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A lecture note from an Introduction to Engineering course focusing on the analysis of wedges and frictional forces on flat belts. It includes objectives, applications, analysis of a wedge, and analysis of a belt. Students will learn how to determine forces on a wedge and tension in a belt, as well as the relationship between belt tension, applied force, and torque in the design of a band brake.
Typology: Lecture notes
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In-Class Activities:
Today’s Objectives:
Students will be able to:
a) Determine the forces on a wedge.
b) Determine tension in a belt.
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.
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.
ANALYSIS OF A WEDGE (continued)
To determine the unknowns, we must apply EofE, Fx = 0 and Fy = 0, to the wedge and the object as well as the impending motion frictional equation, F = μS 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; and, 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 are less than or equal to the number of E-of-E and frictional equations.
ANALYSIS OF A WEDGE (continued)
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.
Detailed analysis (please refer to your textbook) shows that T 2 = T 1 e μ β^ where μ 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 T 2 must be larger than T 1 and the motion resisting friction forces. Hence, T 2 must be greater than T 1.
Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to β radians.
EXAMPLE (continued)
NB = 82.6 lb
F 300 lb B=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 ↑+ F (^) Y = ND cos 15° + 0.3 N (^) D sin 15° – 275.2= 0; ND = 263.7 lb = 264 lb →+ F (^) X = 0.3(263.7) + 0.3(263.7)cos 15° – 0.3(263.7)cos 15° – P = 0; P = 90. 7 lb
W
a _________ weight W.
A) (large, large) B) (small, large) C) (small, small) D) (large, small)
C) T 1 < T 2 D) T 1 = T 2 eμ
A) Be lifted up B) Slide down C) Not be lifted up D) Not slide down
A) B) ←
C) → D)
C) Coefficient of static friction D) Coefficient of kinetic friction
A) The wedge B) The block C) The horizontal ground D) The vertical wall
W
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 μ β^ = 5. 812 e 0.5 ( 0.5^ π^ )^ = 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
P 0.6 N
N
Case b (block B slides over A):
GROUP PROBLEM SOLVING (continued)
