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# Moments and equillibrium

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• 1. ENGINEERING COUNCIL CERTIFICATE LEVEL ENGINEERING SCIENCE C103 TUTORIAL 6 - MOMENTS OF FORCE AND EQUILIBRIUM You should skip this tutorial if you are already familiar with the content On completion of this tutorial you should be able to do the following. • Define moments of force. • Define equilibrium. • Explain the basic concepts of dry friction. • Solve problems involving sliding on an inclined plane. • Explain the use of free body diagrams for static bodies. Note that friction forces do not seem to appear in the exams but this tutorial is a good introduction to the use of free body diagrams. Free body diagrams is also used in later tutorials for dynamic problems. © D.J.Dunn freestudy.co.uk 1
• 2. 1. MOMENTS OF FORCES A moment of a force is the result of multiplying the force by the distance from a given point. The distance must be measured at 90o to the force. The diagram shows a force acting on a body. The moment of the force about point P is F x. The basic units are N m. Often the moment of a force is called Torque but this is more likely to be used in connection with rotating bodies. The distance x is called the moment arm. In order to demonstrate the effect of the moment arm, try suspending a kg mass from a stick and hold it out horizontally. If the mass is suspended from the far end of the stick, it is very difficult to hold up. If the mass is suspended from the end close to the hand, it is easy to hold up. The weight is the same in both cases but the larger the moment, the harder it is to hold it up. WORKED EXAMPLE No.1 If the mass of 1 kg is held 0.6 m from the hand (figure 6), what is the moment of force exerted by the hand on the stick? SOLUTION 1 kg exerts a gravitational force downwards of 9.81 N. This acts 0.6 m measured horizontally from the hand. The moment of force is hence 9.81 x 0.6 = 5.89 N m (anticlockwise). In order to hold the stick horizontally, the hand must exert a moment of 5.89 N m clockwise on the stick. 2. TOTAL EQUILIBRIUM If a body is at rest it is said to have TOTAL EQUILIBRIUM. This must mean that a. all forces in a given direction add up to zero. (Up is positive and left to right is positive). b. all the turning moments about a given point must add up to zero. It follows from this that: i. all the vertical forces upwards (+ve) must equal all the vertical forces downwards (-ve). ii. all the horizontal forces to the right (+ve) must equal all the horizontal forces to the left (-ve) iii. all the clockwise turning moments (+ve) must equal all the anticlockwise turning moments (-ve). © D.J.Dunn freestudy.co.uk 2
• 3. In order to apply these principles, it helps to draw a FREE BODY DIAGRAM. This is a diagram of a body with all the forces and moments shown acting on it which keeps the body in equilibrium. This includes unknown forces and moments which may then be solved by applying the above principles. Free body diagrams are particularly useful when determining the forces and moments which attach it to something. Later on you will find the idea useful in dynamic problems in which dynamic forces are added to the body. WORKED EXAMPLE No.2 Examine the diagram below. Apply the rules and determine the values of F1 and F2. Also determine the distance Y. SOLUTION Balancing horizontal forces we have: Balancing vertical forces we have F1 - 200 = 0 hence F1 = 200 N 100 - F2 = 0 hence F2 = 200N. Now take turning moments about the bottom right hand corner we deduce the distance Y. Remember that the clockwise and anticlockwise moments must add up to zero and that the forces acting through the point do not exert any moment. C.W. moment is F1 x 0.5 = 200 x 0.5 = 100 Nm A.C.W. moment is F2 y = -100 y 100 -100y = 0 so y = 1 m WORKED EXAMPLE No.3 Find the value of force F which just prevents the body shown from revolving about point P Go on to find the force R (magnitude and direction) which keeps the body in equilibrium. SOLUTION The clockwise moment about P is 200 x 0.2 = 40 Nm The anticlockwise moment is -0.5 F For equilibrium 40 - 0.5F = 0 so F = 40/0.5 = 80 N R is the force needed to keep the shape in equilibrium by stopping it from moving vertically or horizontally. Let us now see how to find it. The total downwards force must be 80 N to balance the 80 N up. The total force to the right must be 200 N to balance the 200 N to the left. The force R must be the resultant of these two forces and may be found by vector addition as shown. By scaling or by trigonometry R is 215.4 N at 21.8o to the horizontal. © D.J.Dunn freestudy.co.uk 3
• 4. SELF ASSESSMENT EXERCISE No. 1 1. The body shown in the diagram is in total equilibrium. Calculate the length L and the force R. (Answers 0.14 m, 308 N at 13.1o) 2. Find the value of the force F that holds the lever shown in equilibrium. (Answer 1028.6 N) 3 FRICTION The underlying theory of friction is not covered here and this section is just a revision of the basics. For dry surfaces in contact with each other Coulomb’s laws apply. When a body just slides over a horizontal surface without accelerating, the applied force is balanced by the friction force F. First Law - The force is independent of area. This may be demonstrated by pulling a block with a simple force gauge as shown and determining that the same force is needed to slide it on the large side as on the small side. Second Law - The force is directly proportional to the force R acting normally to the surface such that F = µR and µ is the coefficient of friction. The value of µ depends on the materials. © D.J.Dunn freestudy.co.uk 4
• 5. WORKED EXAMPLE No.4 A block of mass 50 kg rests on a horizontal surface. Calculate the horizontal force needed to just make it slide if the coefficient of friction is 0.32 SOLUTION Mass = 50 kg R = Weight = 50 x 9.81 = 490.5 N F =µ R = 0.32 x 490.5 =156.96 N WORKED EXAMPLE No.5 Calculate the force F acting at 30o as shown that will just make the block slide. Μ = 0.25 SOLUTION Firs resolve F vertically and horizontally. F’ = F cos 30o = 0.866F R’ = F sin 30o = 0.5F Balancing forces vertically R = 400 - 0.5 F Balancing forces horizontally µR = 0.866 F Substitute for R and µ(400 - 0.5 F) = 0.866 F Hence F = 100.9 N WORKED EXAMPLE No.6 Consider a block leaning against a wall as shown. The top end has a roller. A roller cannot exert sideways forces so the reaction force can only be normal to the wall. This makes it easier to solve. The bottom of the block is prevented from sliding by friction. The problem is to calculate the minimum coefficient of friction to stop the bottom sliding out. SOLUTION Balancing vertical forces gives R2 = W Balancing horizontal forces gives µR2= R1 Balancing moments about the bottom gives WA = 4R1 Substitute R2 = W and R1= µR and R2A = 4µR 2 hence µ = A/4 In order to complete this problem we must result to geometry and trigonometry. φ = tan-1 (1/6) = 9.46o θ = sin-1 (4/6) = 41.8o θ + φ = 51.27o 2 2 ½ Length of diagonal = (6 + 1 ) = 6.0828 m A = 3.0414 cos51.27 o =1.9027 m µ = 1.9027/4 = 0.475 © D.J.Dunn freestudy.co.uk 5
• 6. SELF ASSESSMENT EXERCISE No.2 1. Find the value of µ between the floor and the block if it is just on the point of sliding. (Answer 0.44) 2. Assuming W acts at the centre of the ladder, calculate the minimum angle θ which just prevents the ladder from sliding if µ = 0.4 at both points of contact. (Answer 46.4o) 4 FRICTION ON INCLINED PLANES Consider a block on an inclined plane. The weight is no longer normal to the surface but always remains vertical. In order to solve the problem the weight must be resolved normal and parallel to the plane. The reaction force R squeezing the surfaces together is the normal component and the parallel component F1 will tend to make the block slide down the plane. Resolving gives: R = W cosθ and F1 = W sinθ If no force other than the weight is involved, then the block will slide down the plane if F1 is greater than the friction force. In this case it will slide when F1>µ R or F1 > µW cosφ When this is the case, the limiting value of µ is F1/R µ = Wsinφ/Wcosφ = tanφ This is called the angle of Friction. There is fixed symbol for it and β is often used instead of φ. © D.J.Dunn freestudy.co.uk 6
• 7. WORKED EXAMPLE No.7 A block of mass 100 kg rests on a slide. Determine the angle of the slide to the horizontal which would just cause the block to slip if µ = 0.24. SOLUTION Weight of block = Mass x gravity. W = 100 x 9.81 = 981 Newton. The angle of friction is obtained from µ = tan-1β = tan-10.24 = 13.25o SELF ASSESSMENT ASSIGNMENT No. 3 1. A block of mass 15 kg rests on a horizontal floor. The coefficient of friction is 0.33. Calculate the minimum horizontal force to just move it. (48.6 N) 2. A block of mass 20 kg is placed on a slide. The slide is tilted until the block just slides. The angle to the horizontal is then 250. Determine the coefficient of friction. (0.47) 3. A sliding link is pressed against a slideway with a force of 208 N. The coefficient of friction is 0.3. Calculate the friction force. (62.4 N) 4. A mass of 2000 kg is dragged up an incline of 20 o by a rope parallel to the incline. The coefficient of friction is 0.35. Calculate the force in the rope. (13.16 kN) © D.J.Dunn freestudy.co.uk 7
• 8. 5. COMPLEX FREE BODY DIAGRAM PROBLEMS Problems in the exam involve analysing the force and moments of complex static structures by use of free body diagrams. This is best learned through the use of examples. WORKED EXAMPLE No. 8 A block is tipped as shown by a force F at the top corner. The block weighs W and does not slip. Derive a formula relating the force F to the angle θ. At what angle will the block tip over? SOLUTION First we need to determine the dimensions x and h with some Trigonometry. h = L sin θ + B cos θ x = B/2 sin θ – L/2 cos θ Now balance moments about the point of contact. Fh=Wx F (L sin θ+ B/2 cos θ) = W(B/2 sin θ – L/2 cos θ) F= W{Bsinθ − Lcosθ} 2(Lsinθ + B cosθ ) The block tips over when F = 0 hence (B sin θ – L cos θ) = 0 B sin θ = L cos θ sin θ/cos θ = tan θ = L/B θ = tan-1 (L/B) © D.J.Dunn freestudy.co.uk 8
• 9. SELF ASSESSMENT ASSIGNMENT No. 4 1. A uniform block of mass M, length L meters and height t balances as shown on a point x meters from the centre of gravity. Derive the formula that links force F and angle θ assuming it does not slip. Answer F = 2. Mg(2x cos θ − t sin θ) cos θ(L - 2x ) + 2t sin θ A mass M2 (including the pulley and hanger) is balanced by two equal masses M1 (including the hanger) acting as shown. Derive a formula for angle θ that produces perfect static equilibrium. (Answer θ = sin-1 (M2/2M1) © D.J.Dunn freestudy.co.uk 9