This is a unit where you can really go to town on the problems with the vectors. You must read the specification carefully and make sure that you don’t panic if you can’t do all the problems in the post. I have included lots of vector problems which go 360° where the exam board will tend to stick to 90° only to make it easier. However, I am in the business of teaching vectors in 2D and would say I would be a bad teacher to stop just because the exam board don’t do it. You will need the skills found here if you carry on to a Physics type job later on so why not learn now. Most students say it is really hard at first but then the exam is easy and they get it after a bit of graft!
Also you may find it hard to understand the idea that…
- A diagonal vector can be split into two parts the horizontal and vertical parts which can be treated separately.
- Two forces can be added together to make a resultant.
- An object not moving has a sum of all vertical forces as zero and also all horizontal forces as zero.
If you find this is hard keep going as it will click!
6.1 Vectors and Scalars
6.1 Trig Problems (You must have these trig skills before you start the problems)
Unit_7_Statics_Dynamics_Vectors_1 (Easy problems sheet)
6_1_Wires Example (This is an example of resolving – hanging a picture)
Unit_7_Statics_Dynamics_Vectors_2 (Rock Hard Problems Sheet beyond AS/A2)
6.2 Balanced Forces
6_2_Resultant Forces Problems (quick fun problems – no answers)
72_Practical_Coplanar_Forces (practical proving coplanar forces)
6.3 The principle of moments
6_3_Density_of_Ruler_Prac (Use moments to find out density of ruler)
6.4 More on moments
6_4_Bridge_Crane_Prac (Practical showing support forces)
6_4_Putting_Your_Foot_Down_Extension (Nice but more complex sheet on moments in your feet)
Use these quick notes to help you revise each topic from the Chapter. You can also download a PDF with revision cards to print double sided and 2 slides per sheet to get small cards to play revision games… Animated Science Physics AS Revision Cards Chapter 6 but the information is the same as on this post!
Conditions for the Equilibrium of Three Non-Parallel Forces
If we say that an object is under the influence of forces which are in equilibrium, we mean that the object is not accelerating – there is no net force acting. The object may still be travelling – but at a constant velocity – but in most questions the object will be stationary.
The following ideas will help you to solve problems that involve a body acted on by three co-planar forces.
1. The lines of action of the three forces must all pass through the same point.
2. The principle of moments: the sum of all the clock-wise moments about any point must have the same magnitude as the sum of all the anti-clockwise moments about the same point.
3. The sum of all the forces acting vertically upwards must have the same magnitude as the sum of all the forces acting vertically downwards.
4. The sum of all the forces acting horizontally to the right must have the same magnitude as the sum of all the forces acting horizontally to the left.
5. If you resolve all of the forces the ‘ups’ will equal the ‘downs’ and the ‘rights’ will equal the ‘lefts
Vectors and Scalars
Scalars: A scalar quantity is very simply a numerical item such as 7°C or 5kg. It can be in any direction and other examples are mass, density, energy or volume.
Vectors: This is a “scalar” which has a direction or bearing. This means that we can now think about displacement instead of distance travelled. This enables us to think about the motion of objects in more complex ways and define formulae with directions. Other examples are force, velocity, acceleration and momentum.
Vector Equations: When applying Mathematics to Physics problems we find that a scalar such as mass converts one vector to another. F = ma where both “F” and “a” are vectors.
Units: When looking at vector equations such as F=ma the units must balance… Force is measured in Newtons (N),
mass in kilograms (kg), and acceleration in ms-2. The units on both sides must be equal, so 1 N = 1 kgms-2.
Vector Journeys: When we have a journey which consists of two vectors such as displacement we find that it adds up to form a triangle. We can then go direct or indirect.
Resultant: This is the most direct route or “result” of the two vectors. We can find this by simple Pythagoras. If we had two vectors at right angles of 3km and 4km…
R2=42 + 32
tanq = opp/adj = 4/3
Vector Components: Sometimes it is more useful to work out a component of a resultant force in vertical or horizontal direction. This also uses Pythagorean Mathematics…
6cos60° = x = 3 ms-1
6sin60° = y = 5.2 ms-1
Vector Addition: If you have resolved all the forces, speeds or distances into their x and y components you can then sum all the x’s and all the y’s to find the overall motion. It is like adding lots of little effects to get one resultant or one large triangle. We can also do the same by drawing out the problem to scale.
Vector Triangles: We find that if a vector triangle works for distances it can also mirror the situation with the forces involved. For example if we know the geometry of a car on a slope it will enable us to predict the friction or support forces acting on the slope if we know for example the weight of the car.
Block on Desk: For a wooden block sat on a table it is very simple Newton’s 3rd Law tells us that the forces of weight are balanced exactly by the support forces…
0 = S – W
W = S
2D Forces: We can think of this in more detail that the sum of all forces is zero on a object moving at a constant speed or stationary. If we can resolve these directions using trigonometry we can create a balance situation…
Sum of vertical forces = 0
Sum of horizontal forces = 0
Childs Swing: This seems complex but it is not if we think of the problem in balance. Imagine that dad is holding the child back ready to swing. All forces balance according to Newton’s 3rd Law. So the Tension in the rope of the swing must balance both Weight and the Pull force from dad. However, tension acts at an angle so we resolve the forces (or split them) into vertical and horizontal components.
F = Tcosθ (horizontal)
W = Tsinθ (vertical)
Inclined Plane: In this situation the block (or any other object) acts in a similar way to being on a
table but we have to make a triangle of forces. Label them in order…
1) Weight (Vertical)
2) Support Force (at 90° to inc plane)
3) Friction (in line with inc plane)
A common mistake is to get the placement of the 90° corner of the triangle. Also you should notice that there are three similar triangles on this diagram which can all be used to solve problems.
Hanging Weight: We can use the idea of equilibrium for a weight hanging from a string held in two places. It is simple to balance.The problem should be tackled at two triangles and then everything summated…
Left = Right
T1sinθ1 = T2sinθ2
Up = Down
W = T1cosθ1 + T2cosθ2
It simplifies if the angles are equal to…
W = 2Tcosθ
The Principle of Moments
Turning Effects: When a force is applied a distance away from the centre of mass (COM) then we produce a turning effect or “moment”. If you have a situation where an object is in equilibrium then you can solve a problem by equating all the turning effects about any point in the situation…
Clockwise moments = anticlockwise moments
Moments: When a force is applied it has to be perpendicular to the distance to be classed at a moment. To solve problems you sum up all the moments applied “perpendicular” to a point…
Clockwise moments = anticlockwise moments
See-Saw: When we have simple see-saw we can sum up by just balancing from the middle of the see-saw and add any Wd from the pivot point..
W1d1 = W2d2
or if you have two people on one side…
W1d1 = W2d2 + W3d3
Complex Moments: Now what happens if the see-saw is not equal on each side. Now the weight of the see-saw must be taken into account when taking moments. This acts from the COM but looks strange at first…
W1d1 = W2d2 + W3d3
Non- Perpendicular Moments: Often students get caught out by these problems. However, all you have to do is work out the component of the moment which is working at 90° to the slope or perpendicular. In this case the weight of the bar acting half way along the bar is balanced by the vertical component of tension in a cable. Remember the horizontal acts through the point of moments thus isolating it from the balance point. Also note to convert everything to forces! (9.81Nkg-1)
W*d/2 = Tsinθ*d
Couples: These are simple pairs of equal and opposite forces acting on a body which cause a turning force or “torque”. The more “torque” the harder it can push an object. Tractors have a lot of torque to pull loads. Racing cars have lower torque as they don’t need the same pulling force. We can take moments about any point on a couple. However, it is easier to do it about the pivot point if that is possible. It does not matter where you take the moments as the couple always comes out to be Fd Take look at this example taken about “A”..
Moments = Fx – F(d-x)
Moments = Fx + F(d-x)
Moments = Fx + Fd-Fx
Moments = Fd
Single Support Forces: If we think about the pivot point in this prism supporting the weights acting downwards we simply summate the forces regardless of where they are acting. This can be used in conjunction with a moments equation as well.
S = W1 + W2 + W0 – Forces
W1d1 = W2d2 + W3d3 – Moments
Double Support Forces: If we think of a simple table as an example the same principles of forces and moments applies.
Sx + Sy = W0
However, we must set up two equations for the unequal support forces. We have in take moments first from point X, then point Y to give us two equations.
SyD = W0d1
SxD = W0d2
Stable Equilibrium: If an object such as a flower basket or a bookshelf is displaced by a force it will return to that point. This is a situation where the centre of mass must be inside of the pivot point. Fd < Wb/2
Unstable Equilibrium: If it is displaced slightly it will not return to that point and will topple over. This is a situation where the centre of mass is outside the pivot point. Fd > Wb/2
Unstable Equilibrium Calculations: We can use the dimensions of our problems and the theory of moments to setup an equation about the pivot point. Weight acts through the centre of the base and the force acts a distance d up the side of the block.
Fd = Wb/2
Toppling Vehicles: We also find trig and moments useful when deciding at which angle a vehicle will topple over on a slope. If we know the geometry of the centre of mass and middle of the base we can construct a triangle of an angle which creates the ratio of similar forces. The geometry causes the forces to act.
Resolving Forces Parallel to Slope….
hyp x sinθ = opp
F = W x sinθ
Resolving Perpendicular to Slope….
hyp x cosθ = adj
S = W x cosθ
Divide the two equations….
F/s = sinθ/cosθ = tanθ = (b/2)/d
A Simple example would be a lorry has a wheelbase of 2.1 m and centres of mass unloaded of 0.7m from the ground. What is the maximum angle it can drive at before toppling over….
d = 0.7m b/2 = (2.1/2)m = 1.05
opp/adj = tanθ
Angle = tan-1(1.05 / 0.7) = 56.3°