Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9780073398242
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 5.1, Problem 5.29P

The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C.

Chapter 5.1, Problem 5.29P, The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73

(a)

Expert Solution
Check Mark
To determine

The tension in the cable.

Answer to Problem 5.29P

The tension in the cable is TBA=125.3N_.

Explanation of Solution

The diagram for the centre of gravity is depicted below:

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 5.1, Problem 5.29P , additional homework tip  1

Write the equation of x-coordinate of the centre of gravity,

X¯ΣL=Σx¯L (I)

Here, the area is Σx¯L, the x-coordinate is X¯, the total length ΣL.

Write the equation of weight,

W=(m'ΣL)g (II)

Here, the weight is W, the mass per unit length is (m'ΣL), the gravitational acceleration is g.

Write the equation of momentum at equilibrium,

r1[(3/5)TBA][r2W]=0 (III)

Conclusion:

 L(m)x¯(m)x¯L(m2)
11.350.6750.91125
20.60.30.18
30.7500
40.750.20.15
5(π/2)(0.75)=1.17811.077461.26936
Σ4.62810 2.5106

Substitute 4.62810m for ΣL, 2.5106m2 for Σx¯L in Equation (I).

X¯(4.62810m)=2.5106m2X¯=0.54247m

Substitute 4.73kg/m for m', 4.62810m for ΣL, 9.8m/s2 for g in Equation (II).

W=[(4.73kg/m)(4.62810m)](9.8m/s2)=214.75N

Substitute, 1.55m for r1, 0.54247m for r2, 214.75N for W.

(1.55m)[(3/5)TBA][(0.54247m)(214.75N)]=0TBA=125.3N

Thus, The tension in the cable is TBA=125.3N_.

(b)

Expert Solution
Check Mark
To determine

The reaction at C.

Answer to Problem 5.29P

The reaction at C is 137.0N at an angle 56.7°_.

Explanation of Solution

The diagram for the reaction is depicted below:

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 5.1, Problem 5.29P , additional homework tip  2

Refer fig 2.

Write the equation of reaction along x-axis.

Cx35(TBA)=0 (IV)

Here, the reaction along x-axis is Cx.

Write the equation of reaction along y-axis.

Cy+45(TBA)W=0 (V)

Here, the reaction along y-axis is Cy.

Write the expression for the magnitude of the reaction at C,

C=Cx2+Cy2 (VI)

Here, C is the magnitude of reaction.

Write the equation of the direction of the reaction,

θ=tan1(CyCx) (VII)

Conclusion:

Substitute, 125.264N for TBA in equation (IV),

Cx35(125.64N)=0Cx=75.158N

Substitute, 125.264N for TBA, (275/25)P for R in equation (II)

Cy+45(125.264N)(214.75N)=0Cy=114.539N

Substitute, 75.158N for Cx, 114.539N for Cy in equation (VI),

C=(75.158N)2+(114.539N)2=137.0N

Substitute, 75.158N for Cx, 114.539N for Cy in equation (VII)

θ=tan1(114.539N75.158N)=56.7°

Thus, the reaction at C is 137.0N at an angle 56.7°_.

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Chapter 5 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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