PORTMORE COMMUNITY COLLEGE ASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY RESIT EXAMINATIONS SEMESTER 2 July 2012 COURSE NAME: Mechanical Engineering Science CODE: GROUP: ADET 1 DATE: July 3, 2012 TIME: DURATION: 9:00 am 2 hrs INSTRUCTIONS: 1. This paper consists of SIX questions. 2. Candidates must attempt ANY FOUR questions on this paper. 3. All working MUST be CLEARLY shown. 4. Keep all parts of the same question together. 5. The use of non-programmable calculators is permitted. 6. Use g = 9.81 m/s 2 1
[Question 1] DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Instructions: Answer any FOUR (4) questions. (a) State the two laws for a body in equilibrium. (b) Calculate the vertical reactions at the supports A and B of the framework shown in Fig. 1. The framework is made up of equilateral triangles of sides 1 m long. [9 marks] 1 kn 3 kn A 5 kn B Fig. 1 (c) (i) Explain the meaning of the term centroid. (ii) Find the position of the centroid of the lamina shown in Fig. 2, from the x-axis and y-axis. All dimensions are in centimetre. [12 marks] y 40 60 120 Fig.2 100 40 x [Question 2] (a) A loaded truck of mass 16 tonnes accelerates from rest to 35 km/h in 20 seconds on a level road against a constant tractive resistance of 100 N/tonne. (a) Sketch a diagram to show all the forces acting on the truck. (b) Find the force required to accelerate the truck. [8 marks] 2
(c) If the truck then moves at a constant velocity of 60 km/h, find the power required. [7 marks] (b) Find the force required to move with a constant velocity of 32 km/h up an incline of 1 in 20 (sin = 1/20). Assume that the frictional resiatance is 100 N/tonne. [8 marks] (c) A carriage A, of mass 10 t, moving at 5 m/s, collides with a similar 10 t stationary carriage B. After colliding, carriage A moves at 2 m/s. (i) Find the velocity of carriage B after the collision. (ii) Describe the type of collision and state if kinetic energy lost in this collision? (iii) Find the coefficient of restitution. [4 marks].[question 3] (a) Explain the meaning of the following terms: (i) static friction (ii) dynamic friction, (iii) angle of friction. [5 marks] (b) A box of mass 80 kg is pulled at a constant velocity along a level floor with a rope inclined at 40 0 to the horizontal. The coefficient of kinetic friction between the box and the floor is 0.25. (i) Sketch a diagram showing all the forces acting on the box. (ii) Determine the force required to pull the box at constant velocity. (iii) At what angle should the rope be inclined to the horizontal, so that the pulling force is a minimum. [8 marks] [1 mark] (c) The diagram Fig. 3 shows a handle which is used to turn a wheel at constant angular velocity. The force applied to the handle is 100 N. 0.8 m Fig. 3 100 N 3
Detemine the: (i) torque applied to the handle. (ii) work done if the wheel makes 25 revolutions in 2 minutes. (iii) power developed. (Question 4) (a) With the aid of load extension graph explain: (i) Elastic limit (ii) Proportional limit (iii) Maximum load (b) The stress at the limit of proportionality of a specimen of mild steel is 300N/mm 2 and the modulus of elasticity is E = 206 N/mm 2. [9 marks] (i) Calculate the maximum elastic strain. [5marks] (ii) If the elongation of 50 mm length of the same steel is 13mm at its maximum load, calculate the strain. (iii) What is the ratio of strain at maximum load to strain at elastic limit? (c) A steel specimen 15 mm in diameter and 20 cm long, reaches its limit of proportionality at a load of 50 kn. Determine the extension given that the modulus of elasticity is 2.06 x 10 8 Pa. [7 marks] (Question 5) (a) Define the following terms and give a formula for each: (i) mechanical advantage (ii) velocity ratio (iii) efficiency [6 marks] (b) A set of pulley blocks has two pulleys in the upper block and two pulleys in the lower block. (i) Sketch a diagram of the pulley blocks and show the arrangement of the rope around the pulleys. [4 marks] 4
(ii) What is the velocity ratio of the system of pulleys. Give a reason. (c) A lifting tackle uses an effort of 120 N to raise a load of 35 kg, and an effort of 200 N to raise a load of 70 kg. (i) Determine the law of the machine. (ii) What is the efficiency when the load is 70 kg, given that the V.R. is 4. (iii) Calculate the work input when the 70 kg load is raised 1 meter upwards. [7 marks] (Question 6) (a) Explain the meaning of the terms: (i) gauge pressure (ii) absolute pressure. [4 marks] (b) A 10 m tall cylindrical steel tank has a pressure gauge installed 0.5 m from the base. Oil of density 0.85 g/cm 3 is filled to 1.5 m from the top of the tank. (i) Sketch a diagram of the arrangement and show the dimensions stated above. (ii) Calculate the pressure indicated by the pressure gauge. [4 marks] (c) A hydraulic ram has a 5.0 cm diameter piston. Oil at a gauge pressure of 15 MPa pushes on one side of the piston which moves through its whole working stroke of 30.0 cm. Determine: (i) the force exerted by the oil on the piston, (ii) the work done by the oil on the piston. (iii) the power developed if the stroke is completed in 3.5 sec. (d) Using a labelled diagram, explain the operation of the manometer. [4 marks] [5 marks] ****END OF PAPER***** 5
ASSOCIATE DEGREE IN ENGINEERING SOLUTIONS SEMESTER 2 2012 May COURSE NAME: CODE: GROUP: DATE: TIME: DURATION: Mechanical Engineering Science [8 CHARACTER COURSE CODE] AD-ENG I "[EXAM DATE]" "[TIME OF PAPER]" Two hours 6
Solutions [Question 1] (a) Conditions of equilibrium: 1. The resultant force must be zero. 2. The resultant moment about any point must be aero. [2] (b) Take moments about A: CW A = ACW A 1 x 0.5 + 5 x 1 + 3 x 1.5 = R B x 2 0.5 + 5 + 4.5 = 2 R B 10 = 2 R B R B = 5 kn Take moments about B: CW B = ACW B R A x 2 = 1 x 1.5 + 5 x 1 + 3 x 0.5 2 R A = 1.5 + 5 + 1.5 = 8 7
(c) R A = 8/2 = 4 kn (i) The centroid is the c.g. of an area. [1] (ii) y 1 60 2 120 100 40 x A 1 = 20 x 100 = 2000 cm 2 A 2 = 40 x 100 = 4000 cm 2 x 1 = 10 cm x 2 = 70 cm (A 1 + A 2 )x = A 1 x 1 + A 2 x 2 (2000 + 4000) = 2000 x 10 + 4000 x 70 6000 x = 20 000 + 280 000 = 300 000 x = 300 000/6000 x = 50 cm (A 1 + A 2 )y = A 1 y 1 + A 2 y 2 (2000 + 4000) = 2000 x 50 + 4000 x 20 6000 x = 100 000 + 80 000 = 180 000 x = 180 000/6000 x = 30 cm [Question 2] (a) (i) N f F mg 8
(ii) f = 16 x 100 = 1600 N [1] v = 35/3.6 = 9.72 m/s. [1] a = 9.72/20 = 0.486 m/s 2 [1] (b) F = ma F - 1600 = 16000 x 0.486 [3] F = 1600 + 7776 F = 9376 N [2] (iii) 60 km/h = 16.7 m/s [1] F = ma F - 1600 = m x 0 F = 1600 N [4] P = F v = 1600 x 16.7 P = 26700 W (26.7 kw) [2] N mg sin F f mg cos [3] F = ma F - 1600 - mg sin = m a [3] F - 1600-16000 x 10 x (1/20) = 0 F - 1600-8000 = 0 F = 9600 N [2] [Question 3] (a) (i) Static friction acts when an object is at rest until it starts to slip. [2] (ii) Dynamic friction acts when an object is sliding on a surface. [1] (iii) This is the angle between the reaction R and the normal reaction N. [2] 9
N R f pull (b) N P f 40 0 The angle of friction is: = tan -1 (0.25) = 14 0 14 0 W R [2] 800 N 116 0 50 0 P Consider the vector diagram with the three forces W, R and P. 800/sin (116 0 ) = P/sin (14 0 ) P = sin (14 0 ) 800/sin (116 0 ) = 215 N [8] (ii) Angle of inclination for minimum pull = angle of friction = 14 0 [1] (b) (i) = 100 x 0.8 = 80 N m [3] (ii) W = = 80 x 25 x 2 = 12570 J [3] (iii) P = W/t = 12570/120 = 105 W [3] [Question 4] (a) (i) the limit after which load is no longer proportional to extension. [2] 10
(ii) Elastic limit is the point beyond which the material after further loading will not return to the original length. [2] (iii) The highest point of a load extension curve showing the maximum load a material can carry. [2] (b) U Force P Y [7] Ext. (b) (i) = 0.0014563 [5] (ii) [2] (iii) [2] (c) Use E = F L/(A e) or e = F L/(A E) A = π (0.015) 2 /4 = 1.76 x 10-4 m 2 e = 50 x 10 3 x 0.20 1.76 x 10-4 x 206 x 10 9 e = 2.8 x 10-4 m (0.28 mm) 11
(c) (i) = F/A [1] A = (0.006) 2 [1] A = 1.131 x 10-4 m 2 = 5.5 x 10 3 /1.131 x 10-4 = 4.86 x 10 7 Pa [2] (ii) E = / [1] = /E = 4.86 x 10 7 /200 x 10 9 = 2.43 x 10-4 [2] (iii) = e/l [1] e = L = 2.43 x 10-4 x 3 e = 7.29 x 10 4 (0.729 mm) [2] (d) (i) Shear stress is the ratio of force to area where the area is parallel to the force. (ii) In single shear one cross-sectional area is resisting the shear force. (iii) In double shear two cross sectional areas of the material are resisting the load. [Question 5] (a) (i) M.A. is the ratio of the load to the effort. M.A. = Load/Effort [2] (ii) V.R is the ratio of the distance moved by the effort to the distance moved by the load. V.R. = distance moved by effort/distance moved by load [2] (iii) Efficiency is the ratio of the work output to the work input. = Work output/work input of = M.A./V.R. [2] (b) (i) effort 12
[4] (c) (i) (ii) V.R. = 3 When the load is raised by 1 m the three ropes connecting the pulleys are shortened by 1 total of 3 m [2] Load/N Effort/N 350 120 700 200 (ii) (iii) Law is E = a L + b Gradient = a = (200-120)/(700-200) = 80/500 a = 0.16 Substitute E = 120 N and L = 350 N in the equation E = 0.16 L + b 120 = 0.16 (350) + b 120 = 56 + b b = 64 N Law of the machine: E = 0.16 L + 64 N [7] = M.A./V.R. = (700/200)/4 = 0.86 (86 %) [3] = Work output/work input Wo = 700 x 1 = 700 J 0.86 = 700/Wo Wi = 700/0.86 = 814 J 13
[Question 6] (a) (i) Gauge pressure is the excess pressure above atmospheric pressure. [2] (ii) Absolute pressure is the sum of atmospheric pressure and the gauge pressure. [2] (b) (i) 1.5 m 12 m 8.0 m [3] gauge (ii) P = g h = 850 x 9.8 x 8 = 66600 Pa [4] (c) (i) A = d 2 /4 = (5) 2 /4 = 19.63 cm 2 A = 19.63 x 10-4 m 2 [1] (iv) (v) F = P x A = 15 x 10 6 x 19.63 x 10-4 F = 294 x 10 2 N [3] W = F x = (294 x 10 2 )(0.3) W = 8800 J [2] P = W/t 14
= 8800/3.5 P = 2514 W [3] (d) - before the cas is turned the eater levels are the same in each arm. - when the gas supply is turned on the water level in the right arm is pushed upwards. The liquid level in the left arm is pushed down by the gas pressure. - the gas pressure is given by p = g h where h = h 1 h 2. glass U-tube rubber tube liquid gass from supply h 1 h 2 table [5] ****END OF SOLUTIONS***** 15
Examination Paper Analysis Associate of Applied Science Degree Date March 27, 2012 Module: Mechanical Engineering Science B Examiner: Noel Brown Syllabus Objectives Question A B C D E F G H I J K 1 X 2 X X 3 X X 4 X 5 X 6 X Question 1 Question is adequate for this level. Question 2 Question has 35 marks, please reduce to 25 in keeping with the other questions. 16
Question 3 Question is adequate for this level. Question 4 Question is adequate for this level. Question 5 Question is adequate for this level. Question 6 Question is adequate for this level. The quality of the questions are in keeping with the level of the module, in addition, the paper provides good coverage of the syllabus. I have included some formatting and some redistribution of marks, these are suggestions, feel free to accept or reject. Please ensure that the paper is proofread and all track changes removed before it is sent for printing. The time allotted for the paper is adequate as students are required to do four questions from six in two hours. All the questions can be completed in 30 minutes if students are prepared. This paper is a good choice for the backup paper. 17