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