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An in-depth exploration of dry friction characteristics and their application to solve related problems. It covers key concepts, experimental determination of the coefficient of static friction, and a step-by-step approach to solving equilibrium problems. Detailed examples and exercises reinforce the understanding of these fundamental engineering principles. By studying this document, students will gain the ability to analyze and solve real-world problems where dry friction is crucial, such as in brake systems, towing and pushing scenarios, and impending tipping versus slipping conditions. This comprehensive coverage makes this document a valuable resource for students in introductory engineering courses.
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In-Class Activities:
Today’s Objective:
Students will be able to:
a) Understand the characteristics of dry friction
b) Draw a FBD including friction.
c) Solve problems involving friction.
For an applied force on the bike tire brake pads, how can we determine the magnitude and direction of the resulting friction force?
In designing a brake system for a bicycle, car, or any other vehicle, it is important to understand the frictional forces involved.
(continued)
What physical factors affect the answer to this question?
The rope is used to tow the refrigerator.
In order to move the refrigerator, is it best to pull up as shown, pull horizontally, or pull downwards on the rope?
(continued)
The maximum friction force is attained just before the block begins to move (a situation that is called “impending motion”). The value of the force is
called the coefficient of static friction. The value of μs depends on the two materials in contact. Once the block begins to move, the frictional force typically drops and is
(coefficient of kinetic friction) is less than μs.
(continued)
It is also very important to note that the friction force may be less than the maximum friction force. So, just because the object is not moving, don’t assume the friction force is at its maximum of Fs = μs N unless you are told or know motion is impending!
If the block just begins to slip, the
where μs is the coefficient of static friction.
Thus, when the block is on the verge of sliding, the normal force N and frictional force F (^) s combine to create a resultant Rs.
From the figure, tan φs = ( Fs / N ) = (μs N / N ) = μs
Using these two equations, we get μs = (W sin θs ) / (W cos θs ) = tan θs This simple experiment allows us to find the μS between two materials in contact.
A block with weight w is placed on an inclined plane. The plane is slowly tilted until the block just begins to slip.
The inclination, θs , is noted. Analysis of the block just before it begins to move gives (using Fs = μs N):
IMPENDING TIPPING versus SLIPPING (continued) Assume: Slipping occurs Known: F = μs N
Solve: x, P, and N
Check: 0 ≤ x ≤ b/ Or Assume: Tipping occurs
Known: x = b/
Solve: P, N, and F Check: F ≤ μs N
a) Draw a FBD of the box. b) Determine the unknowns. c) Make your friction assumptions. d) Apply E-of-E (and friction equations, if appropriate ) to solve for the unknowns. e) Check assumptions, if required.
Given : Crate weight = 250 lb, μs = 0.
Find : The maximum force P that can be applied without causing movement of the crate.
Plan: ??
EXAMPLE (continued)
X
1.5 ft 1.5 ft
250 lb
0 F
FBD of the crate
P
N
3.5 ft
4.5 ft
There are four unknowns: P, N, F and x.
First, let’s assume the crate slips. Then the friction equation is F = μs N = 0.4 N.
Solution:
EXAMPLE (continued)
X
1.5 ft 1.5 ft
250 lb
O F
FBD of the crate
P
N
3.5 ft
4.5 ft
→ F (^) X = P – 0 .4 N = 0
↑ F (^) Y = N – 250 = 0 Solving these two equations gives: P = 100 lb and N = 250 lb
M (^) O = -100 (4.5) + 250 (x) = 0 Check: x = 1. 8 ≥ 1. 5 : No slipping will occur since x > 1. 5
C) P(C) D) Can not be determined
P(A) P(B) P(C)
100 lb
A
B
A) B)
C) ← D) ↑ A
B
C) 2 lb D) 3 lb
μ (^) S = 0. 2 lb
a) Draw FBDs of the dresser and man. b) Determine the unknowns. d) Apply E-of-E to solve for the unknowns.
Given : Dresser weight = 90 lb, man’s weight = 150 lb. μs = 0.25.
Find : The smallest magnitude of F needed to move the dresser if θ = 30°.
Also determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. Plan:
GROUP PROBLEM SOLVING (continued)
FBD of the man FBD of the^ dresser Dresser : +→ F (^) X = F (cos 30) – 0.25 N = 0
