VECTOR MECH....F/ENGNRS-STATICS -CONNECT
VECTOR MECH....F/ENGNRS-STATICS -CONNECT
12th Edition
ISBN: 9781260689495
Author: BEER
Publisher: MCG CUSTOM
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Chapter 7.5, Problem 7.147P

The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.

Chapter 7.5, Problem 7.147P, The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the

Fig. P7.147

Expert Solution & Answer
Check Mark
To determine

Find the distance a.

Answer to Problem 7.147P

The distance a is 3.50ft_.

Explanation of Solution

Given information:

The length of the cable AB is L=10ft.

The value of angle θ is 30°.

The collar at A is slides freely and the collar at B is prevented from the moving.

Calculation:

Show the free-body diagram of the cable assembly as in Figure 1.

VECTOR MECH....F/ENGNRS-STATICS -CONNECT, Chapter 7.5, Problem 7.147P

Refer the Equation 7.16 in the textbook.

Write the equation of the catenary cable as follows;

y=ccoshxc

Differentiate the equation with x;

dydx=sinhxc

The slope at point A is;

tanθ=|dydx|A=sinhxAcxAc=sinh(tan(90°θ))xA=csinh1(tan(90°θ)) (1)

The length of the portion AC is as follows:

AC=csinh(xAc)

The length of the portion CB is as follows:

CB=csinh(xBc)

Find the distance xB using the relation.

L=AC+CB

Substitute 10 ft for L, csinh(xAc) for AC, and csinh(xBc) for CB.

10=csinh(xAc)+csinh(xBc)csinh(xBc)=10csinh(xAc)sinh(xBc)=10csinh(xAc)xB=csinh1[10csinh(xAc)] (2)

Find the distance (yA) using the relation.

yA=ccosh(xAc) (3)

Find the distance (yB) using the relation.

yB=ccosh(xBc) (4)

Consider the triangle ABD;

Find the value of tanθ using the relation.

tanθ=OppositesideAdjacentside=yByAxB+xA (5)

Find the distance a using the relation.

a=yByA (6)

Use the trial and error procedure to find the value of a.

Consider the value of c and for the given value of θ=30°, find the angle θ in the equation (5). The calculated value of angle θ and the given value of θ=30° should be equal.

Trial 1:

Consider a trial value of 1.60 ft for c.

c=1.60ft

Substitute 1.60 ft for c and 30° for θ in Equation (1).

xA=1.60×sinh1(tan(90°30°))=2.107ft

Substitute 1.60 ft for c and 2.107 ft for xA in Equation (2).

xB=1.60×sinh1[101.60sinh(2.1071.60)]=3.541ft

Substitute 1.60 ft for c and 2.107 ft for xA in Equation (3).

yA=1.60×cosh(2.1071.60)=3.20ft

Substitute 1.60 ft for c and 3.541 ft for xB in Equation (4).

yB=1.60×cosh(3.5411.60)=7.404ft

Substitute 2.107 ft for xA, 3.541 ft for xB, 3.20 ft for yA, and 7.404 ft for yB in Equation (5).

tanθ=7.4043.203.541+2.107θ=36.658°

The calculated value of θ=36.658° is not equal to the given value of θ=30°

Trial 2:

Consider a trial value of 1.70 ft for c.

c=1.70ft

Substitute 1.70 ft for c and 30° for θ in Equation (1).

xA=1.70×sinh1(tan(90°30°))=2.239ft

Substitute 1.70 ft for c and 2.239 ft for xA in Equation (2).

xB=1.70×sinh1[101.70sinh(2.2391.70)]=3.622ft

Substitute 1.70 ft for c and 2.239 ft for xA in Equation (3).

yA=1.70×cosh(2.2391.70)=3.40ft

Substitute 1.70 ft for c and 3.622 ft for xB in Equation (4).

yB=1.70×cosh(3.6221.70)=7.257ft

Substitute 2.239 ft for xA, 3.622 ft for xB, 3.40 ft for yA, and 7.257 ft for yB in Equation (5).

tanθ=7.2573.403.622+2.239θ=33.352°

The calculated value of θ=33.352° is not equal to the given value of θ=30°

Trial 3:

Consider a trial value of 1.803 ft for c.

c=1.803ft

Substitute 1.803 ft for c and 30° for θ in Equation (1).

xA=1.803×sinh1(tan(90°30°))=2.374ft

Substitute 1.803 ft for c and 2.374 ft for xA in Equation (2).

xB=1.803×sinh1[101.803sinh(2.3741.803)]=3.694ft

Substitute 1.803 ft for c and 2.374 ft for xA in Equation (3).

yA=1.803×cosh(2.3741.803)=3.606ft

Substitute 1.803 ft for c and 3.694 ft for xB in Equation (4).

yB=1.803×cosh(3.6941.803)=7.110ft

Substitute 2.374 ft for xA, 3.694 ft for xB, 3.606 ft for yA, and 7.11 ft for yB in Equation (5).

tanθ=7.113.6063.694+2.374θ=30°

The calculated value of θ=30° is equal to the given value of θ=30°

Therefore, the value of c is 1.803 ft.

Substitute 3.606 ft for yA, and 7.11 ft for yB in Equation (6).

a=7.113.606=3.50ft

Therefore, the distance a is 3.50ft_.

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Chapter 7 Solutions

VECTOR MECH....F/ENGNRS-STATICS -CONNECT

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