EP MODIFIED MASTERING ENGINEERING WITH
EP MODIFIED MASTERING ENGINEERING WITH
14th Edition
ISBN: 9780133941357
Author: HIBBELER
Publisher: PEARSON CO
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Chapter 3.3, Problem 13P

A nuclear-reactor vessel has a weight of 500(103) lb. Determine the horizontal compressive force that the spreader bar AB exerts on point A and the force that each cable segment CA and AD exert on this point white the vessel is hoisted upward at constant velocity.

Chapter 3.3, Problem 13P, A nuclear-reactor vessel has a weight of 500(103) lb. Determine the horizontal compressive force

Prob. 3–13

Expert Solution & Answer
Check Mark
To determine

The horizontal compressive force that the spreader bar AB exerts on point A.

The force that each cable segment CA and AD exerts on point A.

Answer to Problem 13P

The horizontal compressive force that the spreader bar AB exerts on point A (FAB) is 433×103lb_.

The force that each cable segment CA and AD exert on point A, FCA=500×103lb_ FAD=250×103lb_.

Explanation of Solution

Given information:

The weight of a nuclear reactor vessel (W) is 500×103lb.

Show the free body diagram of the cable about point C as in Figure 1.

EP MODIFIED MASTERING ENGINEERING WITH , Chapter 3.3, Problem 13P , additional homework tip  1

Show the free body diagram of the cable about point A as in Figure 2.

EP MODIFIED MASTERING ENGINEERING WITH , Chapter 3.3, Problem 13P , additional homework tip  2

Determine the horizontal compressive force of each cable segment CA and AD by applying the equation of equilibrium.

At point C

Along the horizontal direction:

Fx=0FCBcosθFCAcosθ=0FCBcosθ=FCAcosθFCB=FCA

Here, the force acting on cable CB is FCB and the force acting on cable CA is FCA.

Along the vertical direction:

Fy=0WFCAsinθFCBsinθ=0 (I)

At point A

Along the horizontal direction:

Fx=0WcosθFAB=0FAB=Wcosθ (II)

Here, the force acting on cable AB is FAB.

Along the vertical direction:

Fy=0WsinθFAD=0FAD=Wsinθ (III)

Here, the force acting on cable AD is FAD.

Conclusion:

Substitute 30° for θ and FCB for FCA and 500×103lb for W in Equation (I).

500×103FCAsin30°FCAsin30°=02FCAsin30°=500×103FCA=500×1032sin30°FCA=500×103lb

Substitute 30° for θ and 500×103lb for W in Equation (II).

FAB=500×103cos30°=433×103lb

Substitute 30° for θ and 500×103lb for W in Equation (III).

FAD=500×103sin30°=250×103lb

Thus, the horizontal compressive force that the spreader bar AB exerts on point A is 433×103lb_.

The force that each cable segment CA and AD exerts on point A is 500×103lb_ and 250×103lb_.

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