Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 4.3, Problem 4.129P

Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A.

Chapter 4.3, Problem 4.129P, Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm,

Fig. P4.129 and P4.130

Expert Solution & Answer
Check Mark
To determine

The tension in each cable and the reaction at A.

Answer to Problem 4.129P

The tension in each cable and the reaction at A are 625N for TDG, 600N for TCF and 975N for TBE and A=(2100N)i+(175.0N)j(375N)k respectively.

Explanation of Solution

The tension in the cable CF can be represented as TCF, in the cable BE can be represented as TBE and in the cable DG can be represented as TDG. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.129P

Figure 1

The position vector of the point B is given as,

rB=(0.48m)i

The position vector of the point H is given as,

rH=(0.48 m)i(0.15m)k

The position vector of the point C is given as,

rC=(0.48m)i(0.14m)k

The position vector of the point D is given as,

rD=(0.48m)i(0.3m)k

The vector BE is given as BE=(0.48 m)i+(0.2m)j, vector DG is given as DG=(0.48 m)i+(0.14m)k and CF is given as CF=(0.48 m)i.

The tension across the cable BE as,

TBE=TBEBEBE

Here, TBE is the tension across the cable BE, BE is the vector along the path of the points, BE is the magnitude of the vector.

Substitute (0.48 m)i+(0.2m)j for BE and 0.50m for BE to get TBE.

TBE=TBE0.5m((0.48 m)i+(0.2m)j)=TBE25(24i+7j)

The tension across the cable DG as,

TDG=TDGDGDG

Here, TDG is the tension across the cable DG, DG is the vector along the path of the points, DG is the magnitude of the vector.

Substitute (0.48 m)i+(0.14m)k for DG and 0.52m for DG to get TDG.

TDG=TDG0.52m((0.48 m)i+(0.14m)k)=TDG13(12j+5k)

The tension across the cable CF as,

TCF=TCFCFCF

Here, TCF is the tension across the cable CF, CF is the vector along the path of the points, CF is the magnitude of the vector.

Substitute (0.48 m)i for CF and 0.48m for CF to get TCF.

TCF=TCF0.48m((0.48 m)i)=TCFi

The weight is given as,

W=mgj

Here, W is the weight, m is the mass and g is the acceleration due to gravity.

Substitute 350N for mg.

W=(350N)j

The force on the point is zero. This means that the sum of the moments of the force will also be zero.

rH×W+rB×TBE+rD×TDG+rC×TCF=0

Substitute the vector values and determine the cross product.

|ijk0.4800.1503500|+|ijk0.48002470|TBE25+|ijk0.4800.30125|TDG13+|ijk0.4800.14100|TCF=0(0.0875TDG52.5Nm)i+(0.1846TBE+(2425×TDG×0.3m))j+(168Nm+(725TDG×0.48m)+(0.14m)TCF)k=0 (I)

Since the force at the point is zero, the x component is given as,

AxTCF1213TBE2425TDG=0 (II)

Here, Ax is the x coefficient of the reaction at A.

Since the force at the point is zero, the y component is given as,

Ay+725TDGW=0 (III)

Here, Ay is the y coefficient of the reaction at A.

Since the force at the point is zero, the z component is given as,

Az+513(TBE)=0 (IV)

Here, Az is the z coefficient of the reaction at A.

Conclusion:

Equate the coefficients of i in (I) and solve for TDG.

0.0875TDG52.5Nm=0TDG=625N

Equate the coefficients of j in (I) and solve for TBD by substituting 625N for TDG.

0.1846TBE+(2425×(625N)×0.3m)=0TBE=975N

Equate the coefficients of k in (I) and solve for TCF by substituting 625N for TDG.

168Nm+(725(625N)×0.48m)+(0.14m)TCF=0TCF=600N

Substitute 625N for TDG, 600N for TCF and 975N for TBE in (II) to get Ax.

Ax(600N)1213(975N)2425(625N)=0Ax=2100N (V)

Substitute 625N for TDG and 350N for W in (III) to get Ay.

Ay+725(625N)(350N)=0Ay=175.0N (VI)

Substitute 975N for TBE in (IV) to get Az.

Az+513(975N)=0Az=375N . (VII)

From (V), (VI) and (VII), the reaction on A is given as,

A=(2100N)i+(175.0N)j(375N)k

Therefore, the tension in each cable and the reaction at A are 625N for TDG, 600N for TCF and 975N for TBE and A=(2100N)i+(175.0N)j(375N)k respectively.

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