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lOMoARcPSD| 3264513
Faculty of Engineering and Information Sciences Student to complete: Family name Other names Student number Room
ENGG102 Fundamentals of Engineering Mechanics Wollongong Campus
Mid Session Examination Paper Autumn Session 2017 Exam duration
2 hour 30 minutes
Items permitted by examiner
UOW Approved Calculator
Aids supplied
Nil
Directions to students
Attempt all 10 multi-choice questions in Part A. Each question is worth 2.5 marks. No marks will be deducted for wrong multi-choice questions. Part A - Answers must be entered on the score sheet provided. Choose ONE answer only for each multi-choice question. If you choose more than one answer, you will receive no marks for that question. Answer both questions in Part B. Each question is worth 25 marks. Part B - Start each question on a new page. Answer each question in the booklet provided writing on both sides of the paper.
This exam paper must not be removed from the exam venue See Page 7 for USEFUL FORMULAS AND DATA.
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 1 of 7
lOMoARcPSD| 3264513
Part A – Multichoice. Answer all questions ANSWERS MUST BE ENTERED ON THE SCORE SHEET PROVIDED Any work shown for multi-choice questions in the exam booklet will not be marked. Question A1. If you were to move into outer space far from any stellar objects, a) your mass would change, but your weight would not change. b) both your weight and mass would change. c) neither your weight nor your mass would change. d) your weight would change, but your mass would not change. e) since you are floating in space you are weightless and therefore have no mass. Question A2. An apple sits at rest on a horizontal table top. The gravitational force on the apple (its weight) is one half of an action-reaction pair. What force is the other half? a) b) c) d) e)
the force of the Earth’s gravity on the apple the upward force that the table top exerts on the apple the upward force that the apple exerts on the Earth the downward force that the apple exerts on the table top the frictional force between the apple and the table top
Question A3. Consider the expression F = ma and its use to engineers. A load of mass M rests on the floor of a lift which is accelerating upwards. The magnitude of the lift’s acceleration is a and the magnitude of acceleration due to gravity is g. What is the magnitude of the resultant force on the mass. a) Ma b) Mg c) Ma + Mg d) zero Question A4. Considering the static structure shown in Figure Q A4, what will happen if the angle the cable makes with the beam, , is reduced.
1
wall
3
1
a) The force applied by the wall to the beam will reduce.
b) The tension force in the beam will reduce.
c) The force applied by the sign to the beam will increase. d)
4
The force applied by the beam to the cable will increase.
2017 March
ENGG102 Mid Session
Wollongong
2
sign
Figure Q A4.
Autumn 2017
Page 2 of 7
cable
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Question A5. Considering the cable in Figure Q A4 above, labelled 3, identify the most correct free body diagram for this cable from those identified.
Question A6.
The value of Young’s modulus, E, for balsa wood is in the range: a) b) c) d) e)
1 – 5 GPa 1 – 5 MPa 1 – 5 N/mm² 10000 MPa 101 kPa
Question A7.
The magnitude of the resultant vector sum of the two vectors in Figure QA7 is
Figure QA7 a) b) c) d) e)
5.1 cm. 9.6 cm. 8.7 cm. 10.0 cm. 6.3 cm
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 3 of 7
lOMoARcPSD| 3264513
Question A8.
In Question A7, the angle the resultant vector makes with the x-axis is: a) b) c) d) e)
640 540 890 900 680
Question A9. A shopkeeper wants to mount a sign on a wall outside his shop. To calculate how strong the diagonal wire needs to be and the force of the rod on the side of the wall, he first draws a diagram for the structure followed by a FBD as shown.
Using the FBD shown, he then derives the equations for static equilibrium for forces. Which are the correct equations:
a) Fx + Tx = 0 and Fy – Ty - (Mg – mg) = 0 b) Fx - Tsin= 0 and Fy + Tcos- Mg - mg = 0 c) Fx - Tcos= 0 and Fy + Tsin- Mg - mg = 0 d) Fx - Tcos= 0 and -Fy - Tsin+ Mg + mg = 0 e) -Fx + Tcos= 0 and Fy + Tsin- Mg - mg = 0 Question A10. The shopkeeper then realises he needs another equation to solve his problem and thus derives the equation for static equilibrium for moments. Which is the correct equation:
a) L Tsin- mgcosL/2 - MgL = 0 b) L Tsin- mgL/2 - MgL = c) L Tcos- mgL/2 – MgcosL = 0 d) L Tsin - mgL/2 – MgL/2 = 0 e) L Tcos- mgL/2 – MgsinL = 0
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 4 of 7
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Part B – Answer both questions. Question B1 (25 Marks) Consider the heavy frame supporting a load, FEy, as shown in Figure QB1. The distributed weight of the frame is W N/m. You may assume that the total weight of the frame is less than |F Ey|. a) Draw the free body diagram for the frame, labelling all the relevant forces and axes. (5 marks) b) Estimate the directions of all the reactions at B and D. Give reasons for your decisions. (5 marks) c) Establish, stating reasonable assumptions, equations for the resultant forces in the x and y directions at the supports, B and D. State the theory to be adopted and express the x and y components of each reaction in terms of L1, L2, L3, L4, W, FEy as appropriate. (5 marks) d) Solve using Data Case 1, verify your solution and draw the final FBD Case 1: L1=2.00 m, L2=1.00 m, L3=1.00 m, L4=2.00 m, W = 150 N/m and FEy= +1100 N. (5 marks) e) Solve using Data Case 2, verify your solution and draw the final FBD Case 2: L1=2.00 m, L2=1.00 m, L3=1.00 m, L4=2.00 m, W = 150 N/m and FEy= -1100 N. (5 marks)
Hint: Treat the weight of the vertical part of the frame separately from the horizontal part.
A
y +VE
L3
x
B
L4
FEy D C
E
L1
L2
Figure QB1 Heavy Frame supporting a load
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 5 of 7
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Question B2 (25 Marks) Figure QB2 shows a heavy crate is suspended from a cable at D. This cable passes over a pulley at C and is tied to a ring at B. The segment BC is at an angle of to the horizontal. The Cable BE is connected to a known suspended counterbalance mass M1. A third cable AB is horizontal and is tied off at A as shown. Acceleration due to gravity is g=9.81 m/s2. The cables can be considered zero mass and only allow in-line tension forces. a) Using the 6 step engineering method as outlined below, determine the mass of the crate as a function of M1 and . Then determine the numerical answer for the data in (a)v. i. ii. iii. iv. v.
Define the problem. Draw a free body diagram of the ring at B and gather the data. State the relevant theory and assumptions. Estimate the magnitude of the crate’s mass relative to M1, giving reasons. Solve the problem as a function of M1 and and then get the mass of the crate if M1 = 150 kg and = 55o.
(1 Mark) (3 Marks) (2 Marks) (1 Mark) (4 Marks)
vi. Verify your solution.
(1 Mark)
b) Given the same angles as Figure QB2 and = 55o, a new crate is suspended at D. If M1 = 240 kg, find the mass of the new crate. Hence find magnitude and direction of the new reaction force at the pulley, C. Apply the six step method for the reaction at C. i. ii. iii. iv.
Define the current problem. Draw the free body diagram for pulley for this scenario. State the relevant theory and assumptions. Estimate the direction of the reaction force and then estimate its magnitude, giving reasons. v. Solve the problem for the reaction at C. vi. Verify the solution for the reaction.
y
B
(2 Marks) (4 Marks) (1 Mark)
Frictionless, low mass pulley
C
A
(1 Mark) (3 Marks) (2 Marks)
+ve D
x
90.0°
E M1
CRATE
Figure QB2 Static masses connected over a frictionless pulley 2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 6 of 7
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Useful formulas and data
sin= A/C cos= B/C tan= A/B sin²+cos²= 1
Moment = Force x Distance, where Distance refers to the perpendicular distance from the point about which the moment is to be calculated to the line of action of a force. A anticlockwise moment is +ve.
+ve BM
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 7 of 7
Faculty of Engineering and Information Sciences Student to complete: Family name Other names Student number Room
ENGG102 Fundamentals of Engineering Mechanics Wollongong Campus
Mid Session Examination Paper Autumn Session 2017 Exam duration
2 hour 30 minutes
Items permitted by examiner
UOW Approved Calculator
Aids supplied
Nil
Directions to students
Attempt all 10 multi-choice questions in Part A. Each question is worth 2.5 marks. No marks will be deducted for wrong multi-choice questions. Part A - Answers must be entered on the score sheet provided. Choose ONE answer only for each multi-choice question. If you choose more than one answer, you will receive no marks for that question. Answer both questions in Part B. Each question is worth 25 marks. Part B - Start each question on a new page. Answer each question in the booklet provided writing on both sides of the paper.
This exam paper must not be removed from the exam venue See Page 7 for USEFUL FORMULAS AND DATA.
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 1 of 7
lOMoARcPSD| 3264513
Part A – Multichoice. Answer all questions ANSWERS MUST BE ENTERED ON THE SCORE SHEET PROVIDED Any work shown for multi-choice questions in the exam booklet will not be marked. Question A1. If you were to move into outer space far from any stellar objects, a) your mass would change, but your weight would not change. b) both your weight and mass would change. c) neither your weight nor your mass would change. d) your weight would change, but your mass would not change. e) since you are floating in space you are weightless and therefore have no mass. Question A2. An apple sits at rest on a horizontal table top. The gravitational force on the apple (its weight) is one half of an action-reaction pair. What force is the other half? a) b) c) d) e)
the force of the Earth’s gravity on the apple the upward force that the table top exerts on the apple the upward force that the apple exerts on the Earth the downward force that the apple exerts on the table top the frictional force between the apple and the table top
Question A3. Consider the expression F = ma and its use to engineers. A load of mass M rests on the floor of a lift which is accelerating upwards. The magnitude of the lift’s acceleration is a and the magnitude of acceleration due to gravity is g. What is the magnitude of the resultant force on the mass. a) Ma b) Mg c) Ma + Mg d) zero Question A4. Considering the static structure shown in Figure Q A4, what will happen if the angle the cable makes with the beam, , is reduced.
1
wall
3
1
a) The force applied by the wall to the beam will reduce.
b) The tension force in the beam will reduce.
c) The force applied by the sign to the beam will increase. d)
4
The force applied by the beam to the cable will increase.
2017 March
ENGG102 Mid Session
Wollongong
2
sign
Figure Q A4.
Autumn 2017
Page 2 of 7
cable
lOMoARcPSD| 3264513
Question A5. Considering the cable in Figure Q A4 above, labelled 3, identify the most correct free body diagram for this cable from those identified.
Question A6.
The value of Young’s modulus, E, for balsa wood is in the range: a) b) c) d) e)
1 – 5 GPa 1 – 5 MPa 1 – 5 N/mm² 10000 MPa 101 kPa
Question A7.
The magnitude of the resultant vector sum of the two vectors in Figure QA7 is
Figure QA7 a) b) c) d) e)
5.1 cm. 9.6 cm. 8.7 cm. 10.0 cm. 6.3 cm
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 3 of 7
lOMoARcPSD| 3264513
Question A8.
In Question A7, the angle the resultant vector makes with the x-axis is: a) b) c) d) e)
640 540 890 900 680
Question A9. A shopkeeper wants to mount a sign on a wall outside his shop. To calculate how strong the diagonal wire needs to be and the force of the rod on the side of the wall, he first draws a diagram for the structure followed by a FBD as shown.
Using the FBD shown, he then derives the equations for static equilibrium for forces. Which are the correct equations:
a) Fx + Tx = 0 and Fy – Ty - (Mg – mg) = 0 b) Fx - Tsin= 0 and Fy + Tcos- Mg - mg = 0 c) Fx - Tcos= 0 and Fy + Tsin- Mg - mg = 0 d) Fx - Tcos= 0 and -Fy - Tsin+ Mg + mg = 0 e) -Fx + Tcos= 0 and Fy + Tsin- Mg - mg = 0 Question A10. The shopkeeper then realises he needs another equation to solve his problem and thus derives the equation for static equilibrium for moments. Which is the correct equation:
a) L Tsin- mgcosL/2 - MgL = 0 b) L Tsin- mgL/2 - MgL = c) L Tcos- mgL/2 – MgcosL = 0 d) L Tsin - mgL/2 – MgL/2 = 0 e) L Tcos- mgL/2 – MgsinL = 0
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 4 of 7
lOMoAR cPSD| 3264513
Part B – Answer both questions. Question B1 (25 Marks) Consider the heavy frame supporting a load, FEy, as shown in Figure QB1. The distributed weight of the frame is W N/m. You may assume that the total weight of the frame is less than |F Ey|. a) Draw the free body diagram for the frame, labelling all the relevant forces and axes. (5 marks) b) Estimate the directions of all the reactions at B and D. Give reasons for your decisions. (5 marks) c) Establish, stating reasonable assumptions, equations for the resultant forces in the x and y directions at the supports, B and D. State the theory to be adopted and express the x and y components of each reaction in terms of L1, L2, L3, L4, W, FEy as appropriate. (5 marks) d) Solve using Data Case 1, verify your solution and draw the final FBD Case 1: L1=2.00 m, L2=1.00 m, L3=1.00 m, L4=2.00 m, W = 150 N/m and FEy= +1100 N. (5 marks) e) Solve using Data Case 2, verify your solution and draw the final FBD Case 2: L1=2.00 m, L2=1.00 m, L3=1.00 m, L4=2.00 m, W = 150 N/m and FEy= -1100 N. (5 marks)
Hint: Treat the weight of the vertical part of the frame separately from the horizontal part.
A
y +VE
L3
x
B
L4
FEy D C
E
L1
L2
Figure QB1 Heavy Frame supporting a load
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 5 of 7
lOMoAR cPSD| 3264513
Question B2 (25 Marks) Figure QB2 shows a heavy crate is suspended from a cable at D. This cable passes over a pulley at C and is tied to a ring at B. The segment BC is at an angle of to the horizontal. The Cable BE is connected to a known suspended counterbalance mass M1. A third cable AB is horizontal and is tied off at A as shown. Acceleration due to gravity is g=9.81 m/s2. The cables can be considered zero mass and only allow in-line tension forces. a) Using the 6 step engineering method as outlined below, determine the mass of the crate as a function of M1 and . Then determine the numerical answer for the data in (a)v. i. ii. iii. iv. v.
Define the problem. Draw a free body diagram of the ring at B and gather the data. State the relevant theory and assumptions. Estimate the magnitude of the crate’s mass relative to M1, giving reasons. Solve the problem as a function of M1 and and then get the mass of the crate if M1 = 150 kg and = 55o.
(1 Mark) (3 Marks) (2 Marks) (1 Mark) (4 Marks)
vi. Verify your solution.
(1 Mark)
b) Given the same angles as Figure QB2 and = 55o, a new crate is suspended at D. If M1 = 240 kg, find the mass of the new crate. Hence find magnitude and direction of the new reaction force at the pulley, C. Apply the six step method for the reaction at C. i. ii. iii. iv.
Define the current problem. Draw the free body diagram for pulley for this scenario. State the relevant theory and assumptions. Estimate the direction of the reaction force and then estimate its magnitude, giving reasons. v. Solve the problem for the reaction at C. vi. Verify the solution for the reaction.
y
B
(2 Marks) (4 Marks) (1 Mark)
Frictionless, low mass pulley
C
A
(1 Mark) (3 Marks) (2 Marks)
+ve D
x
90.0°
E M1
CRATE
Figure QB2 Static masses connected over a frictionless pulley 2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 6 of 7
lOMoAR cPSD| 3264513
Useful formulas and data
sin= A/C cos= B/C tan= A/B sin²+cos²= 1
Moment = Force x Distance, where Distance refers to the perpendicular distance from the point about which the moment is to be calculated to the line of action of a force. A anticlockwise moment is +ve.
+ve BM
2017 March
ENGG102 Mid Session
Wollongong
Autumn 2017
Page 7 of 7
