will cancel out with the result that we have multiplied each side of the equation by the denominator of the other side and have removed the denominators from the fractions. This is most easily show by example

ie)

becomes

with intermediate step

We can then solve the equation using normal methods (note that if the result is a quadratic you may have to use the quadratic formula)

Also see

• First Order Differential Equations (6)
• Finding the Inverse of a Function (0)
• Find equation of tangent to a curve (0)
• Inequalities (0)
• Why the proof 2=1 is wrong (0)
• Simultaneous Equations (0)

The look and feel of Trevor Pythagoras Books will probably change greatly in the near future as I seek to make it fit in with the look of the rest of the site and make the navigation between the sections of the site as smooth as possible.

Also see

• Integration by Parts (0)
• Cartesian Equation of a Circle (0)
• Arithmetic Series and Progression (2)
• Confidence Intervals (8)
• Area of a Triangle (3)
• Proof by Contradiction (0)

There are two solutions to sin(x)=0.65 we get solution (1) when we take the inverse but we need a way of finding (2) from (1)

The four sections of the CAST are cos,all,sin and tan starting by labelling the bottom right and working round in an anti-clockwise direction.

Use the following method to work out the values for the inverse between 0 and 2pi or 360

1. find the value of the inverse of the positive between 0 and 90 or pi/2 using a normal method (eg a calculator).
   So if we want to find sin^-1(0.6) we calculate 0.6435 rads or 36.87 deg and if we want sin^-1(-0.8) we calculate sin^-1(0.8) to get 53.13 deg or 0.927 rads
2. draw in the four lines on the CAST diagram (shown in green) that represent the angle. Do this by measuring from the horizontal the angle calculated in 1
3. If the value we are finding the inverse for is negative (eg we are finding sin^-1(-0.6)) consider the quadrants that don’t include the name of the function. (so if we are finding inverse sin only consider Tan and Cos, if we are finding inverse cos only consider sin and tan and if we are finding inverse tan only consider sin and cos)
4. If the value we are finding the inverse for is positive (eg we are finding sin^-1(0.6)) only consider the quadrants with the name of the function and all. (so if we are finding inverse sin we only consider sin and all, if we are finding inverse cos only consider only consider cos and all and if we are finding inverse ta only consider tan and all)
5. Calculate the angle from the zero line anti-clockwise to the lines in the quadrant’s we are considering. The angle labels I’ve put on the axes should make this easier. These values are your solutions.

All four lines representing the angle x drawn but we will only be interested in two of them

You can check that this works by putting the values back into you calculator and if you want to check that you’ve got all of the solutions check against the graph of the function.

Example

Find all the solutions of cos(x) = 0.7 between 0 and 360^o or rads.

We can use a calculator to find the value between 0 and 90 deg or pi/2 rads
so write y = cos^-1(0.7) = 45.57 deg (we’ll work in degrees for the example to avoid repeating all the calculations)

We no draw this on the cast diagram and choose the two lines we need (in red), in this case the lines in the cos and “all” quadrants.

We then calculate the anti-clockwise angles to these lines from the horizontal:
so we get 45.57 deg for the in the “all” quadrant and 360-45.57=314.43 for the cos quadrant
so our set of solutions for x between 0 and 360 of cos(x)=0.7 are 45.57 and 314.43

Also see

• Sec, Cosec, Cot (0)
• Tan = sin/cos (6)
• Sin, Cos and Tan (15)
• Integrate cos and sin squared using double angles (2)
• The Chain Rule (0)
• Trigonometry Identities (11)

The dot product is defined to be

things to notice:

1. if a and b are parallel so the dot product is simply the product of their lengths
2. if and and b are perpendicular so the dot product is 0
3. 
4. if a is a unit vector, is |a|= 1 then the dot product gives the magnitude of the component of b in the direction of a

Dot product in component form (with i,j,k’s)

The dot product is very easy to use in component form because of 1 and 2. Since i,j and k are all perpendicular to each other:

and since any vector is parallel to itself and i,j and k are all unit vectors

hence we can find the dot product of with
Now we simply take the dot product of each term in a with each term of b, in a similar way to how you multiply out brackets, to get

Popular Posts

• Area And Circumference of a Circle : pi (38)
• Equations of Motion (17)
• Derive Quadratic Formula (16)
• Sin, Cos and Tan (15)
• Understand the Sine and Cosine Rules (11)
• Trigonometry Identities (11)

These are derived from the formula for double cos as follows

by subsitituion of

and similarly for sin²x but using the substitution for cos²x instead of sin²x.

To use these identities we simply substitute them into the integral and find the integral as normal since we know

so we get

and

Also see

• CAST Diagarams for finding values of sin, cos and tan between 0 and 360 or 2 pi (0)
• The Chain Rule (0)
• Sec, Cosec, Cot (0)
• Trigonometry Identities (11)
• Tan = sin/cos (6)
• Sine and Cos Graphs Differentiating sin and cos (5)

F=ma

where f= (resultant) force, m=mass,a=acceleration.

Example

So for example if gravity is pulling down on a body of mass 2kg and is causing it to accelerate at 9.81ms^-2 then we can work out that the force of gravity acting down on the body (what we refer to as its weight) is
2 x 9.81 = 19.62N

When more than one force is acting

If multiple forces are acting on a body along one line (ie we don’t need to worry about the angles between the forces) then each one is causing the body to accelerate at different rates but obviously the body can only have a single rate of acceleration. To understand this problem you need consider the resultant force and resultant acceleration. These are formed by summing all the forces and acceleration respectively. So

where the ’s are the different forces acting on the body, so gives the resultant force and the ’s are the different accelerations associated with these forces, so gives the resultant acceleration. Also note that both the forces and accelerations have signs, you must take one direction as positive and the other as negative and give signs accordingly.

Multiple Force Example

Suppose the the body from the above example (of mass 2kg) is again falling under gravity of 19.62 newtons but it is encountering air resistance on 2 newtons against its fall. Find the acceleration of the body.
SO applying the formula (taking downwards as positive)
19.62 – 2 = 2 x a
which gives

Also see

• Equations of Motion (17)

f(x)	f ‘(x)
c (constant)	0
xn	nxn-1
ln(x)	1/x
ex	ex
eax	aeax
sin(x)	cos(x)
cos(x)	-sin(x)
tan(x)	sec2(x)
cosec(x)	-cosec(x)cot(x)
sec(x)	sec(x)tan(x)
cot(x)	-cosec2(x)
sin-1(x)
cos-1(x)
tan-1
sinh(x)	cosh(x)
cosh(x)	sinh(x)
tanh(x)	sech2(x)
cosech(x)	-coth(x)cosech(x)
sech(x)	-tanh(x)sech(x)
coth(x)	-cosech2(x)
sinh-1(x)
cosh-1
tanh-1
a(x)b(x)	a’(x)b(x) + b’(x)a(x)
a(b(x))

If there any more standard derivatives that I have missed out please leave them in the comments section

Also see

• Fundamental Theorem of Calculus (0)
• Differentiation From First Principles (with example) (1)
• Integration by Parts (0)
• Implicit Differentiation (1)
• First Order Differential Equations (6)

These can be defined as follows,
if for all x₁ < x₂

• f(x₁) ≤ f(x₂) then f is increasing
• f(x₁ ) < f(x₁ ) then f is strictly increasing
• f(x₁ ) ≥ f(x₂) the f is decreasing
• f(x₁) < f(x₂) then f is strictly decreasing

Monotonicity and Derivatives

If a function f(x) is increasing then what we mean is that the slope is always positive, so if f is continuous we can relate the the properties of increasing and decreasing to the derivative as shown in the table below.

Its important to note that these rules only work if the function is continuous, for example consider f(x) =1/x, which is discontinuous at 0.
We can differentiate it to get f’(x) = -1/x² which we know is always negative (because the squared term is always positive) so we would expect it to be a decreasing function. However if we consider two point either side o, 1 and -1 say  we find that f is not a decreasing function because whilst -1 < 1, -1/x < 1/x contrary to our definition of decreasing

Also see

• Stationary Points (Maximum and Minimums) and Differentiation (1)
• Integrating Fractions – using the natrual logarithm – Example tan(x) (0)
• Hyperbolic Functions (0)

So for example f(x) = x+2 (from reals to reals) has an inverse as every element y is mapped to only by y-2 but f(x) = x² (from reals to reals) doesnt because both -1 and 1 map to 1 so how would we decide which of these would be the unique result of the inverse applied to 1 (we could of course define the domain to be the positive reals in which case it would have an inverse).

When finding the inverse of a function you are really looking to see what maps to each element of the range or codomain, to find the inverse of f you are looking for the element x in the domain for a given y in the range such that f(x) = y. This basically means you are reversing the process of the function.

Finding the Inverse when the function is a formula

When a function is given by a formula what you need to try and do is apply the operations of that formula backwards. This easiest way of doing this is to let the function f(x) = y. Now you know the formula to get from x to y so substitue this in for f(x). All that now needs to be done is to rearrange this equation so that x is the subject and the resulting rexpression on the otherside is only in terms of y. This function is the inverse of f, to show this lets denote it g so we g(y) = x.

Then g(f(x)) = g(y) = x since we started by letting f(x) = y and created g such that g(y) =x.
and f(g(y)) = f(x) = y

so g is indeed the inverse of f.

Example

This is most easily demostrated through an example.
Let
Then to find the inverse of f we first write

and rearrange as follows

So the inverse of f is given by the formula so we can write

Note that we have replaced the y’s with x’s, this doesnt matter as we can put any variable we like into the function but it important to make sure that you use the same variable as the parameter of the function and in the formula that defines it.

Also see

• Introduction to functions and maps (0)
• Simplifying and Solving equations with Algebraic Fractions (0)
• CAST Diagarams for finding values of sin, cos and tan between 0 and 360 or 2 pi (0)
• Exponential Functions (0)
• Why the proof 2=1 is wrong (0)
• Proof by Mathematical Induction (4)