Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 4.2, Problem 4.90P

Chapter 4.2, Problem 4.90P, Fig. P4.89 4.90 Knowing that for the rod of Prob. 4.89, L = 15 in., R = 20 in., and W = 10 lb,

Fig. P4.89

4.90 Knowing that for the rod of Prob. 4.89, L = 15 in., R = 20 in., and W = 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B.

(a)

Expert Solution
Check Mark
To determine

The angle θ corresponding to the equilibrium condition in the given arrangement.

Answer to Problem 4.90P

The angle θ corresponding to the equilibrium condition in the given arrangement is 59.4°_.

Explanation of Solution

The free-body diagram corresponding to the arrangement in Fig. P4.89 is given in Figure 1. The reaction at A and B is represented by vector A and vector B respectively. In the equilibrium condition, the reaction B must pass through the point D where B and W intersect.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.2, Problem 4.90P , additional homework tip  1

Write the expression for the distances AC and AE.

AC=Lcosθ (I)

AE=Lcosθ (II)

Write the expression for the distances CE, BE, and BC in triangle EBC.

CE=AC+AE (III)

BE=Lsinθ (IV)

BC=R (V)

Write the expression of Pythagoras theorem in the triangle EBC.

(CE)2+(BE)2=(BC)2 (VI)

Conclusion:

Use equation (I) and (II) in (III).

CE=Lcosθ+Lcosθ=2Lcosθ (VII)

Use equation (IV), (V) and (VII) in (VI).

(2Lcosθ)2+(Lsinθ)2=R2 (VIII)

Simplify equation (VIII).

4cos2θ+sin2θ=(RL)24cos2θ+(1cos2θ)=(RL)23cos2θ+1=(RL)2cos2θ=13[(RL)21]

The expression for the equilibrium condition of the rod is,

cos2θ=13[(RL)21] (IX)

Given that the length of the rod is 15in, radius of the cylindrical surface is 20in, and the weight is 10lb.

Substitute 20in for R, 15in for L and 10lb for W in equation (IX) and solve for θ.

cos2θ=13[(20in15in)21]cosθ=13[(20in15in)21]θ=cos1{13[(20in15in)21]}=59.39°

Therefore, the angle θ corresponding to the equilibrium condition in the given arrangement is 59.4°_.

(b)

Expert Solution
Check Mark
To determine

The reaction at A and B.

Answer to Problem 4.90P

The reaction at A is 8.45lb_ and that at B is 13.09lb_.

Explanation of Solution

The free-body diagram corresponding to the arrangement is given in Figure 1.

Write the expression for computing the angle α in the free-body diagram.

tanα=BECE (X)

Use equation (IV) and (VII) in (X).

tanα=Lsinθ2Lcosθ=12tanθ (XI)

At the equilibrium condition, a force triangle can be drawn using W, A, and B.

Conclusion:

Substitute 59.39° for θ in equation (XI) and solve for α.

tanα=12tan59.39°α=tan1(12tan59.39°)=40.2°

The force triangle using W, A, and B is given in Figure 2. Since A is acting horizontally and W is acting vertically, the force triangle is a right triangle.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.2, Problem 4.90P , additional homework tip  2

Write the trigonometric relation for determining A in the force triangle.

A=Wtanα (XII)

Write the trigonometric relation for det4ermoini ng B in the force triangle.

B=Wcosα (XIII)

Substitute 40.2° for α and 10lb for W in equation (XII) and (XIII) to find the reaction at A and B.

A=(10lb)tan40.2°=8.45lb

and,

B=10lbcos40.2°=13.09lb

The reaction at B is acting 49.8° clockwise from the horizontal as shown in the force triangle.

Therefore, the reaction at A is 8.45lb_ and that at B is 13.09lb_.

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Chapter 4 Solutions

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