Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Textbook Question
Chapter 14.1, Problem 14.10P

For the satellite system of Prob. 14.9. assuming that the velocity of the base satellite is zero, determine (a) the position vector r of the mass center G of the system, (b) the linear momentum L of the system, (c) the angular momentum Hg of the system about G. Also, verify that the answers to this problem and to Prob. 14.9 satisfy the equation given in Prob. 14.27.

Chapter 14.1, Problem 14.10P, For the satellite system of Prob. 14.9. assuming that the velocity of the base satellite is zero,

Expert Solution
Check Mark
To determine

(a)

The position vector r of the mass centre G of the system assuming the base satellite velocity is zero

Explanation of Solution

Given information:

vA=(4i2j+2k)

vB=(i+4j)

vC=(2i+2j+4k)

mA=4kg

mB=6kg

mC=8kg

Reference equation from Problem 14.27

Ho=r¯×mv¯+HG

Position vectors is written as below

rA=30jrB=15i+25j+35krC=40i

System mass centre is calculated as below,

r=mArA+mBrB+mCrCmA+mB+mC

Substituting the values we get,

r=4(30j)+6(15i+25j+35k)+8(40i)4+6+8

r=22.78i+15j+11.67k

Therefore, mass centre position vector is r=22.78i+15j+11.67k

Conclusion:

Position vector of mass centre is r=22.78i+15j+11.67k

Expert Solution
Check Mark
To determine

(b)

The linear momentum L of the system assuming the base satellite velocity is zero

Explanation of Solution

Given information:

vA=(4i2j+2k)

vB=(i+4j)

vC=(2i+2j+4k)

mA=4kg

mB=6kg

mC=8kg

Reference equation from Problem 14.27

Ho=r¯×mv¯+HG

Position vectors is written as below

rA=30jrB=15i+25j+35krC=40i

Linear momentum is calculated by using the relation,

L=mAvA+mBvB+mCvC

Substituting the values we get,

L=4(4i2j+2k)+6(i+4j)+8(2i+2j+4k)

L=38i+32j+40k

Therefore, linear momentum of the system is L=38i+32j+40k

Conclusion:

Linear momentum of the system is L=38i+32j+40k

Expert Solution
Check Mark
To determine

(c)

The angular momentum of the system about G assuming the base satellite velocity is zero

Explanation of Solution

Given information:

vA=(4i2j+2k)

vB=(i+4j)

vC=(2i+2j+4k)

mA=4kg

mB=6kg

mC=8kg

Reference equation from Problem 14.27

Ho=r¯×mv¯+HG

Position vectors is written as below

rA=30jrB=15i+25j+35krC=40i

Angular momentum is calculated by using the relation,

HG=(rAr)×mAvA+(rBr)×mBvB+(rCr)×mCvC

Substituting the values we get,

HG={(30j(22.78i+15j+11.67k))×4(4i2j+2k)+(15i+25j+35k(22.78i+15j+11.67k))×6(i+4j)+(40i(22.78i+15j+11.67k))×8(2i+2j+4k)}

=(ijk22.781511.671688)+(ijk7.781023.336240)+(ijk17.221511.67161632)

={i(12093.36559.92480+186.72)j(182.24+186.72139.98+551.04+186.72)+k(182.24240186.7260+275.52+240)}

=(826.56i602.26j+211.04k)kgm2/s

HG=(826.56i602.26j+211.04k)kgm2/s

Therefore, angular momentum of the system is HG=(826.56i602.26j+211.04k)kgm2/s

Assume, mass centre G velocity,

v=vxi+vyj+vzk

Linear momentum equation can be written as,

L=(mA+mB+mC)v

Substituting the values we get,

38i+32j+40k=(4+6+8)(vxi+vyj+vzk)

vxi+vyj+vzk=2.11i+1.78j+2.22k

Therefore, individual vector components are,

vx=2.11vy=1.78vz=2.22

Angular momentum of the system about origin is

Ho=rA×mAvA+rB×mBvB+rC×mCvC

Ho={(3oj)×4(4i2j+2k)+(15i+25j+35k)×6(i+4j)+(40i)×8(2i+2j+4k)}

=(ijk03001688)+(ijk1525356240)+(ijk4000161632)

Ho=(240840)ij(210+1280)+k(480+360150+640)

Ho=600i1070j+370k ................. (Equation A)

Check whether the above relation satisfies with equation in 14.27

Ho=r×(mA+mB+mC)v+HG

Ho={(22.78i+15j+11.67k)×(4+6+8)(2.11i+1.78j+2.22k)+(826.56i602.26j+211.04k)}

=(ijk22.781511.67383240)+(826.56i602.26j+211.04k)

Ho=600i1070j+370k ................ (Equation B)

From the above, we can conclude, equation A satisfies with equation B

Conclusion:

Angular momentum of the system is HG=(826.56i602.26j+211.04k)kgm2/s

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Chapter 14 Solutions

Vector Mechanics For Engineers

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