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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Chapter 14.1, Problem 14.12P

A system consists of three identical 19.32-lb particles A, B, and C. The velocities of the particles are, respectively, vA = vA j, vB = vBi, and vC = vCk, and the magnitude of the linear momentum L of the system is 9 lb · s. Knowing that HG = HO, where HG is the angular momentum of the system about its mass center G and HO is the angular momentum of the system about O, determine (a) the velocities of the particles, (b) the angular momentum of the system about O.

Fig. P14.11 and P14.12

Chapter 14.1, Problem 14.12P, A system consists of three identical 19.32-lb particles A, B, and C. The velocities of the particles

(a)

Expert Solution
Check Mark
To determine

Find the velocity of particles.

Answer to Problem 14.12P

The velocity of particle A is (10.00ft/s)j_.

The velocity of particle B is (5.00ft/s)j_.

The velocity of particle C is (10.00ft/s)j_.

Explanation of Solution

Given information:

The velocity of particles is as follows:

vA=vAjvB=vBivC=vCk

The linear momentum (L) of the system is 9lbs.

Calculation:

The mass of three particles is equal.

mA=mB=mC

Determine the weight of the identical particle.

mA=mB=mC=Wg (1)

Here, W is weight of each particle and g is acceleration due to gravity.

Substitute 19.32lb for W and 32.2ft/s2 in Equation (1).

mA=mB=mC=19.3232.2=0.63lbs2/ft

Write the position vectors for the particles based on the given coordinate system:

rA=3krB=2i+2j+3krC=i+4j

Determine position vector (r¯) of the mass center G of the system using the relation:

r¯=i=1nmirii=1nmi=mArA+mBrB+mCrCmA+mB+mC . (2)

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

Substitute 3k for rA, 2i+2j+3k for rB, i+4j for rc, and 0.6lbs2/ft for (mA,mB,mC),  in Equation (2).

r¯=0.6×3k+(0.6)(2i+2j+3k)+0.6(i+4j)1.8=i+2j+2k

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 3k for rA, and i+2j+2k for r¯ in Equation (3).

rA=3k(i+2j+2k)=i2j+k

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 2i+2j+3k for rB and i+2j+2k for r¯ in Equation (4).

rB=2i+2j+3ki+2j+2k=i+k

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 i+4j for rC, and i+2j+2k for r¯ in Equation (5).

rC=i+4ji+2j+2k=2j2k

Express the angular momentum about point O as follows:

HO=rA×(mAvA)+rB×(mBvB)+rC×(mCvC)

Calculate the angular momentum about point G using the relation:

HG=rA×(mAvA)+rB×(mBvB)+rC×(mCvC)

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.

Subtract the angular momentum HOHG as follows:

HOHG=(rA×(mAvA)+rB×(mBvB)+rC×(mCvC)rA×(mAvA)+rB×(mBvB)+rC×(mCvC))=(rArA)×(mAvA)+(rB)×mBvB+rC×(mCvC)0=r¯×(mAvA)+r¯×(mBvB)r¯×(mCvC)=r¯(mAvA+mBvB+mCvC)

L is parallel to r¯ as follows:

L=λr¯LL=λ2r¯r¯λ2=LLr¯r¯ (6)

Substitute 9lbs for L and 3 for r¯ in Equation (6).

λ2=9×93×3=819=32λ=±3lbs/ft

Find the velocity of particles using the relation as follows:

mAvA+mBvB+mCvC=λr¯ (7)

Substitute 0.6lbs2/ft for (mA,mB,mC), ±3lbs/ft for λ, i+2j+2k for r¯, (vAj) for vA, (vBi) for vB, (vCk) for vC in Equation (7).

0.6((vAj))+0.6(vBi)+0.6(vCk)=±3(i+2j+2k)

Resolve the components and solve vA,vB,vC.

Find the velocity of particle A using the Equation:

0.6((vAj))=6jvA=10(ft/s)j

Thus, the velocity of particle A is (10.00ft/s)j_.

Find the velocity of particle B using the Equation:

0.6((vBi))=3ivB=5(ft/s)i

Thus, the velocity of particle B is (5.00ft/s)j_.

Find the velocity of particle C using the Equation:

0.6((vCj))=6kvC=10(ft/s)k

Thus, the velocity of particle C is (10.00ft/s)j_.

(b)

Expert Solution
Check Mark
To determine

Find the angular momentum of the system about O.

Answer to Problem 14.12P

The angular momentum of the system about O is (6.00ftlbs)i+(3.00ftlbs)j(6.0ftlbs)k_

Explanation of Solution

Calculation:

Calculate the angular momentum about point H using the relation:

HO=rA×(mAvA)+rB×(mBvB)+rC×(mCvC) (8)

Here, (rA,rB,rC) are positive vector and (mAvA,mBvB,mCvC) are linear momentum.

Substitute 3k for rA, 2i+2j+3k for rB, i+4j for rc, 0.6((vAj)) for mAvA, 0.6(vBi) for mBvB, and 0.6(vCk) for mCvC in Equation (8).

HO=|ijk00300.6vA8.0|+|ijk2230.6vB00|+|ijk140000.6vC|=(1.8vAi)+(1.8vBj1.2vBk)+(2.4vCj0.6vCj)=(1.8vA+2.4vC)i+(1.8vB0.6vC)j+(1.2vB)k=6i+3j6k=(6.00ftlbs)i+(3.00ftlbs)j(6.0ftlbs)k

Thus, the angular momentum of the system about O is (6.00ftlbs)i+(3.00ftlbs)j(6.0ftlbs)k_

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Chapter 14 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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