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7.7 Rolling Resistance
7.7 Rolling Resistance Example 1, page 1 of 4
1. The fertilizer spreader and the fertilizer it contains have a combined
mass of 40 kg and a center of gravity located at point G. If the coefficient
of rolling resistance for the tires is 5 mm, determine the resultant force that
must be applied to the handle to move the spreader at a constant speed.
250 mm
Px
100 mm
Py
G
Radius = 120 mm
850 mm
7.7 Rolling Resistance Example 1, page 2 of 4
1
Locate the point where the resisting force
from the ground acts on the wheel.
4 For later use in calculating moments, find the
vertical distance between A and B.
Rotation
of wheel
A
Direction of motion
of center of wheel
A
100 mm
B
B
(100 mm)2
5 mm
2 As the wheel rolls to R
the left, the ground
exerts a force, R,
opposing the motion.
5 mm ( = coefficient of rolling resistance)
3
R acts at a point B on the
circumference 5 mm
from a vertical line
through the center A of
the wheel.
(5 mm)2
= 99.875 mm
7.7 Rolling Resistance Example 1, page 3 of 4
5
Free-body diagram of spreader.
250 mm
Px
100 mm
6
Equilibrium equations
Py
+
9.81 m/s
G
850 mm
Px = 0
Fy = 0: 2Ry + Py
+
Weight = 40 kg
Fx = 0: 2Rx
2
392.4 N = 0
(1)
(2)
+
= 392.4 N
A
2Ry
Components of force
from ground acting on
two wheels
99.875 mm
B
2Rx
5 mm
MB = 0 : Px (850 mm + 99.875 mm)
+ Py (250 mm + 5 mm)
(392.4 N)(100 mm + 5 mm) = 0
(3)
Three equations, four unknowns; an additional free
body is needed.
7.7 Rolling Resistance Example 1, page 4 of 4
Free body diagram of a wheel and axle
8
Moment equation for wheel
+
7
Forces from spreader
acting on axle
MA = 0: Rx(99.875 mm)
Ry(5 mm) = 0
(4)
Solving equations 1-4 simultaneously gives
Rx = 7.102 N
Ay
Ax
A
Px = 14.205 N
100 mm
Rx
Ry = 141.868 N
99.875 mm
B
The resultant force acting on the handle is
5 mm
Ry
Py = 108.665 N
(Rx2 + Ry2) =
(7.102 N)2 + (141.868 N)2
= 142 N
Ans.
7.7 Rolling Resistance Example 2, page 1 of 4
2. The mass of the upright piano is 200 kg. The coefficient of
rolling resistance of the casters is 0.2 mm. Determine the value
of the horizontal force P required to move the piano at constant
speed. Make the simplifying assumption that the resisting
forces are the same at all four casters.
P
Radius of the
casters = 14 mm
7.7 Rolling Resistance Example 2, page 2 of 4
1
Locate the point where the resisting force
from the floor acts on the wheel.
Rotation
of wheel
Direction of motion
of center of wheel
A
A
B
14 mm
R
0.2 mm ( = coefficient
of rolling resistance)
2 As the wheel rolls to the left,
the floor exerts a force, R,
opposing the motion.
3
R acts at a point B on the
circumference 0.2 mm from a
vertical line through the center
A of the wheel.
B
0.2 mm
(14 mm)2
(0.2 mm)2
= 13.9986 mm
4 For later use in calculating
moments, find the vertical
distance between A and B.
7.7 Rolling Resistance Example 2, page 3 of 4
5
Free-body diagram of piano
Weight = (200 kg)(9.81 m/s2) = 1962 N
P
6
Ry
Ry
+
Fx = 0: 4Rx
P=0
(1)
+
Rx
Rx
Equilibrium equations.
Fy = 0: 4Ry
1962 N = 0
(2)
Rx
Rx
Ry
Ry
Resisting forces from floor acting on
caster are same at each caster
Two equations, three unknowns; an
additional free body is needed.
7.7 Rolling Resistance Example 2, page 4 of 4
7 Free-body diagram of a caster and axle
Forces from piano
acting on axle
Moment equation for caster
+
8
A
MA = 0 : Rx(13.9986 mm)
Ry(0.2 mm) = 0
Solving Eqs.1-3 simultaneously gives
B
13.9986 mm
Rx
Rx = 7.01 N
Ry = 490.5 N
0.2 mm
Ry
P = 28.0 N
Ans.
(3)
7.7 Rolling Resistance Example 3, page 1 of 3
3. If the coefficient of rolling resistance is 0.3 in., determine the
magnitude of the horizontal force F required to push the 300-lb
drum up the inclined plane.
8 in.
1
F
Locate the point where the
resisting force from the inclined
plane acts on the drum.
Rotation of
drum
15°
Direction of
motion of
center of drum
A
B
15°
R
2
3
R acts at a point B on the circumference
0.3 in. from the center A of the drum.
Note that the 0.3 in. is measured parallel
to the inclined plane.
0.3 in. ( = coefficient
of rolling resistance)
As the wheel rolls up
the plane, the plane
exerts a force, R,
opposing the motion.
7.7 Rolling Resistance Example 3, page 2 of 3
For use in calculating moments, find the
distance from A to B in the direction
perpendicular to the inclined plane.
D
(8 in.)2
5
Free-body diagram of drum
300 lb (Weight of drum)
D
(0.3 in.)2
y
8 in.
= 7.994 in.
A
F
x
A
8 in.
B
7.994 in.
C 0.3 in.
Reference line
perpendicular to
inclined plane
6
C
B
Rx
Ry
15°
0.3 in.
Components
of force from
inclined plane
acting on
wheel
Moment equilibrium equation
+
4
MB = 0: (300 cos lb)(0.3 in.)
+ (300 sin lb)(7.994 in.)
(F cos )(7.994 in.)
+ (F sin )(0.3 in) = 0
(1)
7.7 Rolling Resistance Example 3, page 3 of 3
7 Calculate
8
= 90°
D
A
75° = 15°
Inclined
plane
B
C
15°
90°
9
Substitute
15° = 75°
= 15° into Eq. 1:
(300 cos lb)(0.3 in.)
+ (300 sin lb)(7.994 in.)
(F cos )(7.994 in.)
+ (F sin )(0.3 in) = 0
(Eq. 1 repeated)
Solving gives
F = 92.6 lb
Ans.
7.7 Rolling Resistance Example 4, page 1 of 3
4. A 100-lb steel-rim wheel of 24-in. diameter
rolls at constant velocity down the inclined plane.
If the coefficient of rolling resistance is 0.08 in.,
determine the angle .
1
Locate the point where the resisting force
from the inclined plane acts as the wheel.
Rotation of wheel
A
Direction of
motion of the
center of the
wheel
B
0.08 in. ( = coefficient
of rolling resistance)
R
2 As the wheel rolls down the
plane, the plane exerts a force,
R, opposing the motion.
3 R acts at a point B on the circumference
0.08 in. from the center of the wheel,
measured parallel to the inclined plane.
7.7 Rolling Resistance Example 4, page 2 of 3
4
5 Free-body diagram of wheel
y
D
For use in calculating moments, find the distance
from A to B in the direction perpendicular to the
inclined plane.
D
(12 in.)2
(0.08 in.)2
A
x
A
11.9997 in.
= 11.9997 in.
Components of force
from inclined plane
acting on wheel
B
C
Radius = 24/2 in.
0.08 in. Ry
= 12 in.
Equilibrium equations
+
Fx = 0: Rx
(100 lb) sin
+
6
Fy = 0: Ry
(100 lb) cos
+
C
B
0.08 in.
Rx
MA = 0 : Ry(0.08 in.)
=0
=0
(1)
(2)
Rx(11.9997 in.) = 0 (3)
Solving gives
Rx = 0.667 lb
Ry = 99.998 lb
= 0.382°
(4)
7.7 Rolling Resistance Example 4, page 3 of 3
7
Geometry
8
D
But
A
= 0.382°, by Eq. 4 so,
= 0.382°
Ans.
We could have saved same work by
noticing that the wheel is a two-force
body. Thus because the line of action
of the weight is a vertical line through
the center, the line of action of the
resisting force R must also be a vertical
line passing through the center. The
angle can be found from geometry, as
shown below.
Weight
C
A
Complement of
12 in.
B
0.08 in.
R
= sin-1 0.08 in.
12 in.
= 0.382°
7.7 Rolling Resistance Example 5, page 1 of 4
5. If the coefficient of rolling resistance at the top of the
cylinder is 0.4 mm and at bottom of the cylinder is 0.8 mm,
determine the horizontal force P required to start the block
moving to the left. The weight of the cylinder is negligible
compared to the weight of the block.
100 kg
80 mm
P
7.7 Rolling Resistance Example 5, page 2 of 4
1 Locate the points where the resisting forces act on the cylinder.
0.4 mm (= coefficient of rolling resistance)
Rblock
100 kg
P
B
2 The resisting force from the block opposes
the rolling of the cylinder on the block.
Direction of
motion of center
of the wheel
relative to the
ground
Rotation
A
of
wheel
Direction of motion
of center of the wheel
relative to the block
C
Rground
0.8 mm (= coefficient of rolling resistance)
3
The resisting force from the ground opposes
the rolling of the cylinder on the ground.
7.7 Rolling Resistance Example 5, page 3 of 4
4
For use in calculating moments, find the vertical distances
between the center of the wheel and the point of application of
the resisting forces.
0.4 mm
D
(80 mm)2
B
80 mm
(0.4 mm)2 = 79.999 mm
A
80 mm
C
0.8 mm
E
(80 mm)2
(0.8 mm)2 = 79.996 mm
7.7 Rolling Resistance Example 5, page 4 of 4
5
7
Free-body diagram of block and cylinder
Free-body diagram of the cylinder
0.4 mm
Weight = (100 kg)(9.81 m/s2) = 981 N
100 kg
Rblock-y
Rblock-x
B
F
79.999 mm
A
79.996 mm
Rground-x
Rground-x
C
C
Rground-y
E
0.8 mm
Moment equation
+
Rground-y
Equilibrium equations
+
Fx = 0: Rground-x
F=0
(1)
+
6
Fy = 0: Rground-y
981 N = 0
(2)
Two equations, three unknowns: an additional
free-body is needed.
MB = 0: Rground-x(79.999 mm + 79.996 mm)
Rground-y(0.8 mm + 0.4 mm) = 0
Solving Eqs. 1-3 gives
Rgroung-x = 7.358 N
Rground-y = 981 N
F = 7.36 N
Ans.
(3)
7.7 Rolling Resistance Example 6, page 1 of 5
6. The ancient Britons who constructed the prehistoric monument
Stonehenge moved massive stones over twenty miles from a quarry to
the monument site. One possible way that they might have done this is
to roll the stones over logs laid on the ground. Assuming that the
coefficient of rolling resistance for the top of a log is 0.01 ft and the
coefficient is 0.08 ft for the bottom, estimate the horizontal force
required to move the typical stone shown below. Make the simplifying
assumptions that the resisting forces are the same for each log.
24 ft
100,000 lb
Rope
Force of men
(and perhaps
oxen)
F
4 ft
Diameter of
logs = 1 ft.
7.7 Rolling Resistance Example 6, page 2 of 5
1
Direction of
motion of stone
Locate the points where the resisting
forces act on the log.
2 As the log rolls on the
underside of the stone, the
stone exerts a force opposing
the rolling motion
0.01 ft
Portion
of stone
B
Coefficient of
rolling resistance
Rotation
of log
Direction of motion of center
of log relative to ground
A
Direction of
motion of
center of log
relative to
stone
C
0.08 ft
Rg
Coefficient of
rolling resistance
3 As the log rolls on the
ground, the stone exerts a
force opposing the rolling
motion.
7.7 Rolling Resistance Example 6, page 3 of 5
4 For use in calculating moments, find the vertical distances from
the center to the points C and B where the resisting forces act.
0.01 ft
B
(0.5 ft)2
(0.01 ft)2 = 0.4999 ft
0.5 ft
Radius = (1 ft)/2 = 0.5 ft
A
0.5 ft
(0.5 ft)2
(0.08 ft)2
C
0.08 ft
= 0.4936 ft
7.7 Rolling Resistance Example 6, page 4 of 5
Free-body diagram of stone and logs
100,000 lb
F
Rg-x
Rg-y
The same resisting force acts on each of the ten logs.
Equilibrium equations
+
+
5
Fx = 0: F
10Rg-x = 0
Fy = 0: 10Rg-y
100,000 lb = 0
(1)
(2)
Two equations, three unknowns: an additional free-body
is needed.
7.7 Rolling Resistance Example 6, page 5 of 5
6
Free-body diagram of an individual wheel
Rs-x
Rs-y Force components
from the stone
B
0.4999 ft
Moment equilibrium
7
+
A
MB = 0: Rg-y(0.01 ft + 0.08 ft)
Rg-x(0.4999 ft + 0.4936 ft) = 0
(3)
Solving Eqs. 1-3 gives
0.4936 ft
C
Rg-x
0.01 ft
Rg-y
0.08 ft
Rg-x = 906 lb
Rg-y = 10,000 lb
F = 9,060 lb
Ans.
8 The force F required to move the 100,000-lb stone is large, even when the
ground is level. One scholar has estimated that as many as 600 men were
needed to move such a stone up one of the slopes lying between the quarry
and the monument site.