Mechanics Tutorial 2 - Equilibrium for Coplanar Forces

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

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:

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.

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

 

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:

 

 

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.

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 48.6 = 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:

 

Question 1

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.

Answer

Question 2

(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”, the companion sets out to measure the angle at P from the horizontal, which are shown 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.                                                 

 

Answer