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Dept of REFM Topic Equilibrium of Forces and Moments “Mechanical Engineering Science, 2nd Edition” by J. Hannah & M. J. Hillier, Longman Chapter 1 – Statics, Chapter 2 - Moment _____________________________________________________________________ Reference: Force and Newton's First Law If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion). To understand and use the First Law it is helpful to first develop a concept of a state of Equilibrium. Equilibrium exists when all the forces on a particle are in balance. The velocity of a particle does not change, if the particle is in Equilibrium. Interpretations of the First Law i) A body is in Equilibrium if it moves with constant velocity. A body at rest is a special case of constant velocity i.e. v = 0 = constant. ii) For a body to be in Equilibrium the resultant force (meaning the vector addition of all the forces) acting on the body must be zero. iii) A Force can be defined as 'that which tends to cause a particle to accelerate', assuming that the force is not in Equilibrium with other forces acting on the body. Note: A force cannot be seen only the effect of a force on a body may be seen. Force Units The S.I. unit of force is the Newton abbreviated as N. For example 217 N. For most engineering applications forces are of the order of N or kN. Force Vectors - Components in Polar and Rectangular coordinates A Force is a vector quantity since it has direction as well as magnitude. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM fy Fy F Fx fx Figure 1 In the x-y plane the force vector F in the above diagram can be resolved into its components in the x and y directions (rectangular coordinates), where, F = Fx Fy ; where Fx , and Fy are the components of F in the x and y directions. The vector F can also be represented in terms of components in polar coordinates, F and . The components in rectangular (x,y) coordinates can be expressed in terms of the components in polar coordinates as follows: Fx = F.cos; and Fy F.sin; and the inverse relationship as: Fy 2 2 F = Fx Fy and tan -1 Fx Addition of Force Vectors and the Resultant Force Figure 2 shows forces F1 and F2 acting at a point (on a particle). The vector R is called the Resultant and is equal to the vector sum (addition) of force vectors F1 and F2. The resultant force can be found by calculation (Algebraic) method or graphical (triangle of force) method. Force F1 Force F2 Particle Note Resultant R means "equivalent to" Figure 2 ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Algebraic Method The vectors can be added algebraically by first resolving the forces into their rectangular components in the x and y directions, adding the components and then converting back into polar coordinates. F2, y F2 2 F1,x F2, x 1 F1 F1,y Figure 3 R x F1,x F2 ,x F1 cos 1 F2 cos 2 Ry F1, y F2, y F1 sin 1 F2 sin 2 R Rx Ry 2 2 Ry tan 1 Rx Triangle of Forces Method In this method the vectors are drawn so that the tail of one vector joins the head of the preceding vector. The length of each force vector is proportional to the magnitude of the force it represents and is drawn parallel to the direction of the force. The vector R is then drawn to close the triangle. The order in which the vectors are drawn is not important as long as they are drawn head to tail. Figure 4 shows F1 drawn first and Figure 5 shows F2 drawn first. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Fy Fy F2 R F1 R F2 F1 Fx Fx Figure 4 Figure 5 Equilibrium of Forces on a Particle (Equilibrium of Concurrent Forces) F1 Equilibrant E Equilibrant E Force F2 Resultant R Figure 6 From Newton's First Law for a particle to be in equilibrium there can be no resultant force on the particle. The Equilibrant (the force which puts a system of forces into equilibrium) must therefore be equal and opposite to the Resultant R to reduce the force on the body to zero. Equilibrant = - Resultant E R R Fy E F2 F1 Fx Figure 7 ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM The equilibrant may be shown by reversing the direction of the arrow of the resultant. In the triangle of forces method this makes all the arrows on the vector diagram point in the same direction around the triangle. This then provides a test for equilibrium. If all the forces in a triangle of forces diagram are in the same direction then the forces must be in Equilibrium. If there are more than three forces the triangle becomes a polygon of forces. If the diagram does not close then there is a net resultant force i.e. the body (particle) is not in equilibrium and will tend to accelerate. (i.e. F = ma ,) Equilibrium of Components of Force Vectors for Coplanar and Concurrent forces Coplanar forces: force vectors being in one plane- two dimensional system (e.g. forces in x-y plane, no forces in z direction) Concurrent forces: all forces vectors acting at one point or one particle The rectangular components Fx and Fy of the force vectors which are in equilibrium are independent variables. Therefore, if the sum of the vectors add up to zero, then the sum of their components must also add up to zero. Hence, for equilibrium i=n Fi 0; it follows that i=n Fx, i 0; and i=n F i=0 i=0 i=0 y, i 0; for example, for a particle under the action of three forces to be in equilibrium, Fx,1 Fx,2 Fx,3 0; Fy,1 Fy,2 Fy,3 0; In other words for equilibrium all the components of forces in the x direction must add up to zero and all the components in the y direction must add up to zero. Polygon of Forces The same methods can be applied to multiple forces as to three forces but instead of a triangle of forces the diagram becomes a polygon. Consider a particle in equilibrium with five forces acting on it. F3 F2 F5 F1 F4 Figure 8 F3 F2 F5 F1 F4 Figure 9 ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Figure 8 shows the forces acting on a particle which is not a very practical case. Figure 9 shows a more realistic case with the forces acting on a rigid body but the result is the same provided that all the forces intersect at a common point. (i.e. Concurrent) Particle is a material body whose linear dimensions are small enough to be irrelevant. Rigid Body is a body that does not deform (change shape) as a result of the forces acting on it. Figure 10 shows the polygon of forces. If any of the five forces were not present the body would not be in equilibrium and the polygon of forces would not close. For example suppose force F5 is removed, then there would be a net resultant R (Figure 11) acting on the body, equal and opposite to the force F5. F1 F1 F2 F2 F4 F4 Fy F3 F5 F3 y R Fx Figure 10 x Figure 11 In the latter case the unbalanced force or resultant has the same magnitude but opposite direction to the force F5. Any one of the five forces can be considered as the equilibrant to the remaining four forces. R = -F5 Or equilibrant F5 = - Resultant R Reactions, External Forces and Free Body Diagrams Newton's Third Law The forces of action and reaction between bodies in contact have the same magnitude, but opposite in direction.. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Hammer BANG! Solid Surface Figure 12 Free Body Diagram of Hammer Hammer Free Body Boundary Force on Hammer caused by Surface Force on Surface caused by Hammer Solid Surface Free Body diagram of Surface Figure 13 Figure 12 shows a hammer striking a solid surface and Figure 13 shows how the forces between the hammer and the surface can be viewed separately. Isolating one body or group of bodies and considering the external forces on the body is referred to as a free body diagram (the body is free of other connected bodies). An imaginary boundary is drawn around the free body to show which parts and forces are to be included in the analysis. The ‘Force on the Surface caused by the Hammer’ is an Action Force due to the hammer and the ‘Force on the Hammer caused by the Surface’ is the Reaction Force on the hammer. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM In most problems, Reaction Forces are those forces produced on a body by the fixed points supporting the body. In the above case the ‘Force on the Hammer caused by the Surface. Another example is the weight of a body supported by a solid surface. W Free Body W Boundary N Reaction Wa Action Figure 14 Consider a body of weight W supported by a solid surface. The body is in equilibrium since the body's weight W, is balanced by an equal and opposite reaction N. Mathematically, W = N. Similarly there will be a resultant action Wa of the body on the solid surface. In the right hand diagram the dotted box encloses what is known as a free body diagram and shows all the external forces that act on a body, both actions and reactions. Summation of the forces over a free body diagram will determine the resultant force acting on the body which will be zero if the body is in equilibrium. System of Particles or Bodies Two or more bodies or particles connected together are referred to as a system of bodies or particles. External Forces External forces are all the forces acting on a body defined as a free body or free system of bodies, including the actions due to other bodies and the reactions due to supports. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Transmissibility of Force Statics and dynamics of rigid bodies are concerned only with the external forces applied to a body and it is assumed that a body does not deform. It also assumes that it does not matter at which point on the body, along the line of action of the force, the force is applied. This is known as the principle of transmissibility of force. F F F Figure 15 Loads are forces that are applied to bodies or systems of bodies. For example, gravity acting on a mass will produce a load called the weight of a body; pressure inside a hydraulic piston will produce a load on the piston; wind acting on a building will produce a load on the building; water flowing from the nozzle of a turbine will produce a load on the turbine blade, etc. Reactions at points supporting bodies are a consequence of the loads applied to a body and the equilibrium of a body. Tensile and Compressive Forces When a body is in equilibrium under the action of two forces (which must be equal and opposite if the body is in equilibrium) the forces can either, a) Push on the body which is called a compressive force; or, b) Pull on a body which is called a tensile force. Reaction Action Reaction Action Tensile Force Compressive Force Figure 16 The body is said to be in tension or in compression. Procedure for drawing a free-body diagram Since we must account for all the forces acting on the particle, the importance of drawing a free-body diagram before applying the equation of equilibrium to the solution of a problem cannot be over-emphasized. To construct a free-body diagram, the following three steps are necessary. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Step 1: Imagine the particle to be isolated or cut “free from its surroundings. Draw or sketch its outlined shape. Step 2: Indicate on this sketch all the forces that act on the particle. These forces can be applied surface forces, reaction forces and/or force of attraction. Step 3: The forces that known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that unknown. Example 1 Fp Supported pulley Pulling Force Fs F1 Spring F1 Spring Pulling Force Mass Fs Mass mg Example 2 Ceiling Support Ring Weight = 10 N Consider a weight connected to a ring that is hanging from the ceiling. Draw a free body diagram for the weight and the ring showing all external forces on the free body. Also show the force on the ceiling due to the weight. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Ceiling Support Action Load = 10 N Reaction Force = 10 N Ring Weight = 10 N Free Body boundary Gravity Load =10 N Example 3 The diagram shows a barge been towed by two tugs using tow ropes connected to a ring on the barge’s bow. The tow rope from tug No. 1, exerts a force of 10 kN on the barge in line with the rope and the tow rope from tug No. 2 exerts a force of 30 kN. tow rope Tug No.1 25 Barge 60 Ring . Tug No.2 tow rope a) Draw a free body diagram of the ring showing the reaction force due to the barge in terms of its two components. b) Find the resultant force acting on the barge and its components in the x and y directions. c) Find the reaction force on the ring due to the barge. d) Find the resultant R of the two forces in tow ropes No.1 & No. 2 from the components in the x and y directions. Solution ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM R x 10 cos 25 30 cos 60 24.06 kN R y 10 sin 25 30 sin 60 2175 . kN R 24.06 2 2175 . 2 32.43 kN 2175 . tan 1 42.1 24.06 The equilibrant E R E = 32.43 kN 42.1 42.1 R = 32.43 kN The resultant R is the sum of the actions of the tow ropes on the barge. The equilibrant E is the reaction of the barge to the tow ropes. Try yourself by graphical method: a) Draw a triangle of forces with scale of 1cm = 5 kN and find the reaction force on the ring due to the barge and its components in the x and y directions. b) Draw a diagram showing the total reaction force on the ring as well as the tow rope forces; c) What is the resultant force on the barge. d) Compare your result with the calculation method. Moments and Couples The Definition of a Moment When a force acts on a body (not in equilibrium) in addition to tending to cause it to translate (move in a straight line), it can cause it to rotate. The force multiplied by the perpendicular distance to the point of rotation is called a Moment. Moment is a measure of the turning effect about a point (or an axis). Moment = Force X Perpendicular Distance M Fd ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM F A A d Figure 17 (a) M=Fxd Figure 17 (b) Figure 17 (a) shows a force F acting at a perpendicular distance d from a point A which creates a moment M = F.d. The distance d is called the moment arm. Moments are shown diagrammatically by a curved arrow as in Figure 17 (b). If the force acts at a point B which is a distance r from A then the moment must be obtained from the perpendicular distance d and not the distance r. F A r B d M = F.d = F.r.cos Figure 18 Moment = Force X Perpendicular Distance M F r cos The value of the moment M is said to be obtained by taking the moment of the force F about the point A or briefly by taking moments about A. Note that the in the above examples the body is not in equilibrium hence the force F tends to accelerate the body in translation and rotation. Moment Vector Moment is the cross product of the force vector and the distance vector. Therefore moment is also a vector. Its magnitude is equal to the product of force and the perpendicular distance from a point (or axis). The direction of the moment vector is perpendicular to the plane containing the force and distance vectors or along the direction of the axis of rotation. In two dimensional problems, the direction of the moment vector can be described completely by its sense, that is clockwise or anti-clockwise. Summation of moments in a single plane then reduces to the algebraic sum, where a +ve is assigned arbitrarily to one sense or the other. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Resultant (force and moment) of a system of forces Figure 19 (a) shows an arbitrary body subjected to a number of forces F1, F2, F3. The combined effect, or resultant, R, of these three forces is determined by the vector sum, so that R = F1 + F2 + F3 These can be evaluated graphically by scale drawing, or analytically by resolving each force into components, say x and y, summing the components in each direction to give Rx = F1x + F2x + F3x Ry = F1y + F2y + F3y and then recombining to find the resultant force. Figure 19 (b) shows clearly that the two methods are equivalent but, whichever is used, only the magnitude and direction of the resultant can be found. The line of action of the resultant is still unknown. l3 O Rx F3 l2 l1 l R F2 F 3y F3 R Ry F2 F1 F1 F 1y F 1x (a) F 3x F 2y F 2x (b) Figure 19 Resultant of a system of forces The line of action can be obtained by equating the sum of moments of F1, F2 and F3 about some arbitrarily chosen point, to the moment of the resultant, R, about O. Taking anti-clockwise positive, this gives Mo = F1l1+ F2l2 + F3l3 = Rl where l is the moment arm length of the line of action of R from O. Hence the position of R is determined (Figure 19a). ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM In some cases the vector diagram, Figure 19b, may close by itself, so that R = 0; however, this does not necessarily mean that the sum of the moments is also zero. Consider the special case of two equal and opposite forces, F, distance d apart, as shown in Figure 20. Clearly there is no resultant force, but taking moments about O, clockwise positive, gives Mo = F(d + l) - Fl = Fd O l F d F Figure 20 A system such as this is known as a couple. Note that the moment of a couple is the same about every point in its plane. Example 4 The diagrams show a body subjected to four forces; calculate the total (resultant) moment on the body. a) 300 mm 15 N 30 N 170 mm 30 N A 100 mm 15 N Taking moments about the corner A + M 30 017 . 15 0.3 30 0.05 15 01 . 30 0120 . 15 0.2 6.6 Nm ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics 50 mm Dept of REFM Note that the forces form two couples or pure moments 3.6 Nm and 3.0 Nm (resultant force =0, moment is the same about any point) Equilibrium of Moments In previous section, the forces on a body were defined as being in equilibrium when ‘the vector sum of all the forces acting on the body was zero’. In the same way, the sum of all the moments is zero when the body is in moment equilibrium. This can be written mathematically as: n M i 0; M1 M 2 ....... 0 or i 1 It can be proved that if the body is in equilibrium the sum of the moments of all the forces on acting on a rigid body is the same for all points on the body. This means that it does not matter at which point on a rigid body you choose for taking moments about. In Example 4, the bodies are in vertical and horizontal force equilibrium thus the moment does not depend on where you take moments. In general this is not the case. If there is a non-zero force on the body the resultant moment will depend upon where you take moments. Example 5 0.5 m 10 N B 15 N 1.0 m 1.25 m A 5N 2.5 m 3.0 m Figure A For the body shown in Figure A find the resultant moment and the equilibrant moment. Note that in this case the body is not in vertical and horizontal equilibrium. Taking moments about A, gives the sum of the moments i.e. the resultant moment M R 10 0.5 15 10 . 5 2.5 22.5 Nm but if moments are taken about B the result is, + ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM + M R 10 2.5 15 0.25 5 0.5 18.75 Nm This difference is because the body is not in vertical and horizontal equilibrium as was the case in the previous examples. There is therefore no unique value for the resultant moment since the value depends on where the resultant force acts. General Equations of Equilibrium of a Plane (Two Dimensional) Rigid Body (Non-concurrent forces) A plane rigid body is one for which all the force vectors lie in the two dimensional x y plane . For equilibrium, i=n n Fi 0; M i 0 i 1 i=0 and expanding the force vector into its components in the x and y directions, i=n Fx, i 0; and i=n Fy, i 0; M i 0 n i=0 i=0 i 1 for example, with three forces acting on a rigid body, Fx,1 Fx,2 Fx,3 0; Fy,1 Fy,2 Fy,3 0; M 1 M 2 ....... 0 For complete equilibrium of a plane rigid body all three equations must be satisfied. Applications of Equilibrium Types of Beam Supports Many engineering components can be 'approximated' or 'modeled' by a single span beam. The calculation of stresses and deflections in a beam is often preceded by an analysis of equilibrium. In most cases this involves obtaining the reactions i.e. the forces and moments at the supports of the beam. Consider the following cases; ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM A beam simply supported on points or knife-edges A side view or elevation of a simply supported beam A diagrammatic view of a simply supported beam A beam built in to a wall Called a 'built in' beam or 'fixed beam' and sometimes called a ‘Cantilever’ Diagrammatic sketch of built in beam Types of Support and Connections Deciding how a body or a structure is supported (connected to the ground) or connected to an adjacent body is very important when calculating the reactions applied to the body.(free body diagram). The three most common assumptions are: ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Force perpendicular to surface only. Pin Joint supported by a roller Ry Both parallel and Rx perpendicular forces Pin Joint Fixed to the ground Fix Support Ry Two components of force and Rx M a moment Ry Types of Loading on Beams Loading on beams can be divided into two types, ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM a) W Point load W Diagrammatic representation W / unit length Distrubuted load (uniform) b) W / unit length Diagrammatic representation W / unit length Another way of representing Uniformly distributed loads Distributed loading can be either uniform or non-uniform, Non-uniformly distributed loads c) The point supports of a simply supported beam can be replaced by vertical reactions at those points. Point or simple supports allow the beam to rotate at the support hence the moment is zero at this point. Theoretically there could be a horizontal reaction if there was a horizontal load but this is rarely the case for beams. To ensure that no horizontal loads are implied a support which allows horizontal movement can be used. W Beam with right hand support on rollers to allow free horizontal movement The analysis of beam and other statics problems is aided by the use of free body diagrams and the replacement of supports and loads by reaction forces and weights on a free body diagram. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM C A B W Diagrammatic view of simply supported beam with a b a concentrated load W W B A Free body diagram replacing R B loads and supports by forces C RA w N/m A a b B Diagrammatic view of simply supported beam with uniformly distributed load w F= w(b-a) Free body diagram replacing loads and supports by forces (a+b)/2 The uniformly distributed load w has been replaced by a single force acting at the centre of the distribution of load. Example 6 Consider a beam built into a wall as shown in Figure A. When a beam is built into or fixed to a wall so that it cannot rotate at the wall it is called a cantilever beam. The beam is supporting a load of weight W, at a distance d from the wall. Wall Built in End or Fixed End Point Load of weight W d Beam Figure A ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics W Dept of REFM Figure B shows the free body diagram of the beam with a reaction force V and reaction moment MA at the wall A. W Moment Reaction MA C A B d Reaction V Free Body Diagram of Beam Figure B The reactions are the forces that put the beam in equilibrium. Therefore, from force and moment equilibrium and taking moments about A; M M W. d 0 F V W 0 + A y M A W. d VW (The point C on the beam is d/4 from B, what will be the value of MA by taking moments about C. Compare your answer with the one above.) Example 7 Find the reactions at the supports for the beam shown in the diagram. 20 kN 5 kN/m 3.5m 3.5m 7m First draw a free body diagram, showing that the uniformly distributed load can be considered to act at the mid point along the load and supports can be replaced by reaction forces. 20 kN 35 kN RB RA 3.5m 7m 3.5m Free Body Diagram ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM W 5000 7 35 kN Taking moments about A, +ve From vertical equilibrium R B 14 35 10.5 20 35 . 0 R B 3125 . kN R A R B 35 20 R A 23.75 kN Example 8 A steel block of weight W is placed on a beam resting on a pivot (or fulcrum). The block is lifted by applying a force F at a distance a from the fulcrum. If the steel block is a distance b from the fulcrum express F in terms of m, a and b. b a F b a F W W R Illustration Free Body Diagram F = 0; Taking moments about pivot; Thus, F b W; a + + ; R - F - W = 0 M Wb Fa 0 Where the ratio a is referred to as the Mechanical Advantage. b For example if a = 30 cm and b is 10 cm the beam or lever is said to have a mechanical advantage of 3:1 (3 to 1). This means that the force F can lift a weight W three times greater than F. Taking moments about the pivot eliminates the reaction R from the calculation but from the vertical equilibrium equation above: R=F+W Moments can be taken about any point, such as at point of application of F, hence, Wa b Ra 0 Solving the two latter equations to eliminate R gives the same solution as before. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Example 9 Two forces are acting on a beam as shown in figure. If the beam is fixed supported at position A and free supported at position B (that is point B can only take up vertical force). Determine the reaction forces at the supports A and B. (Hint: By considering the sum of force is zero at equilibrium and taking the moment at point A and at point B.) 350N A B 0.15m 30o 200N 0.30 m 0.40 m Solution: Consider the sum of vertical forces and horizontal forces are zero, and since R Ax = 0 as point B can only take up vertical force. RAx = 200 x sin30o = 100 N RAy + RBy = 350 + 200 cos30o = 350 + 173.2 N = 523.2 N Taking moment about A, RBy x 0.3 = 350 x 0.15 + 200 cos30o x 0.4 RBy x 0.3 = 121.8 N RBy = 405.9 N Therefore the reaction at point B is 405.9N upward. RAy = 523.2 - 405.9 = 117.3 N The reaction at point A is 117.3 N upward and 100 N to the left. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Resultant at A is: RA = (117.32 + 1002) = 154.1 N angle = 49.55o 117.3 100 Example 10 The diagram shows a rectangular plate suspended from a wall. Support A is a fixed pinned support and B is pinned but can move freely in the vertical direction. Find the reactions at A & B 55 kN 55 kN HA A VA 750 mm 750 mm HB B 600 mm 600 mm Illustration Free Body Diagram Fy VA 55 0 Vertical Equilibrium: Horizontal Equilibrium Fx H A H B 0 Moment Equilibrium:taking moments about B + M H A 0.75 55 0.6 0 Solving the three equations, VA 55 kN ; H A 44 kN ; H B 44 kN ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Tutorial 1 Find the components of the 500 N force in rectangular coordinates y 500N 147 x (Ans.-419.3 N, 272.3 N) Tutorial 2 A force vector has the components -45 N and 72 N in rectangular coordinates, in the x and y directions respectively, what are the components in polar coordinates. Draw the vector diagram. (Ans. 84.9 N, 122) Tutorial 3 Find the magnitude and direction of the resultant (i.e. in polar coordinates) of the two forces shown in the diagram, a) Using the graphical Method b) Using the algebraic calculation method. 750N 450N 60 45 Ans. 969N, 93.4 from the +ve x axis ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Tutorial 4 There are three forces acting on a particle which is in equilibrium find the value of the forces P and Q. Hint: The inclination of P and Q is given and is fixed but the sign of P and Q may not be as shown in the diagram. Solve the problem by resolving forces into components. Q 370N 15 30 45 P (Ans. Q = -714 N, P = 714 N) Tutorial 5 The diagram shows three forces acting on a particle 45 kN 84 kN 20 30 50 122 kN Find the equilibrant by drawing the polygon of forces and by calculation method. Ans: E = 61 kN at 37o to the x-axis. ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Tutorial 6 The diagram shows four forces acting on a particle. 60 kN 100 kN 60 30 10 80 kN 110 kN Find the equilibrant and the resultant using the polygon of forces. 21 ; (Ans. E = 44 kN R = 44 kN 21 ) Tutorial 7 Two weights are suspended on ropes which are tied to a ring embedded in a solid wall. The ropes run over frictionless pulleys so it may be assumed that the tension force in the ropes is equal on both sides of the pulley. a) Draw a free body diagram for the ring, the weights and the pulley/rope system. Frictionless Pulley Ring 30 Rope Solid Wall 20 50 N Weight Rope Frictionless Pulley Weight 40 N ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Tutorial 8 The diagram shows ship mooring ropes tied to a ring bolt 1570 N 350 N 30 30 1700 N 2550 N What is the magnitude and direction of the resultant force on the ring bolt and the reaction at the anchoring point. Calculate the vertical and horizontal components of the reaction. (Ans. 2967 N, -20, -2788 N, 1015 N) Tutorial 9 The 150 N collar may slide on a frictionless vertical rod and is connected by a rope, that runs over a frictionless pulley, to a 170 N counter weight. Determine the value of h for which the system is in equilibrium. 40 cm frictionless rod h 170 N Weight 150 N Collar (Ans. 75 cm) ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Tutorial 10 Find the reaction forces at A and B. 100 kN 40 kN A B 2m 0.5m 5m 0.5m Ans. 126 kN, 14 kN Tutorial 11 Find the reaction force and moment at the built-in end of the cantilever shown. 20kN/m 1m 10 kN 3m 5.5 m Ans. 30 kN, - 25 kNm ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics Dept of REFM Tutorial 12 The fixed crane shown in the diagram is attached at its support by a pin joint at A and a rolling connection at B. The weight of the jib is 10 kN and acts as shown. Calculate the reactions at A and B due to lifting a 50 kN weight. A 4m 50 kN 10 kN B 6m 10 m (at A,-215 kN,60 kN, at B, 215 kN) ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics