4.3 CablesDescription of the space shuttle tether

When a structural member’s dimension is large compared to its other two dimensions, the structural member is classified as either a beam or as a cable. Thread, rope, wire, string, and chain are classified as cables.

The difference between a cable and a beam is that a cable can’t resist bending. The shear force and the bending moment in a cable are negligibly small. Also, the axial force can only act in tension. Given the comparatively simple internal structure of a cable, one might expect the analysis of a cable to be simpler than the analysis of a beam. However, the simple internal structure of a cable makes it very flexible and that flexibility leads to appreciable deflections. Compare that to a beam. In the last section, the beam’s deflection was so small that it didn’t affect the beam analysis. In a cable, the situation is different. The tension in the cable affects the shape of the cable and the shape of the cable becomes another quantity that needs to be determined.

This section begins by looking at cables acted on by point loads. Then, cables acted on by line loads are considered. The cable problems involving line loads are divided into two types – cable problems in which the line load is much larger that the weight of the cable and cable problems in which the weight load dominates.

Point Loading in Light Cables

When a cable is acted on by point loads that are large compared to the cable’s weight, the cable’s weight can be neglected. In these situations, the cable’s shape becomes a succession of straight lines, as shown in Fig. 4 – 63. Let’s look at one of those segments.

Let n point loads act on a cable. The free-body diagram of the r-th point is shown in Fig. 4 – 63 where r is any number between 1 and n. Summing forces in the x and y directions yields

(4 – 61)

Equations (4 – 61) are 2n equations. These are the equations that need to be solved to analyze the cable. The equations can be solved if there are 2n unknowns. The tensions T1, … , Tn+1 can represent n + 1 unknowns, so these equations can be solved if there are n – 1 additional unknowns. Depending on the problem, the additional unknowns in Eq. (4 – 61) can be any combination of the angles 1, 2, … , n+1 and the loads Qx1, Qy1, Qx2, Qy2, … , Qxn, Qyn. When all of the angles are known, Eq. (4 – 61) represents a set of linear algebraic equations. The solution of linear algebraic equations can be solved relatively easily. If there aren’t too many equations, they can be solved by hand. However, when some of the unknowns are angles, Eq. (4 – 61) represents a set on nonlinear algebraic equations which can’t be solved by hand except in simple cases. Equation (4 – 61), when some of the angles are unknowns, is generally solved by numerical methods using a computer. The examples at the end of this section illustrate these ideas.

Figure 4 – 62: A space shuttle tether Zoe

DK0

Figure 4 – 63: Point loads large compared to the cable’s weight act on a cable.

Line Loading in Light Cables

Figure 4 – 64 shows a cable that is acted on by a vertical line load q. Notice that q is positive downward and that it’s a known function of x. This situation occurs when a cable suspends large objects like roads and gas and water lines. The weight of the cable is small compared to the load it carries. A section of the cable of length dx is shown in Fig. 4 – 65. The axial force in the cable is called the tension force and it’s denoted by T. Notice that the tension force in the cable increases by an amount dT when incrementing by dx. Summing forces in the x and y directions yields

(4 – 62)

Let’s now manipulate these equations a little, for reasons that will soon be evident. In Eq. (4 – 62) the trigonometric identities for the cosine of the sum of two angles and of the sine of the sum of two angles can be used together with the small angle approximations and to take the d out from under the cosine and sine functions. Specifically,

(4 – 63)

Substitute Eq. (4 – 63) into (4 – 62) to get

(4 – 64)

Equation (4 – 64) can also be rewritten as

(4 – 65a,b)

Equation (4 – 65a) asserts that the horizontal component of the tension force is constant. Equation (4 – 65b) asserts that the vertical component of the tension force changes by qdx. Actually, both of these results could have been discovered directly from Fig. 4 – 65.

Next, denote the horizontal component of the tension force by the constant C = Tcos From Eq. (4 – 65a)

(4 – 66)

Substituting Eq.(4 – 66) into (4 – 65b) and dividing by Cdx yields

(4 – 67)

2

Figure 4 – 64: A known line load q(x) acts on a cable.

Figure 4 – 65: An infinitesimal section of the cable

Figure 4 – 66: The origin of the coordinate system is placed at the lowest point on the cable.

Equation (4 – 67) is the differential equation that governs the relationship between the position y of the cable and the line load q. It’s used to find the tension in the cable, the position of the cable, and the length of the cable.

For simplicity, the origin of the coordinate system will be placed at the lowest

point of the cable, as shown in Fig. 4 – 66. Notice that at the origin

of the coordinate system. Later, the location D of the lowest point of the cable will also need to be determined.

Cable Tension

Integrating Eq. (4 – 67) between x1 = 0 and x1 = x yields the slope of the cable

(4 – 68)

The slope is related to the angle by and the tension T is related to the

angle by C = Tcos. The angle can be eliminated from both of these equations to get T = where the trigonometric identity was used. Now, from Eq. (4 – 66) the tension in the cable becomes

(4 – 69)

Cable Position

The position y(x) of the cable is now found. Integrating Eq. (4 – 68) between x1 = 0 and x2 = x yields

(4 – 70)

Cable Length

The length of the cable is also a quantity of interest. In Fig. 4 – 67 the length of a differential element of a cable is denoted by ds. Observe from the free body

diagram and the differential element that Integrating this with

respect to x yields

(4 – 71)

When the loading q in a cable problem is known, Eq. (4 – 69) is first used to find the tension T(x) in the cable. Then T(x) is substituted into Eq. (4 – 71) to determine the length s in the cable.

3

Figure 4 – 67: A differential element and a free body diagram of the cable section between 0 and x.

The Parabolic Cable

A simple and important case arises when the line load q acting on the cable is constant. When this happens, the position of the cable becomes a parabola and closed-form expressions for the tension in the cable and the length of the cable can be found. From Eq. (4 – 69) the tension in the cable and the cable position are

(4 – 72a,b) .

The length of the parabolic cable is found from Eq. (4 – 71)

so from Eq. (4 – 72)

(4 – 73)

Remember that the length of the cable given above is the length between 0 and x.

The Constants C and D

The horizontal component of the tension force is a constant of integration. It appears in the expressions given above of the tension, the position, and the length of a cable. In order to find the tension, the position, or the length, the constant of integration C needs to be found first. The location D of the lowest point on the cable also needs to be determined before the tension, the position, or the length of a cable can be determined (See Fig. 4 – 66).

In a non-uniform light cable, the constants C and D must be determined by numerical methods. However, when the light cable is uniform (the parabolic cable), the constants C and D can be found analytically. Toward this end, look at Fig. 4 – 66. From Eq. (4 – 72b)

(4 – 74a,b) .

Equations (4 – 74) are a set of two equations expressed in terms of the two unknowns C and D. After a little manipulation,

(4 – 75a,b)

4

Notice in Eq. (4 – 75a) when HA = HB that D = L/2, which states that the lowest point

of the cable is in the center of the cable, and from Eq. (4 – 75b) that where

1

Equations (4 – 75) can be used when the lowest point on the cable is between each end of the cable. In sufficiently taut cables, however, the lowest point of the cable can be to the left of the left end of the cable (when HA is smaller than HB) or to the right of the right end of the cable (when HA is larger than HB). In these cases it can be shown that C and D are given by

(4 – 76a,b)

Line Loading in Heavy Cables

In cables in which the weight load dominates, the line load q(x) is not known in advance. The reason that q(x) isn’t known is that the shape of the cable effects the distribution of the weight in the x direction. On the other hand, the distribution of the weight along the length of the cable is known. Mathematically, q is a known function of s instead of being a known function of x. Let’s see how this affects the governing equation found earlier, namely Eq. (4 – 67).

Refer to Fig. 4 – 68 showing a force dQ acting on a differential element of the

cable. The force per unit length along the cable is denoted by . The force per

unit length along the x direction is . The force per unit length w along the cable

and force per unit length q in the x direction are related by

and also notice from Fig. 4 - 68 that so

(4 – 77)

Equations (4 – 67) can be used for heavy cables if Eq. (4 – 77) is substituted into (4 – 67). This gives

(4 – 78)

1 The three bar sign means “is defined as.”

5

Figure 4 – 68: The weight load q is a known function of s.

Let’s now use Eq. (4 – 78) to find the tension, position, and length of a heavy cable. Shortly, it’ll be seen that this can be done analytically if the weight of the cable is uniform. To solve Eq. (4 – 78) analytically, the variables x and dy/dx are separated. Substitute z = dy/dx into Eq. (4 – 78) and place functions of z on one side of the equation and the functions of x on the other side, to get

(4 – 79)

Observe in Eq. (4 – 79) when w is a function of s that the variables in Eq. (4 – 79) are not actually separated. So, our attention is restricted below to heavy cables that are uniform. Integrating both sides of Eq. (4 – 79),

so

.

Solving for z,(4 – 80)

By letting u = wx/C in Eq. (4 – 80), Eq. (4 – 80) becomes .This function is called the hyperbolic sine function, written sinh(u). Similarly, the hyperbolic cosine function is defined as , and it’s written as cosh(u). The hyperbolic sine and cosine functions have properties that are similar to the sine and cosine functions. For example, you can verify from their definitions that2

(4 – 81a,b,c)

Equation (4 – 80) is now rewritten as

(4 – 82)

Cable Tension Recall from the developments just before Eq. (4 – 69) that , so from Eq. (4 – 82) and Eq. (4 – 81a)

(4 – 83)

Cable Position

The position y(x) of the cable is found be integrating Eq. (4 – 82) between x1 = 0 and x2 = x using Eq. (4 – 81c) to get

(4 – 84)

This shape is called a catenary.

2 The terms hyperbolic sine and hyperbolic cosine are used because the properties of these functions are similar to the properties of the sine and cosine functions and because Eq. (4 – 81a) is the equation of a hyperbola if you let x = coshu and y = sinhu.

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Hyperbolic Sine Function, Hyperbolic Cosine Function

Catenary

Cable Length

See in Fig. 4 – 69 that dy/dx = tan = ws/C. Substituting this into Eq. (4 – 82) and solving for s,

(4 – 85)

The Constants C and D

Like in the case of a light cable, the horizontal component C of the tension force and the position D of the lowest point on a heavy cable need to be found before finding the tension, the position, and the length. Of course, when the height of the ends of the cable are the same, D = L/2, in which case it remains to find C. In the case of a heavy cable, C is found by numerical methods. The examples at the end of this section illustrate how to find C.

Taut Cables*

The shape of a taut cable is nearly straight. In Eq. (4 – 74), the force per unit length q in the x direction and the force per unit length w along the cable are related to each other by a factor that is nearly constant. This means that the analysis of a light cable and the analysis of a heavy cable are expected to give the same results when the cable is taut.

The uniform light cable is compared below with the uniform heavy cable. The heights of the ends of the cable are assumed to be the same on each end (See Fig. 4 – 70). To make the comparison general, quantities are non-dimensionalized. The following non-dimensional quantities are defined:

(4 – 86)

Equation (4 – 86) is substituted into Eqs. (4 – 72), (4 – 73), (4 – 83) and (4 – 85), to get

For light cables:

(4 – 87a)

For heavy cables:

(4 – 87b)

(4 – 87c)

Notice that Eq. (4 – 87a) gives closed-form non-dimensional expressions of the tension Tx and the length s0 of the cable. But, for heavy cables, Eq. (4 – 87b) needs to be solved

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Figure 4 – 69: The slope at s is dy/dx = w/C

Non-dimensionalization

Figure 4 – 70: Light cables versus heavy cables

numerically. Once Eq. (4 – 87b) is solved numerically, Cx is substituted into Eqs. (4 – 87c) and Ts and s0 are determined.

The results are displayed in Fig. 4 – 71. The constant Cx is graphed as a function of = H/L for both the light cable and the heavy cable. Notice that they’re indistinguishable when H/L is small. But as H/L increases, they diverge. The difference is below 10% until H/L = 0.3. Next, look at the graph of the non-dimensional length of the cable s0 for the light cable and the heavy cable. They’re practically identical over the full range of H/L. Finally, look at the graph of Tx for the light cable and Ts for the heavy cable. They’re practically identical over the full range of the H/L, too. Both Tx

for light cables and Ts for heavy cables approach 0.5 as H/L gets larger. To summarize, the light cable analysis and the heavy cable analysis yield nearly identical results (differences of less than 10%) when H/L is under 0.3.

Key TermsCable, Catenary, Heavy Cable, Hyperbolic Cosine, Hyperbolic Sine, Light Cable, Non-dimensionalization, Parabola, Taut Cable

Review Questions

1. What is a cable? Give examples of structural members that are classified as cables. 2. Can the cable weight be neglected if it’s large compared to the loading?3. Assume that a cable is acted on by point loads. Are the governing equations linear or nonlinear when the cable angles are unknown? 4. In a light cable acted on by a vertical line load, which component of the tension is constant?5. Describe the conditions for which the shape of a cable is parabolic.6. Assume that a cable is acted on by a vertical line load. Describe the conditions for which the location of the lowest point on the cable can be determined analytically.7. Define the hyperbolic cosine function and the hyperbolic sine functions in terms of exponential functions. 8. Describe the conditions for which the shape of a cable is a catenary.

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Figure 4 – 71: Comparison between light cables (in one color) and heavy cables (in another color).

Examples

4 – 43 Point Loading in Light CablesCable ABCD is acted on by a 200 N force and an unknown force Q, as shown. Find the tension in the cable and the unknown force Q.

SolutionSet-up This is a problem in which a light cable is acted on by point loads. Transition/Equation The governing equations for this problem were already derived. They’re given in Eq. (4 – 61). In those equations(a) n = 2, Qx1 = –200sin30 = –100 N, Qy1 = –200cos30 = –173.21 = –173 N, Qx2 = 0, Qy2 = –Q, 60○, 30○, 60○. Substituting (a) into Eq. (4 – 61) yields

(b)

Answer Equations (b) are a set of 4 linear algebraic equations in terms of the 4 unknowns T1, T2, T3, and Q. The solution is (c) KnowledgeNotice that T1 and T3 both act 60○ relative to the horizontal axis yet T3 is three times larger than T1. The x component of the 200 N force causes T1 to decrease and it causes T3 to increase. What do you think would have happened had the 200 N force acted at –30○ instead of at +30○?

4 – 44 Point Loading in Light CablesCable ABCD is acted on by a 200 N force and a 100 N force, as shown. Find the tension in the cable and the unknown angle 3.

SolutionSet-up This is a problem in which a light cable is acted on by point loads. Transition/Equation The governing equations for this problem are given in Eq. (4 – 61). In those equations(a) n = 2, Qx1 = –200sin30 = –100 N, Qy1 = –200cos30 = –173.21 = –173 N, Qx2 = 0, Qy2 = –100 N, 60○, 30○

Substituting (a) into Eq. (4 – 61) yields

(b)

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Fig. 43a

TipNotice that all of the cable angles were known in advance. As a consequence, the governing equations were linear and could be solved by hand.

Fig. 44a

Answer Equations (b) are a set of 4 nonlinear algebraic equations in terms of the 4 unknowns T1, T2, T3, and 3. Even though the equations are nonlinear, they can be solved relatively easily. The solution is (c) KnowledgeIn this problem, a cable angle was unknown. As a consequence, the governing equations were nonlinear. Since only one angle was unknown, the equations could still be solved by hand. Also, compare this solution with the solution to Example 4 – 43. Notice that Q in Example 4 – 43 was larger than 100 N so Q pulled down on the cable more than the 100 N force in this example. As a consequence, the angle 3 in this example was less than 60○.

4 – 45 Line Loading in Light CablesTwo suspension cables that span 20 ft support an 80,000 lb repeating section of a road, as shown. Find the maximum tension in the cables and the compression loads in the top of the towers. Also, find the length of a cable. Assume that the load is uniformly distributed.

SolutionSet-up The cables are much lighter than the road so this is a problem in which a light cable is acted on by a line load. Also, the load is uniform, so the formulas for a uniform light cable can be used. Transition/Equation The formula for the tension is given in Eq. (4 – 72). But before using it, C needs to be found. Since this is a uniform cable, Eq. (4 – 75) applies. The line load is q = 40,000/20 = 2,000 lb/ft. AnswerFrom Eq. (4 – 75)

(a)

From Eq. (4 – 72)(b)

It follows from the equation C = Tcos that the tension in the cable is the greatest where the slope is the greatest. Therefore, the maximum tension occurs at the top of the towers. From Eq. (4 – 72) the angle of the cable at point A is found from tan = qL/2C = 2000·10/(2·15,000) = 4/3 so sin = 4/5. From Fig. 45b, the compression load at the top of the towers is

(c)

From Eq. (4 – 73), the length of a cable is (c)

KnowledgeThe tension in a cable acted on by vertical forces, whether the cable is light or heavy, is the greatest where the slope of the cable is the greatest. Also, Eq. (c) can be checked by comparing it with the length of line ACB, as shown in Fig. 45c. The length of line ACB is – which, as expected, is just shy of the 24.9 ft length of the cable.

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Fig. 45a

TipThe maximum tension occurs where the angle is the steepest.

Fig. 45c

Fig. 45b

4 – 46 Line Loading in Light CablesTwo suspension cables that span 25 ft support an 80,000 lb road, as shown. Find the location D of the lowest point on the cable, the maximum tension in the cable, and the length of each cable. Assume that the load is uniformly distributed.

SolutionSet-up This is a problem in which a light cable is acted on by a line load. The load is uniform, so the formulas for a uniform light cable can be used. Transition/Equation The formula for the tension in a cable is given in Eq. (4 – 72). Before using it, the constants C and D need to be determined. The line load is q = 40,000/25 = 1,600 lb/ft. AnswerFrom Eq. (4 – 75)

(a)

From Eq. (4 – 72)(b) The maximum tension occurs at the top of the right tower where the slope of the cable is the greatest. From Eq. (4 – 73), the length of a cable is

(c)

KnowledgeEquation (c) can be checked by comparing it with the length of line ACB, as shown in Fig. 46b. The length of line ACB is – which, as expected, is just short of the 37.9 ft length of the cable.

4 – 47 Line Loading in Light CablesTwo taut suspension cables that span 25 ft support an 80,000 lb section of road, as shown. Find the location D of the lowest point on the cable, the maximum tension in the cable, and the length of each cable. Assume that the load is uniformly distributed.

SolutionSet-up A light cable is acted on by a line load and the line load is uniform. Transition/Equation The formula for the tension in a cable is given in Eq. (4 – 72). Before using it, the constants C and D are determined from Eq. (4 – 76), which are the equations that apply to a taut cable. The line load is q = 40,000/25 = 1,600 lb/ft. AnswerFrom Eq. (4 – 76)

(a)

From Eq. (4 – 72)

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Fig. 46a

Fig. 46b

Fig. 47a

(b)

The maximum tension occurs at the point B where the slope of the cable is the greatest. From Eq. (4 – 73), the length of a cable is (c)

KnowledgeIn this and the previous example the height difference between the ends of the cables was 10 ft and the spans were the same, too. The most striking difference is associated with the levels of tension in the cables. The tension at point B in this example is

greater than the level of tension in Example 4 – 46. Also, notice that the length of line CAB in Fig. 47b is , which is just short of the 27.0 ft length of the cable.

4 – 48 Line Loading in Heavy CablesA 2 lb/ft cable spans 20 ft, as shown. Find the maximum tension in the cable and the cable’s length.

SolutionSet-up This is a heavy cable problem because the weight load acting on the cable dominates the problem. Transition/Equation The formula for the tension in a cable is given in Eq. (4 – 83) and the formula for the cable’s length is given by Eq. (4 – 85). Before using them, the constant C is determined numerically from Eq. (4 – 84). AnswerEvaluating Eq. (4 – 84) at x = 10 ft yields the equation

(a)

Equation (a) is a nonlinear algebraic equation expressed in terms of the unknown C. To solve this equation numerically, rewrite it as the function f(C) = 0 as follows:

(b)

Now, graph f(C) versus C. The solution is found where the function crosses the C axis (See Fig. 48b). The solution is C = 16.847 = 16.8 lb. Finally, from Eqs. (4 – 83) and (4 – 85)(c)

KnowledgeIt is insightful to compare this heavy cable with the light cable in Ex. 4 – 45. Although both problems are quite different, in both problems the spans of the cables are the same and the ends of the cables are 80 inches above the lowest points of the cables. Notice that the calculated lengths of the cables in both problems also turned out to be very

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Fig. 47b

Fig. 48a

Fig. 48b

TipA nonlinear algebraic equation can always be solved numerically by (a) writing it in the form of f(x) = 0, (b) graphing f(x) versus x, and (c) picking off the crossing(s) where f(x) = 0.

close to each other. In this problem s = 25.0 ft and in Ex. 4 – 45 s = 24.9 ft. Furthermore, the cables in these problems were not even taut. This comparison illustrates the similarity of the results obtained when a constant force per unit length in the x direction is assumed versus a constant force per unit length along the cable. Of course, the cable tensions in both problems were very different.

4 – 49 Line Loading in Heavy CablesA 10 lb/ft cable spans 80 ft, as shown. Find the maximum tension in the cable and the cable’s length.

SolutionSet-up This is a heavy cable problem because the weight load acting on the cable dominates the problem. Transition/Equation The formula for the tension in a cable is given in Eq. (4 – 83) and the formula for the cable’s length is given by Eq. (4 – 85). Before using them, the constants C and D need to be determined numerically. AnswerThis problem is complicated by the fact that the position D of the lowest point of the cable is not known. Evaluating Eq. (4 – 84) at x = D and x = L – D yields (a)

where HA = 30 ft, HB = 100 ft, and w = 10 lb/ft. Equations (a) are two nonlinear algebraic equations expressed in terms of the two unknowns C and D. They need to be solved numerically. Moreover, they’re coupled, meaning that both unknowns appear in both equations preventing the equations from being solved one at a time. To solve these equations numerically, rewrite the equations as fA(C, D) = 0 and fB(C, D) = 0 as:

(b)

Next, define the function (c)

Notice that the function f in (c) is zero only if fA and fB are both zero. So to solve (b) it’s sufficient to find where f = 0. The solution f = 0 is found here by the steepest descent method. The method proceeds as follows:Step (1): A first estimate of the solution (C and D) is calculated. To calculate a first estimate, let’s treat the cable as being light and use Eqs. (4 – 75). From Eq. (4 – 75)(d)

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Fig. 49a

Steepest Decent Method

Step (2): Next, a Taylor series expansion of (c) is performed to determine how changing C and D affects f in (c). When the changes are infinitesimal, it follows from (c) that

(e)

In the steepest decent method, the changes are not infinitesimal – they’re finite, although very small. So, (e) is rewritten as

(f)

where the subscript i refers to the i-th step in an iteration. The question arises how to

calculate and how to calculate Ci and Di to make f approach

zero.

Step (3): The partial derivatives are calculated numerically

using Eq. (c) as follows:

(g)

Step (4): The Ci and Di are given by

(h)

where h is a convergence parameter that controls the rate of convergence. The direction in (h) in which the unknowns C and D are being updated is called the direction of steepest descent. To see why (h) works, substitute (h) into (f) to get

(i)

Since f is positive (See Eq. (c)), Eq. (i) tells us that f is decreasing at the i-th step. Step (5): Steps 3 and 4 are repeated in a loop until the error f is sufficiently close to zero. In this example, the convergence parameter in (i) was set to

h =fi which theoretically drives the error to zero in one step.

The results are shown in Table 1. As shown, the converged solution is C = 196 lb and D = 49.0 ft. From Eq. (4 – 83) and (4 – 85), the maximum tension and the cable’s length are (j)

KnowledgeAs shown in Table 1, the error in the equation started out as being very large. After 150 steps the error was 10-4. Nonlinear algebraic equations are generally solved this way. A first estimate is made to place you near the answer. Then, a linear approximation is used to create an iteration that converges to the solution. In this example, the steepest decent method was used. In the appendix, the steepest decent method and the Newton

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Table 1

Step i C D f(error)1 134 51.7 4292210 159 44.6 219925 181 47.4 16050 191 48.3 18.7100 196 49.0 0.0794150 196 49.0 0.0001

method are described. They’re the two most popular methods of driving an error function to zero.

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Problems

Point Loading in Light Cables4 – 239 (L)A cable is acted on by a 100 N point load. Use Eq. (4 – 61) to find the tension in each end of the cable. Answer: TA = 100 N, TB = 173 N.

4 – 240 (L)A cable is acted on by a 100 N point load and an unknown load Q, as shown. Find the tension in the cable and Q. Answer: T1 = 125 lb, T2 = 75 lb, T3 = 100 lb, Q = 125 lb.

4 – 241 (L)A cable is acted on by a 50 N point load and an unknown load Q, as shown. Find the tension in the cable and Q. Answer: T1 = 50 N, T2 = 50 N, T3 = 43.3 N, Q = 25 N.

4 – 242 (L)A cable is acted on by a 50 lb weight and an unknown load Q, as shown. Find the tension in the cable and Q. Answer: T1 = 33.3 lb, T2 = 57.7 lb, T3 = 28.9 lb, Q = 66.7 lb.

4 – 243 (L)Two traffic lights and a street sign hang from a cable, as shown. The street sign weighs 25 lb and each of the traffic lights weighs Q lb. Find the tension in the cable and Q. Answer: T1 = 139 lb, T2 = 139 lb, T3 = 143 lb, Q = 12.1 lb.

4 – 244 (L)A 500 lb crate is being towed from a 2000 lb airplane, as shown. The airplane’s L/D (lift over drag) is 1/3. Find the tension TA in the cable and the airplane’s thrust T. Assume that the flight is level and neglect the aerodynamic drag acting on the cable. Answer: TA = 1000 lb, T = 8370 lb.

4 – 245 (M)A 25 lb weight and two forces, labeled Q1 and Q3, act on the cable shown. Find the tension in the cable and Q1 and Q3. Answer: T1 = 44.2 lb, T2 = 38.9 lb, T3 = 29.8 lb, T4 = 29.8 lb, Q1 = 15.4, Q3 = 29.8 lb.

4 – 246 (M)A 200 lb force and two additional forces, labeled Q1 and Q3, act on the cable shown. Find the tension in the cable and Q1

and Q3. Answer: T1 = 346 lb, T2 = 400 lb, T3 = 346 lb, T4 = 300 lb, Q1 = 200, Q3 = 173 lb.

4 – 247 (M)A 100 N force and a 120 N force act on the cable shown. Find the tension in the cable and the angle 3. Answer: T1 = 200 N, T2 = 173 N, T3 = 211 N, 3 = 34.7○.

4 – 248 (M)Two 2500 lb ski cabs hang from the cable shown. Find the tension in the cable and the angle 3. Answer: T1 = 5600 lb, T2 = 6870 lb, T3 = 8690 lb, 3 = 52.7○.

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4 – 249 (M)A 200 lb force and a 75 lb force act on the cable shown. Find the tension in the cable and the angle 3. Answer: T1 = 115 lb, T2 = 115 lb, T3 = 62.9 lb, 3 = 66.6○.

4 – 250 (M)A 150 lb force and a 100 lb force act on the cable shown. Find the tension in the cable and the angle 1. Answer: T1 = 80.7 lb, T2 = 100 lb, T3 = 173 lb, 1 = 81.7○.

4 – 251 (M)A 50 lb force, a 40 lb force, and a force Q act on the triangular cable shown. Find the tension in the cable, the force Q, and the angles and C. Answer: TAB = 23.1 lb, TAC = 23.1 lb, TBC = 32.1 lb, Q = 45.8 lb, = 40.9○, and C = 68.9○.

4 – 252 (H)Three 25 N forces act on the cable shown. Find the tension in the cable and the angle . Hint: Use symmetry to simplify the problem and note that the solution of the nonlinear algebraic equation of the general form 1 = acos + bcos is

Answer: T1 = 14.4 lb, T2 = 17.5 lb, = 8.63○. Line Loading in Light Cables

4 – 253 (M)Two suspension cables that span 100 ft support a 20,000 lb section of a road, as shown. Find the maximum tension in the cables, and the length of each cable. Answer: Tmax = 5150 lb, s = 232 ft.

4 – 254 (M)Two suspension cables that span 8 ft support a 300 lb boardwalk, as shown. Find the maximum tension in the cables, and the length of one cable. Answer: Tmax = 250 lb, s = 8.70 ft.

4 – 255 (M)A suspension cable that spans 100 ft needs to support a 800 lb gas line, as shown. The maximum tension in the cable is 2500 lb. Find the minimum height H of the suspension point of the cable. Answer: H = 4.05 ft.

4 – 256 (M)Two suspension cables that span 100 ft support a 20,000 lb section of road. Find the tension on each side of the cable and the length of each cable. Answer: TA = 3230 lb, TB = 7790 lb, s = 193 ft.

4 – 257 (M)Suspension cables that span 25 ft support 5000 lb sections of a bridge, as shown. Find the compressive force acting at the top of tower B. Answer: F = 5860 lb.

4 – 258 (M)An ancient foot bridge in a castle is supported by two cables that are each tensioned by a counter weight W. The cables span 8 ft and each supports half of the 600 lb bridge. Find the counter weights W needed to tighten the cables, as shown. Answer: W = 292 lb.

4 – 259 (C, H)A cable supports a tapered water line, as shown. Find the location D of the lowest point of the cable. Answer: D = 13.2 ft.

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Line Loading in Heavy Cables4 – 260 (M, C)A 1.2 lb/ft cable is suspended, as shown. Find the maximum tension in the cable and the cable’s length. Answer: Tmax = 20.5 lb, s = 22.0 ft. 4 – 261 (M, C)A 1.5 lb/ft cable that spans 10 ft is wrapped around a pulley with a 75 lb weight, as shown. Find the sag y0 in the cable. Neglect the weight of the vertical section of the cable. Answer: y0 = 5.14 ft. 4 – 262 (M, C)A 0.8 lb/ft cable that spans 12 ft is 20 ft long. Find the sag in the cable. Answer: y = 7.26 ft. 4 – 263 (M, C)A 10 inch 0.2 lb/in power/communication ribbon is attached to a guide, as shown. Find the sag and the maximum tension in the ribbon when x1 = 8 in and when x2 = 9 in. Answer: y1 = 2.65 ft., T1 = 1.21 lb, y2 = 1.91 ft, T1 = 1.50 lb. 4 – 264 (M, C)A 1.4 lb/ft cable, that spans 20 ft across a street corner, sags 2 ft, as shown. How much more does it sag after a 50 lb traffic light is suspended from the center of the cable? Answer: y = 0.307 ft. 4 – 265 (H, C)The right end of a 10 ft long, 2 lb/ft chain is hooked while the left end rests on the floor. The static coefficient of friction between the chain and the floor is = 0.5. Find the maximum distance D for which the chain remains in static equilibrium. Answer: D = 3.25 ft.

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