STATICS OF RIGID BODIES
FRICTION
Friction is the contact resistance exerted by one body when the second body moves
or tends to move past the first body. Friction is a retarding force that always acts
opposite to the motion or the tendency of motion.
Elements of Friction
P
F
O
N R
F=μN
F
tan θ =
N
μ= tan θ
Where:
N – normal reaction
F – frictional force
µ - coefficient of friction
R – resultant of F and N
Θ – angle of friction
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Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
EXAMPLES
1. A 400 kN block is resting on a rough horizontal surface for which the coefficient of
friction is 0.40. Determine the force P required to cause motion to impend if applied to
the block.
a. Horizontally.
b. Downward at 30o with the horizontal.
SOLUTION
a. Horizontally
400
P
F
∑ Fv =0
N-400 kN=0
F=μN
F=0.40 (400)
P=F=160 kN
b. 30o with the horizontal
400
o
30
P F
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Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
∑ Fv =0
N-400 kN-P sin 30° =0
N-P sin 30° =400 kN; eq. 1
∑ Fh =0
F-P cos 30° =0
μN-P cos 30° =0
0.40N-P cos 30° =0; eq.2
Solve eq. 1 and 2
P=240 kN
2. The 200 kN block shown has impending motion up the plane caused by the
horizontal force of 400 kN. Determine the coefficient of static friction between the
contact surfaces.
200
400
0 R
30o
SOLUTION
∑ Fv =0
- 200+R cos(30°+θ) =0
3
Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
R cos(30°+θ) =200; eq. 1
∑ Fh =0
400-R sin(30°+θ) =0
R cos(30°+θ) =400; eq. 2
Solve eq. 1 and eq. 2
θ=33.4°
tan θ= μ
tan 33.4°= μ
μ=0.66
3. The winch in the figure is used to move the 135 kg uniform log. The coefficient of
static friction between the log and the plane is 0.40.
o
60
A B
log
2.4 m
a. Compute the tension T for impending sliding.
b. Compute the tension T for impending tipping.
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Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
c. Compute the maximum tension that can be applied for which the log remains at rest.
d. Compute the normal reaction between the log and plane.
e. Compute the location of the normal reaction from A.
SOLUTION
a. Tension for impending sliding
∑ Fv =0
N+T sin 60° =135(9.81); eq. 1
∑ Fh =0
F-T cos 60° =0
F=μN
0.4N-T cos 60° =0; eq. 2
Solve eq. 1 and eq. 2
T=626 N
b. Tension for impending flipping
∑ Ma =0
T sin 60° (2.4)-135(9.81)(1.2)=0
T=765 N
5
Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
T
135 kg
o
60
A 1.2 m 1.2 m B
c. Maximum tension that can be applied without moving the log.
Tmax =626 N
d. Normal reaction between log and plane.
∑ Fv =0
N+T sin 60° =135(9.81)
N+(626) sin 60° =135(9.81)
N=782.22 N
e. Location of the normal reaction from A:
T
135 kg
o
60
A 1.2 m 1.2 m B
782.22
∑ Ma =0
T sin 60° (2.4)+782.22x-135(9.81)(1.2)=0
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Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
626 sin 60° (2.4)+782.22x-135(9.81)(1.2)=0
x=0.368 m
4. Find the least value of P required to cause the system of blocks shown in the figure
below to have impending motion to the left. The coefficient of friction under each block
is 0.20.
300 kN
P
0
100 kN
o
30
SOLUTION
100
T1 30o
F1
N1
∑ Fv =0
N1 =100 cos 30°
N1 = 86.60 kN
F1 =μN1
F1 =0.20 (86.60)
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Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
F1 =17.32 kN
∑ Fh =0
T1 =100 sin 30° + F1
T1 =50+ 17.32
T1 = 67.32 kN
300
P
0 T 1
F 2
N 2
∑ Fv =0
N2 +P sin θ =300
N2 = 300-P sin θ
F2 =μN2
F2 =μ(300-P sin θ)
∑ Fh =0
T1 + F2 =P cos θ
67.32 + 0.20(300-P sin θ)=P cos θ
8
Department of Civil Engineering. College of Engineering. University of the East - Manila
STATICS OF RIGID BODIES
θ=11.3°
P=125 kN
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Department of Civil Engineering. College of Engineering. University of the East - Manila
