Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 14.1, Problem 14.11P

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. Knowing that the angular momentum of the system about O expressed in ft · lb · s is HO = −1.2k, determine (a) the velocities of the particles, (b) the angular momentum of the system about its mass center G.

Fig. P14.11 and P14.12

Chapter 14.1, Problem 14.11P, 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 velocities of the particles.

Answer to Problem 14.11P

The velocity of particles A is (4.00ft/s)j_

The velocity of particles B is (1.00ft/s)i_

The velocity of particles C is (3.00ft/s)k_

Explanation of Solution

Given information:

The angular momentum about point O is 1.2k

Calculation:

The mass of three particles A, B, and C is equal.

mA=mB=mC

Determine the weight of the identical particle.

mA=mB=mC=Wg (1)

Here, W is weight of each particle, mB is mass of particle B, mA is mass of particle A, mC is mass of particle B, 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 the angular momentum of the system about the origin using the Equation.

Ho=(rA×mAvA)+(rB×mBvB)+(rC×mCvC) (2)

Here, vA is velocity vector of particle A, mB is mass of particle is B, vB is the velocity of particle B, vC is the velocity of particle C.

Substitute 0.6lbs2/ft for (mA,mB,mC), 3k for rA, 2i+2j+3k for rB, i+4j  for rc, (vAj) for vA, (vBi) for vB, (vCk) for vC, and 1.2k for Ho in Equation (2).

1.2k=(3k)×(0.6vAj)+(2i+2j+3k)×(0.6vBi)+(i+4j)×(0.6vck)1.2k=|ijk00300.6vA0|+|ijk2230.6vB00|+|ijk140000.6vC|1.2k=(1.8i)+(1.8vBj1.2vBk)+(2.4vCi0.6vCj)1.2k=(1.8vA+2.4vc)i+(1.8vB0.6vc)j+(1.2vB)k

Equating i, j, k components.

1.8vA+2.4vC=0 (3)

1.8vB0.6vC=0 (4)

1.2vB=1.2 (5)

Find the velocity at point B as follows:

1.2vB=1.2vB=(1.00ft/s)i

Thus, the velocity of particles B is (1.00ft/s)i_

Find the velocity at point C as follows:

Substitute (1.00ft/s)i for vB in Equation (4).

1.8(1.00)0.6vC=01.80.6vC=0vC=(3.00ft/s)k

Thus, the velocity of particles C is (3.00ft/s)k_

Find the velocity at point A as follows:

Substitute (3.00ft/s)k for vC in Equation (3).

1.8vA+2.4((3.00))=01.8vA+2.4((3.00))=01.8vA=7.2vA=(4.00ft/s)j

Thus, the velocity of particles A is (4.00ft/s)j_

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

r¯=i=1nmirii=1nmi=mArA+mBrB+mCrCmA+mB+mC . (6)

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 (6).

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¯ (7)

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 (7).

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¯ (8)

Here, rB is position vector at point B.

Substitute 2i+2j+3k for rB and i+2j+2k for r¯  in Equation (8).

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¯ (9)

Here, rC is position vector at point C.

Substitute i+4j for rC, and i+2j+2k for r¯ in Equation (9).

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

Express the linear momentum of particle A as follows:

mAvA=0.6(4j)=(2.4lbs)j

Express the linear momentum of particle B as follows:

mBvB=0.6(i)=(0.6lbs)i

Express the linear momentum of particle C as follows:

mCvC=0.6(3k)=(1.8lbs)k

(b)

Expert Solution
Check Mark
To determine

Find the angular momentum HG of the system.

Answer to Problem 14.11P

The angular momentum HG of the system is (1.2i+0.6j2.4k)ftlbs_.

Explanation of Solution

Calculation:

Calculate the angular momentum about point G using the relation:

HG=rA×(mAvA)+rB×(mBvB)+rC×(mCvC) (10)

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

Substitute i2j+k for rA, i+k for rB, 2j2k for rC, (2.4lbs)j for mAvA, (0.6lbs)i for mBvB, and (1.8lbs)k for mCvC in Equation (10).

HG=i2j+k×(2.4lbs)j+i+k×(0.6lbs)i+2j2k×((1.8lbs)k)=(2.4i2.4k)+(0.6j)+(3.6i)=(1.2i+0.6j2.4k)ftlbs

Thus, the angular momentum HG of the system is (1.2i+0.6j2.4k)ftlbs_.

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Chapter 14 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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