Determine the magnitude of forces F1, F2, F3, so that the particle is held in equilibrium.
Answer to Problem 7FP
The magnitude of force F1 is 466 N_.
The magnitude of force F2 is 879 N_.
The magnitude of force F3 is 776 N_.
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.
Using Figure (1),
Determine the magnitude of forces using equation of equilibrium.
Force along x direction:
[(35)F3](35)+600−F2=00.36F3−F2=−600 (I)
Force along y direction:
(45)F1−[35F3](45)=00.8F1−0.48F3=00.8F1=0.48F3F1=0.480.8F3
F1=0.6F3 (II)
Force along z direction:
(45)F3+(35)F1−900=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=776 N
Thus, the magnitude of force F3 is 776 N_.
Substitute 776 N for F3 in Equation (II).
F1=0.6×776=466 N
Thus, the magnitude of force F1 is 466 N_.
Substitute 776 N for F3 in Equation (I).
0.36(776)−F2=−600279.36+600=F2F2=879 N
Thus, the magnitude of force F2 is 879 N_.
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