Lecture 4

1. Resultant of Forces System Lecture 4 Resultant: • it is a single force which can replace the original force system without changing its external effect on rigid body. F1 F4 R F2 F3 • If R > 0 , the resultant will accelerate this body. • If R = 0 , the body will remain in Equilibrium or balanced state. Collinear forces: In this system, line of action of all the forces act along the same line. For example, consider a rope being pulled by two players.Then the resultant force, in the direction of pull of first player is: F1 F2 R = (F1 - F2)

2. Resultant of Forces System Lecture 4 Coplanar parallel force: In this system, all forces are parallel to each other and lie in a single plane. Consider a See-Saw. Two children are sitting in the See-Saw and it is in equilibrium. If the weights of both the boys are (W1and W2), then for equilibrium, the fulcrum reaction should be: R = W1 + W2 Coplanar concurrent point: • In this system, line of action of all forces passes through a single point and forces lie in the same plane. An example is the weight suspended by two inclined strings. • Here, all the three forces pass through point A. • The moment of the forces about the point A is zero. W1 R W2 To find the Resultant Force ( R ): Resolve each force to ( Fy , Fx), with the direction of each force. Find : Rx =  Fx, Ry =  Fy Find : Find : A (direction of R ) 2452 N

3. Resultant of Forces System Lecture 4 • Coplanar non-concurrent forces: • All forces do not meet at a single point, but lie in a single plane. F2 F1 F1 F2 • The Resultant Force ( R ) can be obtained by means of parallelogram law [(R ) is either a single force or a couple] : F2 F3 F4 F1 F3 F3 The forces ( F1 , F2) are moved along their lines of action to intersect by the principle of transmissibility to fine the resultant ( R12 ). The resultant ( R12 ) is then combined with (F3) to find the resultant (R123 ). Finally (R1234 ) is determined by combining R123 ) with (F4). F4 F4 R12 • Or the resultant it can be found by: Resolve each force to ( Fy , Fx), with the direction of each force. Find : Rx =  Fx, Ry =  Fy Find : Find : R123 R1234 (direction of R ) • To find the location of ( R ), [ R . d =  Mo], where ( d ) is the perpendicular distance.

4. Resultant of Forces System Lecture 4 • Non coplanar parallel forces: All the forces are parallel to each other, but not in the same plane. An example is forces acting on the table legs to lift it. To find the Resultant Force ( R ) parallel to ( X or Y – axis ) is determined by the following equations: R =  Fxor R =  Fy To find its location if the forces parallel to (x – axis): R . y =  Mz, and R . z =  My To find its location if the forces parallel to (y – axis): R . x =  Mz, and R . z =  Mx Y Y F3 F2 F1 F3 F1 X X Parallel to ( X – axis ) Parallel to ( Y – axis ) Z F2 Z

5. Resultant of Forces System Lecture 4 • Concurrent - Non coplanar forces: • In this system, all forces do not lie in the same plane, but their line of action passes through a single point ( O ). • Any two forces of a concurrent force system determine a plane, their resultant is determined by parallelogram law. • The resultant of any set of concurrent forces must be a force ( not a couple or force and couple ). z x O R y To find the Resultant Force ( R ): Resolve each force to ( Fy , Fxand Fz ), with the direction of each force. Find : Rx =  Fx, Ry =  Fy and Rz =  Fz Find : Find : Z F3 X Y F1 F2

6. Resultant of Forces System Lecture 4 • Non-Concurrent , Non coplanar & Non-parallel forces: • All forces do not lie in the same plane and their lines of action do not pass through a single point. This is the most general force system. • The Resultant of Non-Concurrent , Non coplanar & Non-parallel force system can be a single force or a couple, but in general, it is a force and a couple. In general: • The Resultant of a parallel force system, either coplanar or noncoplanar, and of a nonconcurrentcoplanar force system is either a force or a couple. • It is a force when the sum of the force components in any direction is different from zero (not equal to zero). • And zero or a couple when the sum of the force components is equal to zero in every direction. • The Resultant force can be located by applying the principle of moments with respect to one or more axes. • When the Resultant is a couple, its moment can be determined by summing moments with respect to one or more axes.

7. Force – Couple System Lecture 4 The combination of the force and couplein the shown figure is refered to as a force – couple system. F Equivalent Couple Changing the values of F and ddoes not change a given couple as long the productFd remains the same.

8. Example ( 1 ):(Concurrent coplanar force system) Lecture 4 Obtain the resultant of the concurrent coplanar forces acting as shown in Figure ( A ) below. Solution: ∑ Fx = + 15 Cos 15 – 75 – 45 Sin 35 + 60 Cos 40 ∑ Fy = + 15 Sin 15 + 105 – 45 Cos 35 – 60 Sin 40 = + 33.453 kN 105kN = -40.359 kN 15kN = 40.359 kN 150 75 kN 400 350 R 60 kN 45kN ∑Fy Fig. A + ve + ve f ∑Fx

9. Example ( 2 ):(Concurrent coplanar force system) Lecture 4 If the magnitude of the resultant force acting on the bracket is to be (450 N) directed along the positive (u axis), determine the magnitude of (F1) and its direction ( ϕ. )? Solution: y F1 u f 30o x F2 = 200 N 13 12 5 F3 = 260 N F1(x) = F1 sin F1(y) = F1 cos  F2(x) = 200 N F2(y) = 0 F3(x) = F3 sin 𝛼= 260 (5/13) = 100 N F3(y) = F3 cis 𝛼= 260 (12/13) = 240 N

10. Example ( 2 ):(Concurrent coplanar force system) Lecture 4 FR(x) = FR cos 30 = 450 cos 30 = 389.7 N FR(y) = FR sin 30 = 450 sin 30 = 225 N (  + ) FR(x) =  Fx ; 389.7 = F1 sin + 200 +100 F1 sin  = 89.7……… ( 1 ) (  + ) FR(y) =  Fy ; 225 = F1 cos  - 240 F1 cos  = 465 ……... ( 2 ) Solving equations ( 1 & 2 ), it yields to:  = 10.9oF1 = 474 N Example ( 3 ):(Non-Concurrent coplanar force system) The force (100 lb) is the resultant of the couple and three forces, two of which are shown in the diagram. Determine the third force and locate it with respect to point (A). Solution: The location and the magnitude of the third force are unknown, so we will suppose the location and its direction as shown in figure below:

11. Example ( 3 ):(Non-Concurrent coplanar force system) Lecture 4

12. Example ( 4 ):(Non-Concurrent Non-coplanar force system) Lecture 4 The ( 50 lb ) force ( R ) in ( Figure below) is the resultant of the four forces, three of which are shown. Determine the fourth force and show it on the sketch. Solution: Let the fourth force to be ( F ), then: 50 = F – 45 Taking the moment about the Y – axis : Taking the moment about the X – axis :

13. Example ( 5 ):(Concurrent Non-coplanar force system) Lecture 4 Determine the Resultant force of the tensions of the guy wire shown in the below figure. Solution: It is convenient to solve such problems to find the length of the wire, then the force multiplier for each force, from which the component for each wire tension is calculated as follows: The length of wire of tension ( 12 lb ) is: The length of wire of tension ( 17 lb ) is: The length of wire of tension ( 18 lb ) is: The force multiplier is: For the force of ( 12 lb ) its components are:

14. Example ( 5 ):(Concurrent Non-coplanar force system) Lecture 4 For the force of ( 18 lb ) its components are: For the force of ( 17 lb ) its components are: Rx = - 1.95 - 4.31 + 4.07 = - 2.19 lb Ry = 11.68 + 17.24 + 16.28 = 45.2 lb Rz = - 1.95 + 2.87 + 2.71 = 3.63 lb