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- 1. FORCES
- 2. Key Words <ul><li>Polygon </li></ul><ul><li>Parallelogram </li></ul><ul><li>Equilibrium </li></ul><ul><li>Resultant </li></ul><ul><li>Magnitude </li></ul><ul><li>Diagram </li></ul><ul><li>Bow’s Notation </li></ul><ul><li>Cartesian Plane </li></ul><ul><li>Representing </li></ul><ul><li>Force </li></ul>
- 3. Overview This chapter helps us to understand what forces are and helps us to apply this knowledge to understand how forces act in engineering components. To master this chapter, work step by step and practise as many examples as possible.
- 4. Definition of a force <ul><li>A force is an influence which changes or tends to change the state of rest or uniform motion of a body. </li></ul><ul><li>Remember this definition although it is often more convenient to think of force as a ‘pull’ or a ‘push’. Forces are always measured in newtons (N). </li></ul><ul><li>One newton is the force required to accelerate a mass of 1 kg at 1 m/s2. </li></ul><ul><li>Forces are vector quantities and to specify a force fully, it is necessary to state the following: </li></ul><ul><li>• magnitude (the strength of the force) </li></ul><ul><li>• line of action (which is the straight line along which the vector acts and indicates direction) </li></ul><ul><li>• direction along the line of action of the force. (Is the force pushing or pulling?) </li></ul>
- 5. Pushing or Pulling Forces <ul><li>You can see in the diagrams below that the forces acting on the body are identical in magnitude, line of action and direction. </li></ul><ul><li>The force in the diagram on the left is pushing on the body while the force in the diagram on the right is pulling on the body. </li></ul><ul><li>In both cases the effect on the body is the same. </li></ul>
- 6. Representing Forces <ul><li>When representing a force on paper (as below): </li></ul><ul><ul><li>Draw the line of action of the force. </li></ul></ul><ul><ul><li>Mark the length of the force on the line of action using a suitable scale. (Use as large a scale as possible for accuracy). </li></ul></ul><ul><ul><li>Place an arrowhead on this line to show the direction in which the force acts. </li></ul></ul><ul><ul><li>Write the magnitude of the force on the line. </li></ul></ul><ul><ul><li>5N </li></ul></ul>
- 7. Describing Forces <ul><li>This diagram will help you to understand how to describe several forces acting on a body. </li></ul><ul><li>The diagram represents four forces whose lines of action pass through the origin of a Cartesian plane </li></ul><ul><li>A number of forces acting on a body are known as a system of forces. This diagram is an example of a system of forces. </li></ul>6N 5N 4N 7N 135 0 45 0 30 0 30 0 240 0
- 8. Describing the Directions in which Forces Act. <ul><li>Method 1: Using compass points </li></ul><ul><li>Using the points of the compass, this diagram represents: </li></ul><ul><ul><li>a force of 4 N in an easterly direction </li></ul></ul><ul><ul><li>a force of 5 N in a direction of 30° north of east </li></ul></ul><ul><ul><li>a force of 6 N in a northwest direction </li></ul></ul><ul><ul><li>a force of 7 N in a direction of 60° south of west. </li></ul></ul>6N 5N 4N 7N 45 0 30 0 60 0 E N W S
- 9. Describing the Directions in which Forces Act. <ul><li>Method 2: Using 360° from true north </li></ul><ul><li>Using 360° clockwise from true north, the diagram represents: </li></ul><ul><ul><li>a force of 4 N on a bearing of 90° </li></ul></ul><ul><ul><li>a force of 5 N on a bearing of 60° </li></ul></ul><ul><ul><li>a force of 6 N on a bearing of 315° </li></ul></ul><ul><ul><li>a force of 7 N on a bearing of 210°. </li></ul></ul>6N 5N 4N 7N 90 0 210 0 315 0 0 0 60 0
- 10. System of Forces <ul><li>Resultant </li></ul><ul><ul><li>If a system of forces acts on a body and a single force can be found that has the same effect as the system, that single force is known as the resultant of the system. </li></ul></ul><ul><li>Equilibrium </li></ul><ul><ul><li>When two or more forces act on a body and the body remains at rest, the forces are said to be in equilibrium. </li></ul></ul><ul><li>Equilibrant </li></ul><ul><ul><li>If a system of forces acts on a body but a single force keeps the body at rest, the single force is known as the equilibrant of the system of forces. The equilibrant of a system of forces has the same magnitude and line of action as the resultant but is opposite in direction. </li></ul></ul>
- 11. Parallelogram of Forces <ul><li>their resultant is represented in magnitude and direction by the </li></ul><ul><li>diagonal drawn from that angular point . </li></ul>resultant Force 2 Force 1 at a point If two forces whose lines of action meet are represented in magnitude and direction by the sides of a parallelogram drawn from one of its angular points,
- 12. Working Examples 1 <ul><li>Two pieces of string are attached to a nail. The pull in one string is 4 N in an easterly direction and the pull in the other string is 6 N in a direction of 60° north of east. </li></ul><ul><li>Determine the resultant and equilibrant forces on the nail. </li></ul>Scale 1 cm = 1 N 6 N R 60° 4 N 36,5° 8,75 N Resultant= 8,75N 36,5 0 north of east
- 13. Working Examples 2 <ul><li>Two rods are attached to a pin. One rod is pulling on the pin with a force of 40 N in an easterly direction and the other is pushing on it with a force of 60 N in a direction of 60° south of west. </li></ul><ul><li>Determine the resultant and equilibrant forces on the pin. </li></ul><ul><li>Before you can attempt this problem using the parallelogram of forces, you must convert the 60 N pushing force to a pulling force along the same line of action. This pulling force is represented by the dotted line in the diagram. </li></ul>
- 14. Example 2 <ul><li>Two rods are attached to a pin. </li></ul><ul><li>One rod is pulling on the pin with a force of 40 N in an easterly direction and the other is pushing on it with a force of 60 N in a direction of 60° south of west. </li></ul><ul><li>Determine the resultant and equilibrant forces on the pin. </li></ul>60 N 40 N Scale 1 cm = 10 N 60° 79° 53 N Resultant = 53N 79° south of east Equilibrant = 53N 79° N orth of west
- 15. Example 3 <ul><li>Two rods are pushing on a pin with forces of 200 N and 150 N respectively. </li></ul><ul><li>The 200 N force acts in a westerly direction and the 150 N force acts in a south-easterly direction. </li></ul><ul><li>Determine the resultant of the two forces and the force which keeps the pin at rest, that is the equilibrant. </li></ul>Scale 1 cm = 50 N 200 N 150 N 135° 48,5° 200 N 150 N R=142 N Resultant = 142N 48,5° south of west
- 16. Components of a force <ul><li>If two or more forces have the same effect as a single force, these forces are called the components of the single force. </li></ul><ul><li>If the diagonal of the parallelogram in the following figure represents R newtons and the sides P newtons and Q newtons, then P newtons and Q newtons are the components of R in their respective directions. </li></ul>P R Q
- 17. Components of a force <ul><li>It is often necessary when solving problems to replace a force by its components at right angles to each other. </li></ul><ul><li>In other words, the x and y co-ordinates of force are determined. </li></ul><ul><li>This can be very useful when forces (or vectors) are added together, by taking the sum of their x and y components. </li></ul>
- 18. Example <ul><li>The following diagram shows how to resolve any force into its vertical and horizontal components. </li></ul><ul><li>The force in this case is a 100 N force in a direction of 30° north of east. </li></ul><ul><li>The components are found by constructing a parallelogram whose sides are parallel to the x and y axis of the drawing. </li></ul>
- 19. Example <ul><li>The x and y axes form what is known as a Cartesian plane. </li></ul><ul><li>The x and y components are then simply measured off the x and y axes. </li></ul><ul><li>In this case the x and y components of the 100 N force are x = 87 N and y = 50 N. </li></ul>
- 20. Assessment <ul><li>Page 119 </li></ul>
- 21. Introduction to triangle of forces <ul><li>In the previous section you learnt how to graphically determine the resultant and equilibrant of two forces (whose magnitude and direction were known), by means of the parallelogram of forces. </li></ul><ul><li>Although useful, this method is limited to solving problems with only two forces. </li></ul><ul><li>When a system of forces contains three forces, we make use of a triangle of forces. </li></ul>
- 22. Definition of triangle of forces <ul><li>If three forces, whose lines of action meet at a point, can be represented in magnitude and direction by the sides of a triangle, taken in order, they are in equilibrium. </li></ul><ul><li>The converse of this is also true, that is: “If three forces are in equilibrium their vectors can be put together to form a triangle.” </li></ul>
- 23. Bow’s notation <ul><li>Bow’s notation is a method which can be used to simplify problem solving where three or more forces are applied to a body in a system of forces. </li></ul><ul><li>The following system of forces contains three forces. </li></ul><ul><li>We will use Bow’s notation to construct a triangle of forces and show that they are in equilibrium. </li></ul><ul><ul><li>Part 1: Construct a space diagram </li></ul></ul><ul><ul><li>Part 2: Construct a force diagram </li></ul></ul>
- 24. Construct a space diagram <ul><li>Step 1 We construct a space diagram, depicting the lines of action and direction of all the forces in the system. </li></ul><ul><li>Step 2 Once this is done, we label the spaces between all force lines, using capital letters. (Any letters will do, for example A, B and C). </li></ul>120° 120° 120° 10N 10N 10N B A C
- 25. Construct a force diagram <ul><li>Step 1 Moving clockwise around the system of forces in the space diagram, you will see that all the forces represented between two capital letters can be expressed as vectors in the force diagram with lower case letters at either end. </li></ul><ul><li>The first 10 N force can be referred to as the force between the spaces A and B or the force ab. </li></ul><ul><li>The second and third 10 N forces are bc and ca respectively. </li></ul><ul><li>Construct the forces in the force diagram at the same angles as those in the space diagram. </li></ul>
- 26. Space Diagram to Force Diagram <ul><li>Step 2 The forces are arranged from head to tail in the order in which they are taken. </li></ul><ul><li>They are considered to be in equilibrium because the force diagram’s end point is the same as its starting point. </li></ul>a 120° 120° 120° 10N 10N 10N B A C b c
- 27. Example 1 <ul><li>Two ropes, inclined at 30° and 45° to the vertical, support a load of 200 N. </li></ul><ul><li>Determine the tensions, that is pulls in both ropes. </li></ul>Scale 1cm = 50 N B A C 30° 45° 200N 45° 30° a b c
- 28. Example 1 <ul><li>Measuring lines bc and ca and using the scale, we find that the tension in the ropes are as follows: </li></ul><ul><ul><li>Rope bc represents a pulling force of 104 N in a direction of 45° north of west. </li></ul></ul><ul><ul><li>Rope ca represents a pulling force of 147 N in a direction of 60° north of east. </li></ul></ul>
- 29. Example 2 <ul><li>The following space diagram shows two rods which are joined to a pin. </li></ul><ul><li>The angle between the rods is 60°. </li></ul><ul><li>The horizontal rod is pushing with a force of 1 000 N in a westerly direction. </li></ul><ul><li>The inclined rod is pulling with a force of 800 N in a direction of 60° north of east. </li></ul><ul><li>Determine the force exerted by the pin, that is the equilibrant. </li></ul>
- 30. Example 2 <ul><li>We will now introduce a method to help us find the unknown force, or equilibrant. </li></ul><ul><li>Step 1 </li></ul><ul><ul><li>Draw the space diagram. </li></ul></ul><ul><ul><li>Nothing as yet is known about the third force (the equilibrant), so indicate it in the space diagram as a curly line. </li></ul></ul><ul><ul><li>Do this so that you can label the space diagram. </li></ul></ul>
- 31. Example 2 <ul><li>Step 2 </li></ul><ul><ul><li>Label the spaces L, M and N as follows: </li></ul></ul>800N 1000N 60° N L M
- 32. Example 2 <ul><li>Step 3 </li></ul><ul><ul><li>Construct the force diagram in the same way as in the previous example to determine the unknown force. </li></ul></ul>800N 1000N 60° N L M m l n
- 33. Example 3 <ul><li>Two forces of 50 N and 25 N act on the pin O, as shown in the diagram. </li></ul><ul><li>The line OX is horizontal. </li></ul><ul><li>Determine the resultant force acting on the pin, as well as the horizontal and vertical components of this resultant. </li></ul>50N 25N B A C 45° 30° O X
- 34. Step 1 <ul><li>Construct a force diagram. </li></ul><ul><li>We proceed as in the previous example; bc is the vector of the equilibrant of the two given forces. </li></ul><ul><li>The force diagram has been drawn to a scale of 1 cm = 2 N </li></ul>50N 25N B A C 45° 30° O X c a b
- 35. Step 1 <ul><li>The resultant of the two forces is equal in magnitude to the equilibrant, but opposite in direction. </li></ul><ul><li>The resultant is therefore represented by the force ac. </li></ul>c a b It is measured to be 49 N in a direction 78° north of east.
- 36. Step 2 <ul><li>Determine the horizontal and vertical components of the resultant. </li></ul><ul><li>Construct a Cartesian plane through the origin that is point a. </li></ul><ul><li>Construct two parallel lines from point c to the x and y axis respectively, and measure the horizontal and vertical components. </li></ul>c a b c a 9.7N 47.8N
- 37. Step 2 <ul><li>We will find: </li></ul><ul><ul><li>The horizontal component is 9,7 N. </li></ul></ul><ul><ul><li>The vertical component is 47,8 N. </li></ul></ul><ul><li>You could probably have worked out some of the above examples using the parallelogram of forces. </li></ul><ul><li>However, it is essential to be familiar with the use of the triangle of forces method, and Bow’s notation. </li></ul>
- 38. Assessment <ul><li>Pages 128 & 129 </li></ul>
- 39. Introduction to Polygon of Forces <ul><li>Definition </li></ul><ul><li>If any number of forces whose lines of action meet at a point can be represented in magnitude and direction by the sides of a polygon, taken in order, then they are in equilibrium. </li></ul><ul><li>The polygon of forces corresponds to the triangle of forces. </li></ul><ul><li>It is used in a similar manner, as shown in the examples on the following page. </li></ul>
- 40. Example 1 <ul><li>Four rods, joined to a pin, are in equilibrium. </li></ul><ul><li>The forces in two of the rods are known to be100 N and 50 N, as shown in the following figure. </li></ul><ul><li>Determine the forces in the other two rods. </li></ul>B C A D 45° 90° 75° 50N 100N
- 41. Step 1 <ul><li>Proceed as when using the triangle of forces and label the spaces between the forces in the space diagram. </li></ul>B C A D 45° 90° 75° 50N 100N
- 42. Step 2 <ul><li>Choose a convenient scale (in this case 1 cm = 25 N). </li></ul><ul><li>Construct the vectors ab and bc which are the known forces. </li></ul><ul><li>This will enable you to find points a, b and c. </li></ul>B C A D 45° 90° 75° 50N 100N a b c
- 43. Step 3 <ul><li>Since the vector of the force cd is horizontal and that of da is vertical, construct horizontal and vertical lines through points c and a respectively. </li></ul><ul><li>The point where these lines intersect is d . </li></ul>B C A D 45° 90° 75° 50N 100N a b c d
- 44. Step 4 <ul><li>To complete the vector diagram, the arrows must be introduced from head to tail, forming a polygon of forces. </li></ul><ul><li>It is found that the line cd represents the force in the horizontal rod which acts in a westerly direction with a magnitude of 114 N. </li></ul><ul><li>The line da represents the force in the vertical rod which acts in a southerly direction with a magnitude of 45,7 N. </li></ul>a b c d
- 45. HINTS <ul><li>When using the polygon of forces, always draw the vectors of all the known forces first. </li></ul><ul><li>It is also sometimes useful to convert pushing forces into pulling forces, (as we did when solving problems using the parallelogram of forces method). </li></ul>
- 46. Example 2 <ul><li>Determine the equilibrant of the four forces as shown in the following figure: </li></ul>90° 60° 75° 100N 90N 75N 150N A C B E D a b c d e Scale 1 cm = 25 N By measuring ea, we find the equilibrant = 116 N , 75° north of east. (As a point of interest, the resultant of the four forces is 116 N in a direction 75° south of west.)
- 47. Example 3 <ul><li>The following figure shows four forces in equilibrium acting at a point. </li></ul><ul><li>Determine the magnitude of forces P and Q . </li></ul><ul><li>PROBLEM: In the space diagram we have a force of known magnitude followed by a force of unknown magnitude </li></ul>200N P 90° 45° 75° Q
- 48. Example 3 <ul><li>Force P which is represented by he is then measured and found to be 144 N in the direction of NE. </li></ul><ul><li>Force Q which is represented by gh is then measured and found to be 72 N in a northerly direction. </li></ul>200N P 90° 45° 75° Q E F G H g f h e
- 49. Assessment <ul><li>Pages 134 to 136 </li></ul>

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