Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
11th Edition
ISBN: 9780077687342
Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 14.1, Problem 14.30P

Show that the relation M A = H ˙ A ' where H A ' is defined by Eq. (1) of Prob. 14.29 and where M A represents the sum of the moments about A of the external forces acting on the system of particles, is valid if, and only if, one of the following conditions is satisfied: (a) the frame Axyz1is itself a newtonian frame of reference, (b) A coincides with the mass center G,

Expert Solution
Check Mark
To determine

(a)

ΣMA=H˙A' If the frame Ax'y'z' is itself a Newtonian frame of reference.

Explanation of Solution

Given information:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.30P , additional homework tip  1

HA'=i=1nri'×mivi'

HA'=i=1nri'×mivi'

HA'=i=1n[(rirA)×mi(vivA)]

Now, differentiate the above equation with respect to time:

H˙A'=i=1n[(r˙ir˙A)×mi(vivA)]+i=1n[(rirA)×mi(v˙iv˙A)]

But,r˙i=viv˙i=air˙A=vAv˙A=aA

Hence,

H˙A'=0+i=1n[(rirA)×mi(aiaA)]H˙A'=i=1n[(rirA)×(FimiaA)]H˙A'=i=1n[(rirA)×F]i=1n[mi(rirA)]×aAH˙A'=ΜAm(rrA)×aA

The term ΣMA=H˙A' is satisfied if the term m(rrA)×aA is equals to zero.

Hence, if the frame of A is Newtonian then,

aA=0

Thus, m(rrA)×aA=0 then, ΣMA=H˙A'

Conclusion:

The given condition is satisfied the given equation.

Expert Solution
Check Mark
To determine

(b)

ΣMA=H˙A' If A coincides with the mass centre G.

Explanation of Solution

Given information:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.30P , additional homework tip  2

HA'=i=1nri'×mivi'

HA'=i=1nri'×mivi'

HA'=i=1n[(rirA)×mi(vivA)]

Now, differentiate the above equation with respect to time:

H˙A'=i=1n[(r˙ir˙A)×mi(vivA)]+i=1n[(rirA)×mi(v˙iv˙A)]

But,r˙i=viv˙i=air˙A=vAv˙A=aA

Hence,

H˙A'=0+i=1n[(rirA)×mi(aiaA)]H˙A'=i=1n[(rirA)×(FimiaA)]H˙A'=i=1n[(rirA)×F]i=1n[mi(rirA)]×aAH˙A'=ΜAm(rrA)×aA

The term ΣMA=H˙A' is satisfied if the term m(rrA)×aA is equals to zero.

Hence, if point A coincides with mass centre then,

r=rA

Thus, m(rrA)×aA=0 then, ΣMA=H˙A'

Conclusion:

The given condition is satisfied the given equation.

Expert Solution
Check Mark
To determine

(c)

ΣMA=H˙A' the acceleration aA of A is directed along the line AG.

Explanation of Solution

Given information:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.30P , additional homework tip  3

HA'=i=1nri'×mivi'

HA'=i=1nri'×mivi'

HA'=i=1n[(rirA)×mi(vivA)]

Now, differentiate the above equation with respect to time:

H˙A'=i=1n[(r˙ir˙A)×mi(vivA)]+i=1n[(rirA)×mi(v˙iv˙A)]

But,r˙i=viv˙i=air˙A=vAv˙A=aA

Hence,

H˙A'=0+i=1n[(rirA)×mi(aiaA)]H˙A'=i=1n[(rirA)×(FimiaA)]H˙A'=i=1n[(rirA)×F]i=1n[mi(rirA)]×aAH˙A'=ΜAm(rrA)×aA

The term ΣMA=H˙A' is satisfied if the term m(rrA)×aA is equals to zero.

Hence, if the acceleration of A is directed along the line AG then,

aAisparallelto(rrA)

Thus, m(rrA)×aA=0 then, ΣMA=H˙A'

Conclusion:

The given condition is satisfied the given equation.

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Chapter 14 Solutions

Vector Mechanics for Engineers: Dynamics

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