Sections 8.3, 8.4, & 8.5
ENGR-1100 Introduction to Engineering Analysis
Lecture 27
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 1.1-1.6
1
WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
In-Class Activities:
Reading Quiz
Applications
Analysis of a Wedge
Analysis of a Belt
Concept Quiz
Group Problem Solving
Attention Quiz
Today’s Objectives:
Students will be able to:
a) Determine the forces on a wedge.
b) Determine tension in a belt.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
2
APPLICATIONS
How can we determine the force required to pull the wedge out?
When there are no applied forces on the wedge, will it stay in place (i.e., be self-locking) or will it come out on its own? Under what physical conditions will it come out?
Wedges are used to adjust the elevation or provide stability for heavy objects such as this large steel pipe.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
3
APPLICATIONS (continued)
How can we decide if the belts will function properly, i.e., without slipping or breaking?
Belt drives are commonly used for transmitting the torque developed by a motor to a wheel attached to a pump, fan or blower.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
4
How can you determine the tension in the cable pulling on the band?
Also from a design perspective, how are the belt tension, the applied force P and the torque M, related?
APPLICATIONS (continued)
In the design of a band brake, it is essential to analyze the frictional forces acting on the band (which acts like a belt).
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
5
ANALYSIS OF A WEDGE
It is easier to start with a FBD of the wedge, since you know the direction of its motion.
Note that:
a) the friction forces are always in the direction opposite to the motion, or impending motion, of the wedge;
b) the friction forces are along the contacting surfaces; and
c) the normal forces are perpendicular to the contacting surfaces.
W
A wedge is a simple machine in which a small force P is used to lift a large weight W.
To determine the force required to push the wedge in or out, it is necessary to draw FBDs of the wedge and the object on top of it.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
6
ANALYSIS OF A WEDGE (continued)
To determine the unknowns, we must apply E-of-E, Fx = 0 and Fy = 0, to the wedge and the object as well as the impending motion frictional equation, F = S N.
Next, a FBD of the object on top of the wedge is drawn. Please note that:
a) at the contacting surfaces between the wedge and the object, the forces are equal in magnitude and opposite in direction to those on the wedge
b) all other forces acting on the object should be shown.
ANALYSIS OF A WEDGE (continued)
Now, of the two FBDs, which one should we start analyzing first?
We should start analyzing the FBD in which the number of unknowns is less than or equal to the number of E-of-E and frictional equations.
ANALYSIS OF A WEDGE (continued)
NOTE:
If the object is to be lowered, then the wedge needs to be pulled out. If the value of the force P needed to remove the wedge is positive, then the wedge is self-locking, i.e., it will not come out on its own.
BELT ANALYSIS
Detailed analysis (please refer to your textbook) shows that T2 = T1 e where is the coefficient of static friction between the belt and the surface. Be sure to use radians when using this formula!!
If the belt slips or is just about to slip, then T2 must be larger than T1 and the motion resisting friction forces. Hence, T2 must be greater than T1.
Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to radians.
EXAMPLE
1. Draw a FBD of the crate. Why do the crate first?
2. Draw a FBD of the wedge.
3. Apply the E-of-E to the crate.
4. Apply the E-of-E to wedge.
Given: The crate weighs 300 lb and S at all contacting surfaces is 0.3. Assume the wedges have negligible weight.
Find: The smallest force P needed to pull out the wedge.
Plan:
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
11
EXAMPLE (continued)
+ FX = NB – 0.3NC = 0
+ FY = NC – 300 + 0.3 NB = 0
Solving the above two equations, we get
NB = 82.57 lb = 82.6 lb, NC = 275.3 lb = 275 lb
The FBDs of crate and wedge are shown in the figures. Applying the E-of-E to the crate, we get
NB
300 lb
FB=0.3NB
FC=0.3NC
NC
15º
15º
FC=0.3NC
NC
FD=0.3ND
ND
P
FBD of Crate
FBD of Wedge
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
12
Known forces are indicated by blue color arrow, Unknown forces indicated by yellow color.
The thickness of arrow lines are 2 1/4 pt.
Bolden the Decimal Point
Symbols used in the equations are from the Symbols library.
do not use sentence case on ‘cos’ and ‘sin’, they are to be in lower case.
EXAMPLE (continued)
NB = 82.6 lb
300 lb
FB=0.3NB
FC=0.3NC
NC = 275 lb
15º
15º
FC=0.3NC
NC
FD=0.3ND
ND
P
FBD of Crate
FBD of Wedge
Applying the E-of-E to the wedge, we get
+ FY = ND cos 15 + 0.3 ND sin 15 – 275.2= 0;
ND = 263.7 lb = 264 lb
+ FX = 0.3(275.3) + 0.3(263.7)cos 15 – (263.7)sin 15 – P = 0;
P = 90.7 lb
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
13
Known forces are indicated by blue color arrow, Unknown forces indicated by yellow color.
The thickness of arrow lines are 2 1/4 pt.
Bolden the Decimal Point
Symbols used in the equations are from the Symbols library.
do not use sentence case on ‘cos’ and ‘sin’, they are to be in lower case.
READING QUIZ
W
1. A wedge allows a ______ force P to lift
a _________ weight W.
A) (large, large) B) (small, large)
C) (small, small) D) (large, small)
2. Considering friction forces and the indicated motion of the belt, how are belt tensions T1 and T2 related?
A) T1 > T2 B) T1 = T2
C) T1 < T2 D) T1 = T2 e
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
14
Answers:
1. B
2. C
CONCEPT QUIZ
2. The climber hanging at A in the picture weighs 100 lb and the individual on top on the flat weighs 150 lb. The coefficient of static friction between this individual’s shoes and the ground is 0.6. The climber will ______ ?
A) Be lifted up B) Slide down
C) Not be lifted up D) Not slide down
1. Determine the direction of the friction force on object B at the contact point between A and B.
A) B)
C) D)
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
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Answers:
1. A
2. C The maximum force she can apply before she slides is 150 (0.6) = 90 lb. This will not be enough to lift the boy up as she needs to apply more than 100 lb to lift the boy up. Without knowing the coefficient of friction between the cliff and the rope, we can not say whether the boy will slide down or not.
Wait! What? It is really more complicated.
2. The climber hanging at A in the picture weighs 100 lb and the individual on top on the flat weighs 150 lb. The coefficient of static friction between this individuals shoes and the ground is 0.6. The climber will ______ ?
A) Be lifted up B) Slide down
C) Not be lifted up D) Not slide down
In this problem, there are really 3 possibilities.
The climber is pulled up
The climber stays at A
The climber falls from A
Because we do not know the coefficient of friction at the 90o turn, it is impossible to say if there is enough friction combined with the weight of the holding individual to stop the climber from falling.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
16
Answers:
1. A
2. C The maximum force she can apply before she slides is 150 (0.6) = 90 lb. This will not be enough to lift the boy up as she needs to apply more than 100 lb to lift the boy up. Without knowing the coefficient of friction between the cliff and the rope, we can not say whether the boy will slide down or not.
ATTENTION QUIZ
2. In the analysis of frictional forces on a flat belt, T2 = T1 e .
In this equation, equals ______ .
A) Angle of contact in degrees B) Angle of contact in radians
C) Coefficient of static friction D) Coefficient of kinetic friction
1. When determining the force P needed to lift the block of weight W, it is easier to draw a FBD of ______ first.
A) The wedge B) The block
C) The horizontal ground D) The vertical wall
W
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
17
Answers:
1.A
2.B
Tuesday 12/15/2020; 8:00 – 11:00 am:
Students starting late will not get extra time.
• The test ends at 10:35 am.
• The deadline to upload your exam to LMS is 10:55 am.
You have a 5-minute grace period to upload your exam to LMS without a penalty.
You can upload your exam to LMS after 11:00 am with a penalty of 1 point/minute (e.g., if you submit your exam on 11:15 am your exam grade will be reduced by 15 points!).
Final Exam Information
Final Grade Components
Highest grade of Exams 1, 2 and 3 20%
Other two Exams (15% each) 30%
HomeWorks 20%
In-Class assignments 5%
Final examination 25%
The 2 HW assignments and 4 CA with the lowest grades will be dropped.
4 Problems equally weighted (25 points each):
Problems 1 and 2 (mandatory):
Frames/Machines
Friction (Dry/Belt)
Problems 3 and 4, select 2 problems out of 3:
Test 1 topics
Test 2 topics
Test 3 topics
Final Exam
You will be graded on 4 problems, 25 points per problem. Problems 1 and 2 are mandatory and will be systematically graded.
Select two problems out of problems 3, 4 and 5 and submit only the 2 selected problems. If you submit the three problems (3, 4 and 5), only 2 will be graded at random.
Clearly show all steps and state all assumptions made in order for full credit to be given.
Express your final answers clearly and highlight them in boxes.
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Final Exam
Makeup Exam
Tuesday 12/15/2020, 11:30 am – 2:30 pm
Only for students that have a conflict with the regular final exam
NOTE: Let your instructor know no later than Tuesday 12/08/2020 and specify the course that is conflicting with IEA
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Regular and Extra-Time Submission Deadlines
End of the test Deadline for submission Late submission
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1-point penalty/minute thereafter.
50 % Extra time 11:53 am 12:13 pm 5-minute grace period,
1-point penalty/minute thereafter.
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1-point penalty/minute thereafter.
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The exam is open book and notes.
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You connect from LMS to Webex at least 5 minutes before 8:00 am on 12/15/2020
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GROUP PROBLEM SOLVING
Given: Blocks A and B weigh 50 lb and 30 lb, respectively.
Find: The smallest weight of cylinder D which will cause the loss of static equilibrium.
Plan:
GROUP PROBLEM SOLVING (continued)
Plan:
1. Consider two cases: a) both blocks slide together, and,
b) block B slides over the block A.
2. For each case, draw a FBD of the block(s).
3. For each case, apply the E-of-E to find the force needed to cause sliding.
4. Choose the smaller P value from the two cases.
5. Use belt friction theory to find the weight of block D.
GROUP PROBLEM SOLVING (continued)
Case a (both blocks sliding together):
+ FY = N – 80 = 0
N = 80 lb
+ FX = 0.4 (80) – P = 0
P = 32 lb
P
B
A
N
F=0.4 N
30 lb
50 lb
Case b has the lowest P (case a was 32 lb) and thus will occur first. Next, using a frictional force analysis of belt, we get
WD = P e = 5.812 e 0.5 ( 0.5 ) = 12.7 lb
A Block D weighing 12.7 lb will cause the block B to slide over the block A.
+ Fy = N cos 20 + 0.6 N sin 20 – 30 = 0 N = 26.20 lb
+ Fx = – P + 0.6 ( 26.2 ) cos 20 – 26.2 sin 20 = 0 P = 5.812 lb
20º
30 lb
0.6 N
P
N
Case b (block B slides over A):
GROUP PROBLEM SOLVING (continued)
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