International Edition---engineering Mechanics: Statics, 4th Edition
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN: 9781305501607
Author: Andrew Pytel And Jaan Kiusalaas
Publisher: CENGAGE L
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Textbook Question
Chapter 2, Problem 2.14P

Find the forces Q 1 , Q 2 , and Q 3 so that the two forces systems are equivalent.

Chapter 2, Problem 2.14P, Find the forces Q1,Q2, and Q3 so that the two forces systems are equivalent.

Expert Solution & Answer
Check Mark
To determine

The forces Q1,Q2,Q3.

Answer to Problem 2.14P

The forces in right hand systems are Q1=121.74 lb, Q2=0Q3=202.9 lb

Explanation of Solution

Given Information:

The following system of forces is given,

  International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 2, Problem 2.14P

and the resultant of both force systems are the same.

Calculation:

The forces applied can be written in vector notation as,

  P1=100( (30)i+(00)j+(40)k (30) 2 + (00) 2 + (40) 2 )=60i+0j+80kP2=120( (30)i+(30)j+(40)k (30) 2 + (30) 2 + (40) 2 )=61.74i+61.74j+82.32kP3=60( (00)i+(30)j+(00)k (00) 2 + (30) 2 + (00) 2 )=0i+60j+0k

  Q1=Q1( (30)i+(00)j+(00)k (30) 2 + (00) 2 + (00) 2 )=Q1iQ2=Q2( (03)i+(03)j+(00)k (03) 2 + (30) 2 + (00) 2 )=Q2(0.71i0.71j)Q3=Q3( (00)i+(30)j+(40)k (00) 2 + (30) 2 + (40) 2 )=Q3(0.6j+0.8k)

Since resultant of both the force systems is the same, hence,

  Q1+Q2+Q3=P1+P2+P3Q1i+Q2(0.71i0.71j)+Q3(0.6j+0.8k)=60i+0j+80k+61.74i+61.74j+82.32k+60j

Compare the components to get the following equations,

  Q10.71Q2=60+61.74                  ...(1)0.71Q2+0.6Q3=60+61.74            ..(2)0.8Q3=80+82.32                  ...(3)

Solve equation (1-3) to get Q1=121.74 lb, Q2=0Q3=202.9 lb

Conclusion:

Therefore, the forces in the right hand systems are Q1=121.74 lb, Q2=0Q3=202.9 lb

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Chapter 2 Solutions

International Edition---engineering Mechanics: Statics, 4th Edition

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