cables_uniform.pdf
TRANSCRIPT
6.6 Cables: Uniform Loads
6.6 Cables: Uniform Loads Example 1, page 1 of 4
A B
400 ft
3800 ft
1. Each cable of the center span of the suspension
bridge supports a uniform load of 10 kip/ft along
the horizontal. If the span is 3,800 ft and the sag
400 ft, determine the maximum and minimum
tensions in the cable.
6.6 Cables: Uniform Loads Example 1, page 2 of 4
The maximum tension occurs where the slope is
greatest points A and B.
The minimum tension occurs where the slope is least the
low point of the cable.
2
1
3800 ft
400 ft
BA
6.6 Cables: Uniform Loads Example 1, page 3 of 4
C
B
T By
T Bx
400 ft
T o
(1900 ft)/2 950 ft
(3800 ft)/2 = 1900 ft
(Because of symmetry, we know
that the low point occurs at the
middle of the 3800-ft span.)
C(Resultant acts through middle of
1900 ft span)
Resultant load
(10 kip/ft)(1900 ft)
= 19,000 kip
Tension at low point
(minimum tension)
Tension components at B5
7
8
64
3 Free-body diagram of CB
3800 ft
400 ft
BA
6.6 Cables: Uniform Loads Example 1, page 4 of 4
T max
T By
19,000 kip
B
Equilibrium equation
Fx = 0: T + TBx = 0
Fy = 0: TBy
19,000 kip = 0
MB = 0: (19,000 kip)(950 ft) T (400 ft) = 0
Solving gives
T = 45,125 kip (minimum tension in cable) Ans.
TBx = 45,125 kip
TBy
= 19,000 kip
9
The maximum tension is the resultant of TBx and TBy
:
TBx = 45,125 kip
+
++
T max = (45,125 kip)2 + (19,000 kip)2
= 49,000 kip Ans.
6.6 Cables: Uniform Loads Example 2, page 1 of 4
A
B
C
8 ft
h
50 ft30 ft
2. A length of oil pipeline weighing 3200 lb is supported by a system
of cables as shown. Determine a) the distance h to the lowest point C
on the cable and b) the maximum tension in the cable.
6.6 Cables: Uniform Loads Example 2, page 2 of 4
B x
B y
A y
A x
A
B
C
3200 lb
(80 ft)/2 40 ft
(Weight of pipeline acts through
midpoint of 30 ft + 50 ft 80-ft span,
not through low point, C)
8 ft
2
40 ft
The maximum tension occurs at B, where the
slope is largest, so let's draw a free-body
diagram of the entire system, including the
tension components at B:
1 Equilibrium equation
MA = 0: By(40 ft + 40 ft)
Bx(8 ft) (3,200 lb)(40 ft) = 0 (1)
We could write two additional equilibrium
equation, but they would introduce two
additional unknowns, Ax and Ay, so nothing
would be gained. An additional free-body
diagram involving Bx and By is needed.
3
4
+
6.6 Cables: Uniform Loads Example 2, page 3 of 4
50 ft
8
7 Free-body diagram of BCPass a vertical section through the low point, C, and
draw a free-body diagram of the portion of the
system to the right of C.
5
The 3200-lb weight spread over an 80-ft length
of pipe is equivalent to a uniformly distributed
load of 3200 lb/80 ft = 40 lb/ft.
6+
++
9 Equilibrium equation for free-body BC
Fx = 0: T + Bx = 0 (2)
Fy = 0: By 2000 lb = 0 (3)
MB = 0: (2000 lb)(25 ft) T (h + 8 ft) = 0 (4)
A
B
C
80 ft
B x
B y
A y
A x
50 ft
B xB
C
B y
h + 8 ft
T o
(50 ft)/2
25 ft
Weight distributed load length of pipe
40 lb/ft 50 ft
2000 lb
(T o is horizontal
because C is
the minimum
point on the
curve.)
h
8 ft
6.6 Cables: Uniform Loads Example 2, page 4 of 4
BB
x 4000 lb
B y 2000 lb T
max
Eq. 4 is nonlinear, but Eqs. 1-3 are linear and can be easily
solved to give
Bx = 4000 lb
By = 2000 lb
T = 4000 lb
Using the value of T = 4000 lb in Eq. 4 gives
(2000 lb)(25 ft) T (h + 8 ft) = 0 (Eq. 4 repeated)
Solving gives
h = 4.5 ft Ans.
The maximum tension is the resultant of Bx and By:11
10
Tmax = (4000 lb)2 + (2000 lb)2
= 4470 lb Ans.
6.6 Cables: Uniform Loads Example 3, page 1 of 2
A B
The towel near the middle of the clothesline span will touch the
ground if the sag is 75 in. 72 in. = 3 in. We can now work the
problem either of two ways: 1) assume the sag = 3 in., compute
the corresponding maximum tension, and compare it to the
280-lb breaking strength; or 2) assume the maximum tension is
280 lb, compute the corresponding sag, and compare it to the
maximum permissible sag of 3 in.
3. An 18-lb washer load of wet beach-towels is hung on a clothesline
to dry. Each beach towel is 72-in. long, and the clothesline is tied to
supports located 75 in. above the ground. If the breaking strength of
the clothesline is 280 lb, determine if the clothesline can be tightened
enough to keep the middle towel from touching the ground.
1
75 in75 in
28 ft
6.6 Cables: Uniform Loads Example 3, page 2 of 2
(28 ft)/4 7 ft(28 ft)/4 7 ft
B x 252 lbB
T max
B y 9 lb
B y
B x
T o
3 in
(18 lb)/2
9 lb
Let's arbitrarily choose the first approach assume a sag
= 3 in. Then a free-body diagram of the right half of the
clothesline and hanging towels would appear like this:
Equilibrium equations
Fx = 0: T + Bx = 0 (1)
Fy = 0: 9 lb + By = 0 (2)
MB = 0: 9 lb(7 ft)
T (3 in. 1 ft/ 12 in.) = 0 (3)
Solving Eq. 1-3 gives
Bx = 252 lb
By = 9 lb
T = 252 lb
The maximum tension occurs where the slope is
greatest at B (or A):
= (252 lb)2 + (9 lb)2 = 252.2 lb
2
3
4
+
++
Since the maximum tension, 252.2 lb, is less than
the breaking strength, 280 lb, the clothesline can
be tightened enough to keep the towel from
touching the ground. Note, however, that the
poles supporting the clothesline must be
well-anchored in the ground since they must
resist a horizontal force of Bx = 252 lb acting at
the point where the clothesline is attached.
B
6.6 Cables: Uniform Loads Example 4, page 1 of 4
A
0.5 kg/m0.5 kg/m
3 m s
CB
80 m 60 m
4. Chains AB and BC are attached to a roller support at B
as shown. The chains support beams that have mass per
length of 0.5 kg/m. Determine the maximum tension in
each chain and the sag s in chain BC.
1 The maximum tension in chain AB occurs where the
slope is greatest, points A and B.
6.6 Cables: Uniform Loads Example 4, page 2 of 4
We know the location of the low point of chain AB, so
let's choose a free-body diagram of the right half of
the chain because then we will know the distances
appearing in the moment equation.
2 Equilibrium equation
Fx = 0: T + Bx = 0 (1)
Fy = 0: By 196.2 N = 0 (2)
MB = 0: T (3 m)
+ (196.2 N)(20 m) = 0 (3)
5
Solving Eqs. 1-3 gives
T = 1308 N
Bx = 1308 N
By = 196.2 N
6
+
++
B
3 m
B y
B x
T o
(40 m)/2 20 m
Resultant force
(0.5 kg/m)(9.81 m/s2 )(40 m)
196.2 N
By symmetry, we know the low point occurs at mid span:
(80 m)/2 40 m
4
3 Sag distance is given:
6.6 Cables: Uniform Loads Example 4, page 3 of 4
s
(60 m)/2 30 m
Resultant weight
(0.5 kg/m)(9.81 m/s2 )(30 m)
147.15 N
(30 m)/2 15 m
T'o
B'x
B'y
T max
B x 1308 N
B y 196.2 N
B
Maximum tension in AB
Next consider a free-body diagram of the
left half of chain BC:
Equilibrium equation
Fx = 0: Bx + T = 0 (4)
Fy = 0: By 147.15 N = 0 (5)
MB = 0: T (s) (147.15 N)(15 m) = 0 (6)
+
++
= (1308 N)2 + (196.2 N)2 = 1323 N Ans.
7
8 9
B
6.6 Cables: Uniform Loads Example 4, page 4 of 4
B
147.15 N196.2 N
T o 1308 N T'
o
F B, vertical force
acting on roller support
Eqs. 4-6 are three equations in four unknowns, Bx ,
By , T and s. An additional free-body is needed.
The free-body diagram below shows that no
horizontal force acts on the base of the roller support at B.
10
11
T max
B'x 1308 N
B'y 147.15 N
Equilibrium equation
Fx = 0: 1308 N + T = 0 (7)
Solving Eqs. 4-7 simultaneously gives
T = 1308 N
s = 1.688 m Ans.
Bx = 1308 N
By = 147.15 N
12
+
Maximum tension in BC13
Tmax = (1308 N)2 + (147.15 N)2
= 1316 N Ans.
6.6 Cables: Uniform Loads Example 5, page 1 of 5
40°
A
0.2 kg/m
B
3 m
2 m
5. Cable AB supports a uniformly distributed mass of
0.2 kg/m. The slope of the cable at B is known to be
40°. Determine the maximum tension in the cable and
the length of the cable.
The maximum tension occurs at B, where the slope of
the cable is the greatest.
1
6.6 Cables: Uniform Loads Example 5, page 2 of 5
40°
T max cos 40°
T max sin 40°
2 mA
y
A x
A Weight = (0.2 kg/m)(9.81 m/s2 )(3 m)
= 5.886 N
(3 m)/2 1.5 m
B
3 m
T max
Free-body diagram2
Equilibrium equation
MA = 0: (Tmax sin 40°)(3 m) (Tmax cos 40°)(2 m)
(5.886 N)(1.5 m) = 0 (1)
Solving gives
Tmax = 22.28 N Ans.
+
To compute the length of the cable, we need
to use the equation of the cable curve,
y = (2) wx2
2T
3
4
6.6 Cables: Uniform Loads Example 5, page 3 of 5
Here, w is the distributed load per horizontal meter,
w = (0.2 kg/m)(9.81 m/s2)
= 1.962 N/m (3)
The quantity T is the horizontal component of the cable tension.
Since T is the same at all points, we can evaluate it at support B:
T = Tmax cos 40°
= (22.28 N) cos 40° (4)
Substituting Eqs. 3 and 4 in Eq. 2 gives
y = (Eq. 2 repeated)
or,
y = 0.057478x2 (5)
2T
wx2
1.962 N/m by Eq. 3
(22.28 N) cos 40° by Eq. 4
We will also need the slope,
= 2(0.057478x)
= 0.114956x (6)
dx
dy
5
6.6 Cables: Uniform Loads Example 5, page 4 of 5
A(x A, y
A)
B(x B, y
B)
y
x
40°
dy40°
dx
dy
dx
ds
3 m
y 0.05784 x2
The length of the cable is
sAB = ds
= (dx)2 + (dy)2
= 1 + ( )2 dx
A
B
dy
dx
B
A
0.114956x by Eq. 6
(Change the variable of
integration from s to x)6
dy
dx
Also at B, the slope is known:
= tan 40° (9)
xBxA
xA
xB
Thus the length of the cable can be expressed as
sAB = 1 + (0.114956x)2 dx (7)
To evaluate this integral, we have to find the values of xA
and xB. From the figure, we see that
xA = xB 3 m (8)
7
6.6 Cables: Uniform Loads Example 5, page 5 of 5
Using Eq. 6 to evaluate the left-hand side of Eq. 9 gives
( )B = tan 40° (Eq. 9 repeated)
0.114956xB, by Eq. 6
Solving for xB gives
xB = (tan 40°)/0.114956
= 7.299311 m
and using this result in Eq. 8 gives
xA = xB 3 m
= 7.299311 3
= 4.299311 m
dy
dx
This integral is best evaluated numerically with the
integral function of a calculator.
The result is
sAB = 3.61 m Ans.
8
9
7.299311
4.299311
By Eq. 7, then, the length of the cable is
sAB = 1+ (0.114956x)2 dx
6.6 Cables: Uniform Loads Example 6, page 1 of 5
C
4 ftB
A
48 ft
w lb/ft
6 ft
6. Determine the largest uniform load, w lb/ft, that the cable
can support if it will fail at a tension of 3,000 lb. Also
determine the location of the low point C of the cable.
6.6 Cables: Uniform Loads Example 6, page 2 of 5
+
2 Moment equilibrium equation
MB = 0: (48w)(24 ft) (3000 lb)(sin )(48 ft)
+ (3000 lb)(cos )(4 ft) = 0 (1)
C
B
A
B x
B y
(3000 lb) sin
(3000 lb) cos
Resultant load
w lb/ft 48 ft
(48w) lb
48 ft
24 ft(48 ft)/2 24 ft
4 ft
6 ft
The maximum tension in the cable occurs where the
slope is greatest, point A. The cable will fail if the
tension there exceeds 3,000 lb. A free-body diagram of
the entire system, with a maximum tension of 3,000 lb at
A, would appear as below:
1
The line of action of the weight of the cable does
not pass through the low point C, because the ends
of the cable are at different elevations.
As part of our solution
to the problem, we will
calculate .
6.6 Cables: Uniform Loads Example 6, page 3 of 5
C
B
A
B x
B y
(3000 lb) sin
(3000 lb) cos 4 ft
6 ft
d
(Unknown distance)
We could write two additional equilibrium equation, but they
would introduce two additional unknowns, Bx and By, so there
would be no advantage gained. Instead, we need another
free-body diagram. To get it, consider a free body of the entire
cable, and pass a vertical section through the low point C.
3
6.6 Cables: Uniform Loads Example 6, page 4 of 5
C
A
(3000 lb) sin
(3000 lb) cos
Resultant
w d
4 ft + 6 ft 10 ft
T o
d/2 d/2
d
Free-body diagram of portion to left of section4
5
++
Equation of equilibrium
Fy = 0: (3000 lb)(sin ) wd = 0 (2)
MC = 0: (3000 lb)(cos )(10 ft)
(3000 lb)(sin )d
+ (wd)(d/2) = 0 (3)
We could write a third equilibrium equation, but it
would introduce an additional unknown, T so no
advantage would be gained.
Eqs. 1-3 are three nonlinear equations in three unknowns,
, w, and d. These equations are best solved with a
calculator capable of solving simultaneous nonlinear
equations. Alternatively, proceed as follows. First note
that Eq. 2,
3000 sin wd = 0 (Eq. 2 repeated)
can be solved for d:
d = (3000 sin )/w (4)
This equation can be used to eliminate d from Eq. 3:
3000 cos (10) (3000 sin )d
+ wd2/2 = 0 (Eq. 3 repeated)
(3000 sin )/w 2 (3000 sin )/w
Multiplying through by w and combining terms gives
(30,000 cos )w (3,000 sin )2/2 = 0 (5)
6
6.6 Cables: Uniform Loads Example 6, page 5 of 5
Next note that Eq.1,
(48)(24)w (3,000 sin )(48)
+ (3,000 cos )(4) = 0 (Eq. 1 repeated)
can be solved for w:
w = 125 sin 10.4167 cos (6)
and this equation can be used to eliminate w from Eq. 5:
(30,000 cos ) w (3,000 sin )2/2 = 0 (Eq. 5 repeated)
125 sin 10.4167 cos
Carrying out the multiplication gives
(3.75 106) cos sin (0.312501 106) cos2
(4.5 106) sin2 = 0 (7)
Dividing both sides by 106 gives
(3.75) cos sin (0.312501) cos2
(4.5) sin2 = 0 (8)
Solving this equation by trial and error gives
= 36.48°
Using this value for in Eq. 6 then gives w:
w = 125 sin 10.4167 cos (Eq. 6 repeated)
= 125 sin 36.48° 10.4167 cos 36.48°
= 65.94 lb/ft Ans.
Eq. 4 then gives distance d:
d = (Eq. 4 repeated)
=
= 27.0 ft Ans.
w3000 sin
3000 sin 36.48°
65.94
7 8
6.6 Cables: Uniform Loads Example 7, page 1 of 4
The 100-kg mass produces a cable
tension of 100 kg 9.81 m/s2 = 981 N
at point A.
1
7. The cable system shown supports a uniformly
distributed mass of 5 kg/m along the horizontal.
Determine the tension at B and the length of portion AB
of the cable. Assume that the pulleys are frictionless.
4 m
20 m
B
A
5 kg/m
100 kg
6.6 Cables: Uniform Loads Example 7, page 2 of 4
B B x
B y
(981 N) sin
(981 N) cos
Resultant weight of cable between A and B
(20 m)(9.81 m/s2 )(5 kg/m)
981 N
20 m
10 m
A
981 N
4 m
2 Free-body diagram
Equilibrium equation
Fx = 0: (981 N) cos + Bx = 0 (1)
Fy = 0: (981 N) sin + By 981 N = 0 (2)
MB = 0: (981 N) cos (4 m)
(981 N) sin (20 m)
+ (981 N)(10 m) = 0 (3)
+
++
Eqs. 1-3 are best solved with a calculator that solves
simultaneous nonlinear equations. Alternatively, Eq. 3
involves only one unknown, , and can be solved by
trial-and-error to yield
= 40.67° (4)
Using this value in Eq. 1 and 2 then leads to
Bx = 744.1 N (5)
By = 341.7 N (6)
3
4
6.6 Cables: Uniform Loads Example 7, page 3 of 4
The tension at B is then
Bx2 + By
2 = (744.1 N)2 + (341.7 N)2
= 818.8 N Ans.
To compute the length of the cable, we need to use the
equation of the cable curve,
2T
wx2
y = (7)
Here, w is the distributed load per horizontal meter,
w = (5 kg/m)(9.81 m/s2)
= 49.05 N/m (8)
The quantity T is the horizontal component of the cable
tension. Since T is the same at all points, we can evaluate it
at support B:
T = Bx
744.1 N by Eq. 5 (9)
5
6
Substituting Eqs. 8 and 9 in Eq. 7 gives,
y = (Eq. 7 repeated)
or,
y = 0.03296 x2 (10)
We will need the equation for the slope:
= 2(0.3296 x) = 0.06592 x (11)
wx2
2T
49.05 N/m
744.1 N
dx
dy
The length of the cable is
sAB = ds
= (dx)2 + (dy)2
or
sAB = 1 + ( )2 dx (12)
A
B
(0.06592 x)2, by Eq. 11
ds
dxdy
(Change the
variable of
integration
from s to x)
dx
dy
7
8
B
A
xA
xB
6.6 Cables: Uniform Loads Example 7, page 4 of 4
x
y
B(x B, y
B)
A(x A, y
A)
20 m
y 0.03296x2
40.67°
dx
dy
40.67°
To evaluate the integral for sAB, we have to find the values
of xA and xB.
From the figure, we see that
xB = xA + 20 m (13)
Also at A, the slope is known, so
= tan 40.67°
Using Eq. 11 to evaluate the left hand-side of this equation gives
9
10
dy
dx
Solving for xA gives
xA = 13.03436 m (14)
Using this result in Eq. 13 gives
xB = xA + 20 m (Eq. 13 repeated)
= 13.03436 m + 20 m
= 6.96564 m (15)
By Eq. 12, then, the length of cable is
sAB = 1 + (0.06592 x)2 dx
This integral is best evaluated numerically with the
integral function of a calculator. The result is
sAB = 21.7 m Ans.
6.96564
-13.03436
dy
dx( )A = tan 40.67°
0.06592xA, by Eq. 11
11
6.6 Cables: Uniform Loads Example 8, page 1 of 6
22 kg/m
16 m
A
B
Cx
y
d A
d B
30 m
8. The chain AB supports a horizontal, uniform beam of
mass per length 22 kg/m. If the maximum allowable
tension in the chain is 7 kN, determine distances dA and
dB of the supports above the low point C of the chain.
Also determine the length of the chain.
The maximum tension occurs at support A,
where the slope of the chain is greatest.
1
6.6 Cables: Uniform Loads Example 8, page 2 of 6
A
B B x
B y
A x
A y
d A d
B
(46 m)/2 23 m
30 m + 16 m 46 m
Resultant weight
(22 kg/m)(9.81 m/s2 )(46 m)
9928 N
9.928 kN
3
C
Free-body diagram of entire system.
Equation of equilibrium
MB = 0: Ax(dA dB) Ay(46 m) + (9.928 kN)(23 m) = 0 (1)
We could write two additional equations of
equilibrium, but they would contain two additional
unknowns, Bx and By, so nothing would be gained.
4
2
+
6.6 Cables: Uniform Loads Example 8, page 3 of 6
22 kg/m
A
B
C
B x
B y
A y
A x
C
A
A y
T o
d A
30 m
(30 m)/2 15 m
Resultant weight
(22 kg/m)(9.81 m/s2 )(30 m)
6475 N
6.475 kN
d A
30 m
To obtain an additional free-body diagram, pass a
vertical section through the low point of the chain.
A x
5 Free-body diagram of portion of chain to left of section.6
6.6 Cables: Uniform Loads Example 8, page 4 of 6
Equation of equilibrium
Fx = 0: Ax + T = 0 (2)
Fy = 0: Ay 6.475 kN = 0 (3)
MC = 0: Ax(dA) Ay(30 m) + 6.475 kN)(15 m) = 0 (4)
Equation 3 gives
Ay = 6.475 kN (5)
Since we know that the maximum tension of 7 kN occurs
at A and is the resultant of Ax and Ay, we have
Ax2 + Ay
2 = 7 kN
Substituting Ay = 6.475 kN and solving for Ax gives
Ax = 2.660 kN (6)
Substituting for Ax and Ay in Eqs. 2 and 4 gives
Ax + T = 0 (Eq. 2 repeated)
7
+
++
2.660 kN
Ax dA Ay(30) + (6.475)(15) = 0 (Eq. 4 repeated)
2.660 kN 6.475 kN
Solving gives
T = 2.660 kN (7)
dA = 36.52 m (8) Ans.
Distance dB can now be found by substitution in Eq. 1:
Ax (dA dB) Ay(46) + (9.928)(23) = 0
2.660 kN 36.52 m 6.475 kN
by Eq. 6 by Eq. 8 by Eq. 5
Solving gives
dB = 10.4 m Ans.
8
6.6 Cables: Uniform Loads Example 8, page 5 of 6
To compute the length of the chain, we need to use the
equation of the chain curve:
y = (9)
Here, w is the distributed load per horizontal meter,
w = (22 kg/m)(9.81 m/s2)
= 215.8 N/m
= 0.2158 kN/m (10)
The quantity T has been given in Eq. 7
T = 2.660 kN (Eq. 7 repeated)
Substituting into Eq. 9 gives
0.2158 kN/m by Eq. 10
y = (Eq. 9 repeated)
2.660 kN by Eq. 7
or,
y = 0.040564 x2 (11)
9
2T
wx2
2T
wx2
We will need the equation for the slope,
= 2(0.040564 x) = 0.08113 x (12)dy
dx
(Change the variable of
integration from s to x)
dydx
dsB
A
The length of the chain is
sAB = ds
= (dx)2 + (dy)2
or
sAB = 1 + ( )2 dx (13)dy
dx
10
A
B
xA
xB
6.6 Cables: Uniform Loads Example 8, page 6 of 6
x
y 0.040564 x2
B(x B 16 m)
y
A(x A 30 m)
30 m 16 m
Since the location of the low point of the chain is
known, xA and xB are known:
11 The integral for the cable length, Eq. 13, can now be
written as
sAB = 1 + ( )2 dx (Eq. 13 repeated)
This integral is best evaluated numerically with the integral
function of a calculator. The result is
sAB = 69.2 m Ans.
(0.08113x)2, by Eq. 12
-30
16 dy
dx
12
6.6 Cables: Uniform Loads Example 9, page 1 of 4
CA B
10 kg
h
d
9. A 40-m length of rope has a uniformly distributed
mass of 0.1 kg/m and has one end fixed and the other
end attached to a cart as shown. Determine the distance
d and the sag h when the cable and cart are in
equilibrium under the force supplied by the 10-kg load.
The total weight of the rope is 40 m 0.1 kg/m 9.81 m/s2 =
39.24 N, and the force of the cart acting horizontally on the
rope is 10 kg 9.81 m/s2 = 98.1 N, or more than twice as much.
Thus it seems reasonable to assume that the sag, h, is small
compared to the span d, h d, and the weight of the rope is
well-approximated as a uniformly distributed load along the
horizontal of
w = (0.1 kg/m) (9.81 m/s2) = 0.981 N/m (1)
At the end of the problem, we can check the reasonableness of
these assumptions.
1
6.6 Cables: Uniform Loads Example 9, page 2 of 4
B x
Vertical force from
ground acting on cart
B y
B y
C x
C y
B x T
o
(Horizontal component
of tension)
(10 kg) (9.81 m/s2 )
98.1 N
CB
The equation describing the shape of the rope is
y = (2)
T is the horizontal component of tension in the rope
and can be found by considering free bodies of the
cart and rope.
wx2
2T
2
Free-body diagram of cart3
Free-body diagram of rope5
The equation of the rope is now
0.981 N/m by Eq. 1
y = (Eq. 2 repeated)
98.1 N by Eq. 3
or,
y = 0.005 x2 (4)
We will also need the equation for the slope,
= 2(0.005 x) = 0.01x (5)
2T
wx2
+4
6
Fx = 0: 98.1 N + Bx = 0
Solving gives
Bx = 98.1 N = T (3)
dy
dx
6.6 Cables: Uniform Loads Example 9, page 3 of 4
0.005d + 1 + (0.005d)2
0.005d + 1 + (0.005d)2
d/2d/2
CB
hx
C d/2x B d/2
x
dxdy
C
B
B
C
ds
dxdy
The equation for the length of the slope is
sBC = ds
= (dx)2 + (dy)2
= 1 + ( )2 dx (6)
xC and xB can be expressed in terms of the unknown, d:
7
The best way to solve for d in Eq. 7 is to use the solver on a
calculator and to use the calculator integral function to input
Eq. 7 in the solver. Alternatively, use a table of integrals to
evaluate the integral in Eq. 7 to get
40 = 1 + (0.005d)2
+ (50) ln
Solving this equation by trial-and-error gives
d = 39.74 m Ans. (8)
d
2
xB
xC
8
y
d/2
-d/2
-d/2
d/2
dx
dy
Thus Eq. 6 can be written
sBC = 1 + ( )2 dx (Eq. 6 repeated)
Rope length = 40 m (0.01 x) by Eq. 5
or,
40 = 1 + (0.01x)2 dx (7)
6.6 Cables: Uniform Loads Example 9, page 4 of 4
hB C
y
x
The sag, h, can now be found from Eq. 4:
y = 0.005x2 (Eq. 4 repeated)
h = by Eq. 839.74
2
The result is
h = 1.974 m Ans.
Note that the sag is much smaller than the span,
h = 1.974 m << 39.74 m = d, as we assumed in
the beginning of the problem.
9
d2