### Extra Examples

```Friction
Example 1: A 100-lb force acts as shown on a 300-lb block placed
on an inclined plane. The coefficients of friction between the
block and the plane are ms =0.25 and mk = 0.20. Determine
whether the block is in equilibrium, and find the value of the
friction force.
1. Force required for equilibrium:
Assuming that F is directed down
SFx = 100lb – 3/5 (300lb) – F = 0
F = 80 lb
SFy = N – 4/5(300lb) = 0,
+240 lb
2. Maximum friction force.
N=
The F required to maintain
equilibrium is directed up and to the
right; the tendency of the block is
move down the plane.
Fm = ms N, Fm = 0.25(240 lb) = 60 lb
Since the value of F required to maintain equilibrium (80lb) is larger
than the maximum possible (60lb), the block will slide down the plane
3. Actual value of friction force:
Factual = Fk ( the body is moving)
Factual = Fk = mk N = 0.2(240lb) = 48 lb
The sense of this force is opposite to the
sense of motion.
The forces acting on the block are not
balanced, the resultant is:
3/5 (300lb) – 100lb – 48lb = 32lb
Example2: Determine whether the block shown is in equilibrium and
find the magnitude and direction of the friction force when q = 30o
and P = 50lb.
y
x
q
1. Assume equilibrium:
SFy =N – 250cos30o-50sin30o = 0
N = +241.5 lb
q
N
F
SFx =F– 250sin30o +50cos30o = 0, F =
+81.7 lb
2. Maximum friction force:
Fm = ms N = 0.3 (241.5 lb) = 72.5 lb
Since F > Fm, then the block moves down
Friction force: F = mk N = 0.2(241.5lb) = 48.3 lb
Example 3: The block in the figure has a mass of 100 kg. The
coefficient of friction between the block and the inclined surface
is 0.2. (a) Determine if the system is in equilibrium when P= 600
N
P
20o
30o
y
W= 981 N
30o
P
x
20o
F
30o
N
1. Determination of F and N:
SFy= Psin 20o + N – Wcos 30o = 0, N =
644.36 N
SFx= Pcos 20o – F – Wsin 30o = 0, F =
73.32 N
2. Determination of Fmaximum
Fm =msN= (0.20)(644.36)=128.87N
Since Fm > 73.32, then
the block is in equilibrium.
y
b) Determine the minimum force P to
prevent motion
981 N
30o
Pmin
20o
x
The minimum P will be required when
motion of the block down the incline is
impending.
F must resist this motion as shown.
Equilibrium exists when:
F
30o
N
SFx = Pmin cos 20o + F – 981sin 30o = 0
SFx = Pmin cos 20o + 0.2N – 981 sin 30o = 0
SFy = Pmin sin 20o + N – 981cos 30o = 0
Then N = 724 N, P min = 368 N
c) Determine the maximum force P for which the system is in
equilibrium
y
The maximum force P will be required
when motion of the block up the incline is
impending.
981 N
30o
Pmax
20o
x
F
For this condition, F will tend to resist this
motion as shown. Then:
SFx = Pmax cos 20o – 0.2 N – 981 sin 30o = 0
SFy = Pmax sin 20o + N – 981 cos 30o = 0
30o
N
Solving simultaneously:
N = 626 N
Pmax = 655N
Center of Gravity
Centroids – Simple Example for a
Composite Body
Find the centroid of the given body
Centroids – Simple Example for a
Composite Body
To find the centroid,
x
1
AT
y
1
AT
xA
i

i
y i Ai
A2
A1
Determine the area of the
components
1
m m   60 m m   3600 m m
2
A2   120 m m  100 m m   12000 m m
2
A1 
2
120
Centroids – Simple Example for a
Composite Body
The total area is
x1 
b

3
b
 40 m m
3
y1  h1 
x2 
120 m m

h1
3
 40 m m
3
120 m m
2
y 2  h1 
 60 m m 
60 m m
A1
 60 m m
2
h2
3
 60 m m 
A2
100 m m
2
 110 m m
Centroids – Simple Example for a
Composite Body
To centroid of each component
AT  A1  A2  3600 m m  12000 m m
2
 15600 m m
2
Compute the x centroid
x
1
AT


x i Ai
1
2
A2
A1
  40 m m   3600 m m 2    60 m m  12000 m m 2  
2

15600 m m 
 55.38 m m
Centroids – Simple Example for a
Composite Body
To centroid of each component
AT  A1  A2  3600 m m  12000 m m
2
 15600 m m
2
Compute the y centroid
y
1
AT


y i Ai
1
2
A2
A1
  40 m m   3600 m m 2   110 m m  12000 m m 2  
2

15600 m m 
 93.85 m m
Centroids – Simple Example for a
Composite Body
The problem can be done using a table to represent the
composite body.
Body
Triangle
Square
Area(mm2)
3600
12000
Sum
15600
centroid (x)
centroid (y)
x (mm)
40
60
55.38 mm
93.85 mm
y(mm)
40
110
x*Area (mm3) y*Area (mm3)
144000
144000
720000
1320000
864000
1464000
Centroids – Simple Example for a
Composite Body
An alternative method of computing the centroid is
to subtract areas from a total area.
Assume that area is a large
square and subtract the small
triangular area.
A1
A2
Centroids – Simple Example for a
Composite Body
The problem can be done using a table to represent the
composite body.
Body
Square
Triangle
Area(mm2)
19200
-3600
Sum
15600
centroid (x)
centroid (y)
x (mm)
60
80
55.38 mm
93.85 mm
y(mm)
80
20
x*Area (mm3) y*Area (mm3)
1152000
1536000
-288000
-72000
864000
1464000
Centroids –Example for a
Composite Body
Find the centroid of the given body
Centroids –Example for a
Composite Body
Determine the area of the
components
A1 
1
2
 90
m m   60 m m 
 2700 m m
2
A2
A2   120 m m   90 m m 
 10800 m m

2
A1
 40 m m 
2
 2513.3 m m
2
A3 
2
A3
Centroids –Example for a
Composite Body
The total area is
x1 
b
90 m m

3
3
h1
y1  h1 
x2 
b
 30 m m
 60 m m 
60 m m
3
2
A1
 45 m m
2
y 2  h1 
h2
x3  b 
4r
 60 m m 
3
3
A2
3
90 m m

 40 m m
120 m m
 120 m m
2
 90 m m 
4  40 m m 
3
 73.02 m m
y 3  60 m m  20 m m  40 m m  120 m m
A3
Centroids – Example for a
Composite Body
Body
Triangle
Square
Hemisphere
Area(mm2)
2700
10800
-2513.27
Sum
10986.73
centroid (x)
centroid (y)
x (mm)
30
45
73.02
y(mm)
40
120
120
x*Area (mm3) y*Area (mm3)
81000
108000
486000
1296000
-183528.00
-301592.89
383472.00
34.90 mm
100.34 mm
x
1
AT
xA
i
i

383472.00 m m
10986.73 m m
3
2
 34.90 m m
y
1
AT

y i Ai 
 100.34 m m
1102407.11 m m
10986.73 m m
2
3
1102407.11
Centroids – Example for a
Composite Body
An Alternative Method would be to
subtract to areas
Body
Triangle
Square
Hemisphere
Area(mm2)
-2700
16200
-2513.27
Sum
10986.73
centroid (x)
centroid (y)
x (mm)
60
45
73.02
34.90 mm
100.34 mm
y(mm)
20
90
120
x*Area (mm3) y*Area (mm3)
-162000
-54000
729000
1458000
-183528.00
-301592.89
383472.00
1102407.11
Centroids – Class Problem
Find the centroid of the body
Centroids – Class Problem
Find the centroid of the body
```