VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter 14.1, Problem 14.14P

For the system of particles of Prob. 14.13, determine (a) the position vector r ¯ of the mass center G of the system, (b) the linear momentum m v ¯ of the system, (c) the angular momentum HG of the system about G. Also verify that the answers to this problem and to Prob. 14.13 satisfy the equation given in Prob. 14.27.

Fig. P14.13

Chapter 14.1, Problem 14.14P, For the system of particles of Prob. 14.13, determine (a) the position vector r of the mass center G

(a)

Expert Solution
Check Mark
To determine

Find the position vector of the mass center of the system.

Answer to Problem 14.14P

The position vector of the mass center of the system is (1.867m)i+(1.533m)j+(0.667m)k_.

Explanation of Solution

Given information:

The mass of the particles A is 3kg.

The mass of the particles B is 2kg.

The mass of the particles C is 4kg.

The position vector is r¯.

The mass center is G.

Calculation:

Find the position vectors from point O to each satellite in meters.

Refer to figure P14.13 in the textbook.

Express the position vector point A as follows:

rA=3j

Express the position vector point B as follows:

rB=1.2i+2.4j+3k

Express the position vector point C as follows:

rC=3.6i

Determine the mass center G of the system using the relation:

r¯=i=1nmirii=1nmi=mArA+mBrB+mCrCmA+mB+mC . (1)

Here, (mA,mB,mC) is mass of A, B, and C and (rA,rB,rC) is position vector.

Substitute 3j for rA, 1.2i+2.4j+3k for rB, 3.6i for rc, and 3kg for mA, 2kg for mB,, and 4kg for mC in Equation (1).

r¯=3×3j+(2)(1.2i+2.4j+3k)+4(3.6i)9=6j+2.4i+4.8j+6k+14.4i9=1.86667i+1.53333j+0.6667k

Thus, the position vector of the mass center of the system is (1.867m)i+(1.533m)j+(0.667m)k_.

(b)

Expert Solution
Check Mark
To determine

Find the linear momentum of the system.

Answer to Problem 14.14P

The linear momentum of the system is (12.00kgm/s)i+(28.0kgm/s)j+(14.00kgm/s)k_.

Explanation of Solution

Calculation:

Express the linear momentum of particle A as follows:

mAvA=3(4i+2j+2k)=12.0i+6j+6k

Express the linear momentum of particle B as follows:

mBvB=2(4i+3j)=8i+6j

Express the linear momentum of particle C as follows:

mCvC=4(2i+4j+2k)=8.0i+16.0j+8.0k

Find the linear momentum of the system using the relation:

mv¯=mAvA+mBvB+mCvC (2)

Substitute 12.0i+6j+6k for mAvA, 8i+6j for mBvB, and 8.0i+16.0j+8.0k for mCvC in Equation (2).

mv¯=12.0i+6j+6k+8i+6j+(8.0i+16.0j+8.0k)=12.0i+28j+14k=(12.00kgm/s)i+(28.0kgm/s)j+(14.00kgm/s)k

Thus, the linear momentum of the system is (12.00kgm/s)i+(28.0kgm/s)j+(14.00kgm/s)k_.

(c)

Expert Solution
Check Mark
To determine

Find the angular momentum of the system about G and also verify this problem and to problem 14.13 satisfy the Equation given in problem 14.27.

Answer to Problem 14.14P

The angular momentum of the system about G is (2.80kgm2/s)i+(13.33kgm2/s)j(24.30kgm2/s)k_.

Explanation of Solution

Calculation:

Find the position vector from the particles rA to the center of mass using the relation:

rA=rAr¯ (3)

Here, rA is position vector at point A and r¯ is the mass center.

Substitute 3j for rA, and 1.86667i+1.53333j+0.6667k for r¯ in Equation (3).

rA=3j(1.86667i+1.53333j+0.6667k)=1.86667i+1.4667j+0.6667k

Find the position vector from the particles rB to the center of mass using the relation:

rB=rBr¯ (4)

Here, rB is position vector at point B.

Substitute 1.2i+2.4j+3k for rB and 1.86667i+1.53333j+0.6667k for r¯ in Equation (4).

rB=1.2i+2.4j+3k(1.86667i+1.53333j+0.6667k)=0.66670.86667j+2.3333k

Find the position vector from the particles rC to the center of mass using the relation:

rC=rCr¯ (5)

Here, rC is position vector at point C.

Substitute 3.6i for rC, and 1.86667i+1.53333j+0.6667k for r¯ in Equation (5).

rC=3.6i1.86667i+1.53333j+0.6667k=1.73333i1.53333j0.6667k

Calculate the angular momentum about point G using the relation:

HG=rA×(mAvA)+rB×(mBvB)+rC×(mCvC) (6)

Here, (mA,mB,mC) is mass of particles A, B, C, (rA,rB,rC) are positive vector for particles and (vA,vB,vC) is velocity of particles A, B, C.

Substitute 1.86667i+1.4667j+0.6667k for rA, 1.73333i1.53333j0.6667k for rC, 0.66670.86667j+2.3333k for rB, 12.0i+6j+6k for mAvA, 8i+6j for mBvB, and 8.0i+16.0j+8.0k for mCvC in Equation (6).

HG=[1.86667i+1.4667j+0.6667k×(12.0i+6j+6k)0.66670.86667j+2.3333k×(8i+6j)+1.73333i1.53333j0.6667kC×(8.0i+16.0j+8.0k)]=[|ijk1.866671.466670.66671266|+|ijk0.666670.866672.3333860|+|ijk1.733331.53330.66678168|]=(12.8i+3.2j28.8k)+(14i+18.6667j-10.9333k)+(1.6i-8.5333j+15.4667k)=2.8i+13.3333j24.2667k=(2.80kgm2/s)i+(13.33kgm2/s)j(24.30kgm2/s)k

Thus, the angular momentum of the system about G is (2.80kgm2/s)i+(13.33kgm2/s)j(24.30kgm2/s)k_.

Find the value of r¯×mv¯ as follows:

Substitute 1.86667i+1.53333j+0.6667k for r¯ and (12.00kgm/s)i+(28.0kgm/s)j+(14.00kgm/s)k for mv¯.

r¯×mv¯=|ijk1.866671.53330.6667122814|=i(21.466218.6676)j(26.13338.0000)+k(52.26667618.3996)=2.8i18.33j+33.8667k=(2.8kgm2/s)i(18.133kgm2/s)j+(33.8667kgm2/s)kHG+r¯×mv¯=[(2.80kgm2/s)i+(13.33kgm2/s)j(24.30kgm2/s)k+(2.8kgm2/s)i(18.133kgm2/s)j+(33.8667kgm2/s)k]=(4.8kgm2/s)j+(9.6kgm2/s)k

Show that the HO=HG+r¯×mv¯.

Express the angular momentum about point O as follows:

HO=rA×(mAvA)+rB×(mBvB)+rC×(mCvC) (7)

Substitute 3j for rA, 1.2i+2.4j+3k for rB, 3.6i for rc, 12.0i+6j+6k for mAvA, 8i+6j for mBvB, and 8.0i+16.0j+8.0k for mCvC in Equation (7).

HO=|ijk0301266|+|ijk1.22.43860|+|ijk3.6008168|=(18i36k)+(18i+24j12k)+(28.8j+57.6k)=(4.8kgm2/s)j+(9.6kgm2/s)k

Hence, the HO=HG+r¯×mv¯ is proved.

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Chapter 14 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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