friction additional examples. example 1: a 100-lb force acts as shown on a 300-lb block placed on an...
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Friction
Additional Examples
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, N = +240 lb
The F required to maintain equilibrium is directed up and to the right; the tendency of the block is move down the plane.
2. Maximum friction force.
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.
q
qN
F
xy 1. Assume equilibrium:
SFy =N – 250cos30o-50sin30o = 0
N = +241.5 lb
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
20o
30o
P
20o
30o
P xy
W= 981 N
F
N
30o
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.
20o
30o
Pmin x
y981 N
F N
30o
b) Determine the minimum force P to prevent motion
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:
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
20o
30o
Pmax x
y981 N
F
N
30o
The maximum force P will be required when motion of the block up the incline is impending.
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
Solving simultaneously:
N = 626 N
Pmax = 655N
Center of Gravity
Additional Examples
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,
i iT
i iT
1
1
x x AA
y y AA
Determine the area of the components
21
22
1120 mm 60 mm 3600 mm
2
120 mm 100 mm 12000 mm
A
A
1A
2A
Centroids – Simple Example for a Composite Body
The total area is
1
11 1
2
22 1
120 mm40 mm
3 360 mm
60 mm 40 mm 3 3120 mm
60 mm2 2
100 mm60 mm 110 mm
3 2
bx
hy h
bx
hy h
1A
2A
Centroids – Simple Example for a Composite Body
To centroid of each component
Compute the x centroid
1A
2A
2 2T 1 2
2
3600 mm 12000 mm
15600 mm
A A A
i iT
2 22
1
140 mm 3600 mm 60 mm 12000 mm
15600 mm55.38 mm
x x AA
Centroids – Simple Example for a Composite Body
To centroid of each component
Compute the y centroid
1A
2A
2 2T 1 2
2
3600 mm 12000 mm
15600 mm
A A A
i iT
2 22
1
140 mm 3600 mm 110 mm 12000 mm
15600 mm93.85 mm
y y AA
Centroids – Simple Example for a Composite Body
The problem can be done using a table to represent the composite body.
Body Area(mm2) x (mm) y(mm) x*Area (mm3) y*Area (mm3)
Triangle 3600 40 40 144000 144000Square 12000 60 110 720000 1320000
Sum 15600 864000 1464000
centroid (x) 55.38 mmcentroid (y) 93.85 mm
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 isAssume that area is a large square and subtract the small triangular area.
1A
2A
Centroids – Simple Example for a Composite Body
The problem can be done using a table to represent the composite body.
Body Area(mm2) x (mm) y(mm) x*Area (mm3) y*Area (mm3)
Square 19200 60 80 1152000 1536000Triangle -3600 80 20 -288000 -72000
Sum 15600 864000 1464000
centroid (x) 55.38 mmcentroid (y) 93.85 mm
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
1
2
2
2
2
3
2
190 mm 60 mm
2
2700 mm
120 mm 90 mm
10800 mm
40 mm2
2513.3 mm
A
A
A
1A
2A 3A
Centroids –Example for a Composite Body
The total area is
1
11 1
2
22 1
3
3
90 mm30 mm
3 360 mm
60 mm 40 mm 3 390 mm
45 mm2 2
120 mm60 mm 120 mm
3 24 40 mm4
90 mm 73.02 mm3 3
60 mm 20 mm 40 mm 120 mm
bx
hy h
bx
hy h
rx b
y
1A
2A 3A
Centroids – Example for a Composite Body
Body Area(mm2) x (mm) y(mm) x*Area (mm3) y*Area (mm3)
Triangle 2700 30 40 81000 108000Square 10800 45 120 486000 1296000
Hemisphere -2513.27 73.02 120 -183528.00 -301592.89
Sum 10986.73 383472.00 1102407.11
centroid (x) 34.90 mmcentroid (y) 100.34 mm
3
i i 2T
1 1102407.11 mm
10986.73 mm
100.34 mm
y y AA
3
i i 2T
1 383472.00 mm
10986.73 mm
34.90 mm
x x AA
Centroids – Example for a Composite Body
An Alternative Method would be to subtract to areas
Body Area(mm2) x (mm) y(mm) x*Area (mm3) y*Area (mm3)
Triangle -2700 60 20 -162000 -54000Square 16200 45 90 729000 1458000
Hemisphere -2513.27 73.02 120 -183528.00 -301592.89
Sum 10986.73 383472.00 1102407.11
centroid (x) 34.90 mmcentroid (y) 100.34 mm
Centroids – Class Problem
Find the centroid of the body
Centroids – Class Problem
Find the centroid of the body
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