Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9780073398242
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 2.4, Problem 2.93P

Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.

Chapter 2.4, Problem 2.93P, Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and

Fig. P2.93 and P2.94

Expert Solution & Answer
Check Mark
To determine

The direction and magnitude of the resultant forces exerted at A by two cables.

Answer to Problem 2.93P

The direction and magnitude of the resultant forces exerted at A by two cables is given by θx=48.2°_, θy=116.6°_, θz=53.4°_, and R=913lb_.

Explanation of Solution

The tension in the cable AB is 425lb.

The tension in the cable AC is 510lb.

Write the equation to find the magnitude of the vector AB.

|AB|=ABx2+ABy2+ABz2 (I)

Write the equation to find the unit vector along AB.

λ=AB|AB| (II)

Write the equation of FAB in terms of its rectangular components.

FAB=TABλ (III)

Here, TAB is the tension in the cable AB, and λ is the unit vector along AB.

Write the equation for force FAB in component form.

FAB=Fxi+Fyj+Fzj (IV)

Write the equation to find the magnitude of the vector AC.

|AC|=ACx2+ACy2+ACz2 (V)

Write the equation to find the unit vector along AC.

λ=AC|AC| (VI)

Write the equation of FAC in terms of its rectangular components.

FAC=TACλ (VII)

Here, TAC is the tension in the cable AC, and λ is the unit vector along AC.

Write the equation for force F in component form.

FAC=Fxi+Fyj+Fzj (VIII)

Refer figure P2.93.

Write the equation to find the resultant of the two vectors.

R=FAB+FAC (IX)

Here, R is the resultant of the two vectors FAB and FAC

Write the equation to find the magnitude of the vector R.

|R|=Rx2+Ry2+Rz2 (X)

Write the equation to find the direction of vector R.

cosθn=RnR (XI)

Here, Rn is the component of the vector corresponding to the direction n, R is the magnitude of the vector R, and θn is the angle corresponding to the direction.

Conclusion:

Refer Fig.P2.93 and calculate the vector coordinates of the vector AB.

AB=(40in)i(45in)j+(60in)k

Substitute 40in for ABx, 45in for ABy, and 60in for ABz in the equation (I).

AB=(40in)2+(45in)2+(60in)2=85in

Substitute (40in)i(45in)j+(60in)k for AB, and 85in for AB in equation (II).

λ=(40in)i(45in)j+(60in)k85in=0.47i0.52j+0.70k

Substitute 425lb for TAB, and 0.47i0.52j+0.70k for λ in equation (III).

FAB=(425lb)(0.47i0.52j+0.70k)=(200lb)i(225lb)j+(300lb)k

Substitute (200lb)i(225lb)j+(300lb)k for FAB in equation (IV).

(200lb)i(225lb)j+(300lb)k=Fxi+Fyj+Fzj

Refer Fig.P2.93 and calculate the vector coordinates of the vector AC.

AC=(100in)i(45in)j+(60in)k

Substitute 100in for ACx, 45in for ACy, and 60in for ACz in the equation (V).

AC=(100in)2+(45in)2+(60in)2=125in

Substitute (100in)i(45in)j+(60in)k for AC, and 125in for AC in equation (VI).

λ=(100in)i(45in)j+(60in)k125in=0.8i0.36j+0.48k

Substitute 510lb for TAC, and 0.8i0.36j+0.48k for λ in equation (VII).

FAC=(510lb)(0.8i0.36j+0.48k)=(408lb)i(183.6lb)j+(244.8lb)k

Substitute (408lb)i(183.6lb)j+(244.8lb)k for FAC in equation (VIII).

(408lb)i(183.6lb)j+(244.8lb)k=Fxi+Fyj+Fzj

Substitute (200lb)i(225lb)j+(300lb)k for FAB and (408lb)i(183.6lb)j+(244.8lb)k for FAC in equation (IX).

R=(200lb)i(225lb)j+(300lb)k+(408lb)i(183.6lb)j+(244.8lb)k=(608lb)i(408.6lb)j+(544.8lb)k

Substitute 608lb for Rx, 408.6lb for Ry, and 544.8lb for Rz in the equation (X).

R=(608lb)2+(408.6lb)2+(544.8lb)2=912.92lb

Substitute x for n, 912.92lb for R, and 608lb for Rx, in equation (XI).

cosθx=RxR=608lb912.92lb=0.66599θx=48.2°

Substitute y for n, 912.92lb for R, and 408.6lb for Ry, in equation (XI).

cosθy=RyR=408.6lb912.92lb=0.44757θy=116.6°

Substitute z for n, 912.92lb for R, and 544.8lb for Rz, in equation (XI).

cosθz=RzR=544.8lb912.92lb=0.59677θz=53.4°

Therefore, the direction and magnitude of the resultant forces exerted at A by two cables is given by θx=48.2°_, θy=116.6°_, θz=53.4°_, and R=913lb_.

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Chapter 2 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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