VECTOR MECHANICS FOR ENGINEERS: STATICS
VECTOR MECHANICS FOR ENGINEERS: STATICS
12th Edition
ISBN: 9781260912814
Author: BEER
Publisher: MCG
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Chapter 7.4, Problem 7.99P

Knowing that dc = 9 ft, determine (a) the distances dB and dD (b) the reaction at E.

Chapter 7.4, Problem 7.99P, Knowing that dc = 9 ft, determine (a) the distances dB and dD (b) the reaction at E. Fig. P7.99 and

Fig. P7.99 and P7.100

(a)

Expert Solution
Check Mark
To determine

The distances dB and dD.

Answer to Problem 7.99P

The distance dB is 5.20ft. The distance dD is 12.60ft.

Explanation of Solution

Refer Fig P7.99.

The figure 1 below shows the free body diagram of the portion ABC.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 7.4, Problem 7.99P , additional homework tip  1

The total moment about the point C is zero.

Refer the free body diagram and write the equation for the moment about point C.

9Ax12Ay+(1kip)(6ft)=0

Here Ax is the horizontal reaction at point A, Ay is the vertical reaction at point A.

Re-write the above equation to get an expression for Ax .

Ax=43Ay23 (I)

The figure 2 below shows the free body diagram of the entire cable.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 7.4, Problem 7.99P , additional homework tip  2

The moment about point E is zero.

Refer the free body diagram of the entire cable and write the equation of the moment about point E.

12Ax30Ay+(1kip)(18ft)+(1kip)(24ft)+(2kips)(9ft)=0

Simplify the above equation.

12Ax30Ay+60=0 (II)

Since the system is in equilibrium the total vertical and horizontal components will be zero.

Refer figure 2 and write the equation for total horizontal force.

Ax+Ex=0 (III)

Here Ex is the horizontal reaction force at point E.

Refer figure 2 and write the equation for the total vertical force.

Ay+Ey1kip1kip2kip=0 (IV)

The figure 4 below shows the free body diagram of the portion AB.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 7.4, Problem 7.99P , additional homework tip  3

The moment about point B is zero.

Refer figure 4 and write the equation for the moment about point B.

Ay(6ft)+AxdB=0 (V)

The figure 5 below shows the free body diagram of the portion DE.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 7.4, Problem 7.99P , additional homework tip  4

Refer figure 5 and write the formula for the distance h.

h=(9ft)tan3.8° (VI)

Here h is the vertical distance between point D and E.

Refer figure 5 and write the formula for distance dD.

dD=12ft+h (VII)

Conclusion:

Substitute equation (I) in equation (II).

12(43Ay23)30Ay+60=016Ay30Ay8+60=014Ay=52Ay=3.7143kips

Substitute 3.7143kips for Ay in equation (I) to get

Ax=43(3.7143kips)23=4.257kips

Substitute 4.257kips for Ax in equation (III) to determine Ex.

Ex=4.257kips

Substitute 3.7143kips for Ay in equation (IV) to determine Ey.

Ey=3.7143kips+1kip+1kip+2kip=0.2857kips

Substitute 4.257kips for Ax, 3.7143kips for Ay in equation (V) to determine dD.

dB=3.7143kips(6ft)4.257kips=5.20ft

Calculate h from equation (VI).

h=(9ft)tan3.8°=0.599ft

Substitute 0.599ft for h in equation (VII) to determine dD.

dD=12ft+0.599ft=12.60ft

The distance dB is 5.20ft. The distance dD is 12.60ft.

(b)

Expert Solution
Check Mark
To determine

The reaction at point E.

Answer to Problem 7.99P

The reaction at point E is 4.30kips making an angle 3.81° with the horizontal.

Explanation of Solution

Refer Fig P7.99.

The figure 1 below shows the free body diagram of the portion ABC.

The total moment about the point C is zero.

Refer the free body diagram and write the equation for the moment about point C.

9Ax12Ay+(1kip)(6ft)=0

Here Ax is the horizontal reaction at point A, Ay is the vertical reaction at point A.

Re-write the above equation to get an expression for Ax .

Ax=43Ay23 (I)

The figure 2 below shows the free body diagram of the entire cable.

The moment about point E is zero.

Refer the free body diagram of the entire cable and write the equation of the moment about point E.

12Ax30Ay+(1kip)(18ft)+(1kip)(24ft)+(2kips)(9ft)=0

Simplify the above equation.

12Ax30Ay+60=0 (II)

Since the system is in equilibrium the total vertical and horizontal components will be zero.

Refer figure 2 and write the equation for total horizontal force.

Ax+Ex=0 (III)

Here Ex is the horizontal reaction force at point E.

Refer figure 2 and write the equation for the total vertical force.

Ay+Ey1kip1kip2kip=0 (IV)

Write the formula for the magnitude of the reaction at point E.

E=Ex2+Ey2 (V)

Here E is the magnitude of the reaction at point E.

Write the formula for the angle made by the reaction at point E with horizontal.

θ=tan1(EyEx) (VI)

Here θ is the angle made by the reaction at point E with horizontal.

Conclusion:

Substitute equation (I) in equation (II).

12(43Ay23)30Ay+60=016Ay30Ay8+60=014Ay=52Ay=3.7143kips

Substitute 3.7143kips for Ay in equation (I) to get

Ax=43(3.7143kips)23=4.257kips

Substitute 4.257kips for Ax in equation (III) to determine Ex.

Ex=4.257kips

Substitute 3.7143kips for Ay in equation (IV) to determine Ey.

Ey=3.7143kips+1kip+1kip+2kip=0.2857kips

Substitute 0.2857kips for Ey, 4.257kips for Ey in equation (V) to determine E.

E=(4.257kips)2+(0.2857kips)2=4.30kips

Substitute 0.2857kips for Ey, 4.257kips for Ey in equation (VI) to determine θ.

θ=tan1(0.2857kips4.257kips)=3.81°

Thus the reaction at point E is 4.30kips making an angle 3.81° with the horizontal.

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Chapter 7 Solutions

VECTOR MECHANICS FOR ENGINEERS: STATICS

Ch. 7.1 - A semicircular rod is loaded as shown. 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