Mechanics Tutorial 2 - Equilibrium for Coplanar Forces

 Contents

Free body diagrams

Real objects are complicated things.  In mechanics, we model them as point masses, which makes things rather simpler.  We do this as a free body diagram which shows all the forces acting on an object as acting on the centre of mass.

Consider this diagram here:

becomes:

We treat all objects as if they are point masses, so that we don’t have to do a very complicated diagram.  All forces act on a single point.

Forces in equilibrium mean that they are balanced.  Coplanar forces act in the same plane.  Two balanced forces are equal in magnitude but opposite in direction to the other.

We can see easily from the free body diagram that the resultant force is zero

 This does NOT mean that there are no forces.  It means that the directions and values of the forces add up to zero.    This is true for any number of forces.   If an object is at rest, it means that it is subject to forces that add up to zero.

If the resultant force is zero, the system is in equilibrium.

If there is an overall resultant force, movement, or a turning effect will result.  Strictly speaking, acceleration will result.

Triangle of Forces

If we are considering three coplanar forces in equilibrium, use the triangle of forces rule:

If 3 forces acting at a point can be represented in size or direction by the sides of a closed triangle, then the forces are in equilibrium, provided their directions can form a closed triangle.

This means that the forces can follow each other round a triangle

Notice how:

• The forces form into a closed triangle.

• The directions of the forces go round the triangle.

Consider this:

The forces F1 and F2 make a closed triangle, but they are not going nose-to-tail.  These forces are NOT in equilibrium.  F1 and F2 make up a resultant FR.

Now let’s reverse the directions of F1 and F2, but keep the magnitudes the same:

If we move F1 we find that the three forces form a closed triangle and the arrows go nose-to-tail.

This is what the three forces look like when acting on a single point:

This means that the system is in equilibrium

The resultant of Forces 1 and 2 will be of the same magnitude as Force 3, but in the opposite direction. This means that the forces will be balanced and the resulting force is zero.  If any force vector polygon forms a closed loop, the forces are in equilibrium. This is the polygon of forces rule and it's true for any number of forces. The study of forces in equilibrium is called statics, and is an important consideration when structures are designed.

In statics, remember that all forces add up to zero.  That does not mean that there are no forces;  the forces balance each other out.   These situations are for external forces acting on an object, but within an object, all forces sum to zero.  Otherwise the object would fly apart.

In statics problems, we need to know how to resolve forces, which can be done by:

• accurate drawing;

• use of trigonometry.

Resolving the Vectors by Accurate Drawing

Consider three forces acting in equilibrium:

We see that the forces (from the weights) form a closed triangle and the directions form a closed loop.  Therefore the forces are balanced.  We can show this as a triangle of forces.

1. Choose a scale (e.g. 2 cm = 1 N)

2. Use graph paper.

3. Use a sharp pencil.

4. Use a compass.

5. Use a protractor if angles are mentioned.

6. Draw the arrows in the direction specified.

Here is the equilibrium situation above represented by accurate drawing:

The angles are measured with a protractor to give the values shown.

Resolving Vectors by Trigonometry

The problem with accurate drawing is that of accuracy.  If you get the answer to the nearest degree, you're doing well.  And accurate drawing is not easy.  If you are challenged by measuring, resolution of vectors using trigonometry is the answer.

Any vector in any direction can be resolved into vertical and horizontal components at 90 degrees to each other.

For three forces in equilibrium we can draw a force vector diagram:

For this situation we know that weight always acts vertically downwards.  We can resolve the two other vectors into their horizontal and vertical components:

1. T1 resolves into T1 cos q1 (horizontal) and T1 sin q1 (vertical);

2. T2 resolves into T2 cos q2 (horizontal) and T2 sin q2 (vertical).

We know that the three forces add up to zero, so we can say:

• T1 cos q1 + - T2 cos q2 = 0.  This means that the forces are equal and opposite.

• Ž T1 cos q1 = T2 cos q2

• T1 sin q1 + T2 sin q2 + - W = 0 Weight is acting downwards and downwards is, by convention, negative.

• Ž T1 sin q1 + T2 sin q2 = W

 Be careful that you don't assume that W is split evenly between T1 sin q1 and T2 sin q2.  This is only true when the weight is half way between the ends.  Many students write T1 sin q = mg which is wrong.

Forces acting symmetrically

Consider a mass m that is suspended, so that it hangs freely, half-way between two points X and Y.  Its weight will be mg and it will cause a tension, T1 in the  left hand part of the string and a tension T2 in the right hand part of the string, as shown:

We can show that this obeys the triangle of forces rule:

An isosceles triangle is formed.     We can resolve the forces T1 and T2.  This is a symmetrical situation. T1 = T2:

The horizontal components add up to zero because they are equal in magnitude, and in opposite directions.

The vertical components add up to mg, the vertical downwards force:

Therefore:

 Many students write T1 sin q = mg which is wrong.

 Worked Example: A 6.0 N weight is attached to a light string which is then tied to the midpoint of a second string of length 0.8 m.   This string is suspended from two fixed points which are on the same horizontal line 0.60 m apart.  The arrangement is shown below:   What is the angle between the two halves of the string?   What is the tension in each half of the string? First of all, draw the forces and the directions: The weight vector splits the set up into two right-angled triangles. The horizontal components are equal and opposite, so cancel each other out. The weight vector can be considered as the resultant of two downward vectors which add up the total downward force.  Since this is a symmetrical system, each vector is 3 N. Each closed triangle looks like this: Now we can find q, the angle with the vertical.  It is the opposite, and we know the hypotenuse, so we use the sine function.  0.4 sin q = 0.3  sin q = 0.3/0.4 = 0.75 q = sin-1 0.75 = 48.6o We now can say that the angle between the two strings is 2 x 48.6o = 97.2o We can now work out the tension: The angle q is the adjacent, so we use the cosine function. T cos 48.6 = 3N T = 3N ÷ cos 48.6 = 3 N ÷ 0.661 = 4.54 N

At AS level, you are most likely to encounter symmetrical systems like this.  If the system is NOT symmetrical, don't panic.  Remember:

• The weight (downwards force) can be split into two force vectors.  These add up to the weight.

• The horizontal force vectors add up to zero.

• If you are completely stuck, use accurate drawing!  Whatever you do, don't leave a blank!

 A ship is in a dock and is secured to bollards on the harbour walls by two thick ropes.  It is facing into a very strong wind which is putting a force of 28 000 N on the ship.  This is shown in the diagram below:     (a) Calculate the tension T in the top rope. (b) Calculate the force at 90 degrees to the wind that is acting on the bottom bollard.  Which direction is it acting in? (c) What is the overall force acting on the ship?  Explain your answer. (d) The bottom bollard was not very well put in, and it gets ripped out of the harbour wall.  Calculate the resultant force acting on the ship immediately after the bollard gives way, and state the direction relative to the wind.  0o is parallel to the wind.

Non-symmetrical Equilibrium (Extension)

This is more challenging, but is not that difficult.  The key things to remember are that:

• The tension T1 and T2 will be different.

• The angles q1 and q2 will be different.

Consider a mass, m, which is being suspended freely between two points X and W, but its point of suspension is to the left of the midpoint.  The weight is mg.

We can apply these rules:

• The two horizontal components add up to zero;

• The two vertical components add up to the weight.

The vertical components:

The horizontal components:

Since there are can be two unknowns, simultaneous equations may be needed.

 Worked Example Three forces are shown below are in equilibrium.  The diagram is NOT to scale. Calculate the magnitude of the Force F and the angle . Answer This may appear quite difficult at first sight as there are two unknowns. The key thing to remember is that the resultant force is 0. Therefore the force F is equal in magnitude and opposite in direction to the force F' (pronounced “eff-prime”). The force F' is the vector sum of the 30 N force and the 45 N forces.  Therefore the solution to this is to resolve the vectors into vertical and horizontal components.   Resolve the 30 N and the 45 N forces into vertical and horizontal components.  Upwards is positive. Vertical Components are added:   Horizontal components are added:   The minus sign is there because the horizontal component of the 30 N force is from right to left, i.e. in the opposite direction to the horizontal component of the 45 N force. So we can work out the magnitude of the force F':   We can work out the angle that the force makes with the horizontal: So our force in equilibrium looks like this:

 (Challenge)   A rock climber of mass 68 kg is making a traverse horizontally across a cliff face 20 metres wide between the fixed points P and Q.  When he has got 12.5 metres across the rock face, he loses his footing and dangles in mid air.  While the climber is shouting at him to “get a life and do something useful”, the companion, who is a keen physics student, sets out to measure the angle at P from the horizontal, which he finds to be 20o as in the diagram.   a.       What is the weight of the climber?                                                           b.      Show that the angle q is about 31o.                                                          c.       Show that force T2 is 1.1 × T1.                                                                  d.      Calculate the values of forces T1 and T2.

 Be careful that you don't assume that W is split evenly between T1 sin q1 and T2 sin q2.  This is only true when the weight is half way between the ends.  Many students write T1 sin q = mg which is wrong.   Avoid using bearings (popular in Maths Mechanics) when calculating the angle between vectors - unless, of course, you are asked about planes or ships.