The easiest way to deal with this is to resolve forces into perpendicular components. For example if we are working in a 2D plane it might make sense to consider the horizontal and vertical parts of a force separately. We can then do two sets of calculations, one vertical and the other horizontal, to find out the affect of the force.

Consider the example below,

It shows how we can split a force into its two components and what they’re relative sizes would be. To work out the magnitude of the components we must know the magnitude of the force we are resolving and the angle between it and one of the two perpendicular component. We can then use trigonometry.

If we know the angle between the force and its component then the magnitude of the component is the magnitude of the force multiplied by the cos of the angle and the magnitude of the other, perpendicular component, is the magnitude of the force multiplied by the sin of the angle.

So in the above example where the magnitude of the force is M and that of the horizontal and vertical components is H and V:

V = Msin(a) and H = Mcos(a)
or
V = Mcos(b) and H = Msin(b)

Finding a resultant force from components

Once you have done calculations in two perpendicular directions you are likely to end up with two perpendicular forces which you want to combine to find the resultant force. To do this we must use pythagoras’s theorem and more trigonometry.

Again consider the example above. Suppose you know H and V and want to find M and the angles a and b. To find M we use Pythagoras so we get

and to find the angles a and b we must use the tan function to get
and

Example – finding the resultant force

Find the direction and magnitude of the resultant force below.

First find all the horizontal components (taking forces acting right as positive).
The horizontal component of the:
3N force is 3cos(30) = 2.598
2N force is 2cos(90) = 0 — note that this is expected because a vertical force has no horizontal component as they are perpendicular
4N force is -4cos(45) = -2.828

Then to find the horizontal component H of the resultant force we add these up to get
H = 2.598 + 0 – 2.828 = -0.23

And then we do the same in the vertical direction. The vertical components of the:
3N force is 3sin(30) = 1.5
2N force is 2sin(90) = -2
4N force is 4sin(45) = 2.828

And adding these up we find that the vertical component of the resultant force is
V = 1.5 -2 + 2.828 = 1.328

We now combine these to find the magnitude and direction of the resultant force.
The magnitude M is found using Pythagoras’s to get

and the angle, measured above the right horizontal, is
or 80.17 above the left horizontal

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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

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

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

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• Equations of Motion (17)

The obvious answer is yes but it would appear that it doesn’t because of the following:
1) The resistive force is some function of the objects velocity and is probably either proportional to its velocity or the square of its velocity.
2)When an object is at rest it experiences no resistive force, if it did it would start moving again.

This means that as a objects velocity approaches 0 the force that is causing it to slow down is also approaching 0. This means that the velocity of the object wont actually reach 0 but will become infinitely small.

You may therefore think that there is some sort of minimum friction, in addition to the component which is proportional to the objects velocity but this would mean that there would still be friction when the object is at rest so the object would then begin to move backwards.

Any comments or further thoughts would be appreciated to help explain this.

By David Woodford

Also see

• Static Friction (0)

The most common example of moments used in on a see saw. Here there a two levers, each side of the see saw, acting from the central pivot. if a person sits on the see saw there is a moment about the pivot because of there weight. While the seesaw is still horizontal this moment is the product of their weight (mass times gravity) and the distance they are still from the pivot because the distance a long the see saw to the pivot is perpendicular to their weight.

Like forces if something is in equilibrium, velocity isn’t changing (usually but not always at rest), the moments have to be balanced. Because moments are turning forces this means that all the moments acting clockwise have to equal all the moments acting anti clockwise. This means that if two people are sitting on the seesaw provided the product of their weight and distance is the same the seesaw wont turn, even if one person is much heavier than the other. For example if a heavy person sits close to the centre they can be balanced by a light person sitting further away.

Just to note – moments don’t have to be calculated about a pivot or turning point, they can be found about any point, so we cud take them about one of the people or the end of the seesaw. IF you do this though, remember that the pivot is providing an upward force on the seesaw that is equal to its mass plus that of the two people times gravity. This sort of method can often be useful in more complex systems as by taking moments about a point then we can ignore all forces acting through it (as their distance is 0) and simplify our equations.

However we can also calculate the moments caused by forces that aren’t perpendicular to their distance, for example the moment cause by the person on the seesaw when it is titled or on the ground. Here we have to find the component of the force that is in the direction that is perpendicular to the distance, by resolving using sin and cos.

This is done by multiplying the product either by the sin of the angle between force and the direction of the distance between the force and pivot or the cos of the angle between them force and the direction perpendicular to the distance.

for example if the seesaw is tilted by 30 degrees from the horizontal we can have to multiply by

cos(30)=sin(60) =√3 /2

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The 4 equations of motion deal with an object which is travelling with constant acceleration (which can be 0 and therefore constant speed).

The equations are as follows:

(1) v = u + at
(2) s= t(v+u)/2
(3) v²=u²+2as
(4) s = ut + at²/2

Where
a = acceleration – this must be constant for equations to hold
u = initial velocity, ie at the start of the journey
v = final velocity, ie at the end of the journey
s = displacement which is a vector quantity for the distance of the object from its starting point
t = time taken for journey
Any of these values can be found using the equations if at least 3 of the other values are known.

Equation (1) comes from the definition of acceleration as acceleration is the rate of change of velocity and therefore for constant acceleration
a = (v-u)/t which you can rearrange to make
v = u + at

Equation 2 can be found by either considering a distance time graph or using the average speed. (v+u)/2 gives the average speed during the journey as a is constant and buy multiplying this by t we find the displacement. Or from the graph we can find it as the area under the graph is a trapezium of height t and sides u and v so using the area of a trapezium formulae we find 2.

Equation (4) can be formed by substituting 1 into 2 so
s = t(u+at+u)/2
s = t(2u + at)/2
s= 2ut/2 + at²/2
s = ut + at²/2

Equations 3 can also be found using 1 and 2 by rearranging 2 to get an expression for t we find
t=2s/(u+v)

If we substitute this into v = u+at we get
v=u+2as/(v+u)
And by bringing up the v+u we find
v²+vu = u²+vu + 2as
And because we have vu on both sides we can cancel these to find 3

v²=u²+2as

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• Force, Mass and Acceleration (0)

any finite number divided by 0 = infinite

so 3 / 0 = infinite

which can be rearranged so

0 x infinite = 3

if a miracle is defined as something with probability 0 (ie it is impossible and would require divine intervention to occur) and we assume the universe is infinite in either time or space or both.

the expected number of success of n trials is the number of trials x the probability of success (eg if you threw a dice 12 times you’d expect 2 sixes). So if the universe is infinite then there are infinite trials and if a miracle if a miracle has probability 0 the expected number of miracle = infinite x 0 and as shown above this equals 3.

so to recap

3/0 = infinite

so 0 x infinite = 3

p(miracle) = 0

E(miracle) = n p = 0 x infinite = 3

so in all of time and space 3 miracles are expected. However as any finite number divided by 0 = infinite then all we can say is that in all of infinite time and space a finite number of miracles are expected.

From this we can calculate the probability of one miracle occurring using the poission distribution

if X = number of miracles

X ~ P_o(3) as above

P(X = 1) = (e^-33¹)/1! = 0.0498 x 3 = 0.149

So there you go, when the universe is infinite

The probability of a  single miracle is 0.149

or

P(miracle) = 0.149

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