Engineering Mechanics: Statics, Student Value Edition (14th Edition)
Engineering Mechanics: Statics, Student Value Edition (14th Edition)
14th Edition
ISBN: 9780134056388
Author: Russell C. Hibbeler
Publisher: PEARSON
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Textbook Question
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Chapter 3.4, Problem 7FP

Determine the magnitude of forces F1, F2, F3, so that the particle is held in equilibrium.

Chapter 3.4, Problem 7FP, Determine the magnitude of forces F1, F2, F3, so that the particle is held in equilibrium.

Expert Solution & Answer
Check Mark
To determine
The magnitude of forces F1, F2, and F3 for equilibrium.

Answer to Problem 7FP

The magnitude of force F1 is 466N_.

The magnitude of force F2 is 879N_.

The magnitude of force F3 is 776N_.

Explanation of Solution

Given information:

The given force values are 600 N and 900 N.

Explanation:

Show the free body diagram of the forces acting on the particle as in Figure 1.

Engineering Mechanics: Statics, Student Value Edition (14th Edition), Chapter 3.4, Problem 7FP

Using Figure (1),

Determine the magnitude of forces using equation of equilibrium.

Force along x direction:

[(35)F3](35)+600F2=00.36F3F2=600 (I)

Force along y direction:

(45)F1[35F3](45)=00.8F10.48F3=00.8F1=0.48F3F1=0.480.8F3

F1=0.6F3 (II)

Force along z direction:

(45)F3+(35)F1900=00.8F3+0.6F1=900 (III)

Conclusion:

Substitute 0.6F3 for F1 in Equation (III).

0.8F3+0.6(0.6F3)=900F3(0.8+0.36)=900F3=9001.16F3=776N

Thus, the magnitude of force F3 is 776N_.

Substitute 776N for F3 in Equation (II).

F1=0.6×776=466N

Thus, the magnitude of force F1 is 466N_.

Substitute 776N for F3 in Equation (I).

0.36(776)F2=600279.36+600=F2F2=879N

Thus, the magnitude of force F2 is 879N_.

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Chapter 3 Solutions

Engineering Mechanics: Statics, Student Value Edition (14th Edition)

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