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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Chapter 18.2, Problem 18.102P
To determine

(a)

The couple M0 .

Expert Solution
Check Mark

Answer to Problem 18.102P

The couple M0 is 0.2115ftlb.

Explanation of Solution

Given information:

Angular velocity of disk in z-direction is 60rad/s, radius of the disk is 3in ,length of BC member is 4in.

The figure is represented below.

<x-custom-btb-me data-me-id='1725' class='microExplainerHighlight'>Vector</x-custom-btb-me> Mechanics For Engineers, Chapter 18.2, Problem 18.102P

Figure (1)

Write the equation for the mass of the disk.

m=Wg .........(1)

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

Write the expression for the angular momentum about point A.

HA=I¯xωxi+I¯yωyj+I¯zωzk .........(2)

Here, mass moment of inertia about the x-axis is I¯x, mass moment of inertia about the y-axis is I¯y, mass moment of inertia about the z-axis is I¯z, angular velocity of disk about x axis is ωx, angular velocity of disk about y axis is ωy .and angular velocity of disk about z axis is ωz.

Write the expression for the angular velocity of disk in x direction.

ωx=0 .........(3)

Substitute 0 for ωx, ω2 for ωy, and ω1 for ωz .in Equation (2).

HA=I¯x×0×i+I¯yω2j+I¯zω1kHA=I¯yω2j+I¯zω1k .........(4)

Here, ω2 is the angular velocity in y-direction and ω1 is the angular velocity in z-direction.

Write the expression for angular velocity in vector form.

Ω=ω2j ..........(5)

Write the expression for rate of angular velocity of the reference frame Axyz.

(H˙A)xyz=I¯yω˙2j+I¯zω˙1k .........(6)

Write the expression for rate of total angular velocity.

H˙A=(H˙)Ax'y'z'+Ω×HA .........(7)

Substitute (I¯yω2j+I¯zω1k) for HA, I¯yω˙2j+I¯zω˙1k for (H˙A)xyz and ω2j for Ω in Equation (7).

H˙A=I¯yω˙2j+I¯zω˙1k+ω2j×(I¯yω2j+I¯zω1k) .........(8)

Write the expression for Matrix multiplication of the vector product for Equation (8).

H˙A=(I¯yω˙2j+I¯zω˙1k)+I¯zω1ω2i=I¯zω1ω2i+I¯yω˙2j+I¯zω˙1k .........(9)

Write the expression for the mass moment of inertia about the y-direction.

I¯y=14mr2 .........(10)

Here mass of the disk is m and radius of the disk is r.

Write the expression for the mass moment of inertia about the z- direction.

I¯z=12mr2 .........(11)

Substitute 14mr2 for I¯y and 12mr2 for I¯z in Equation (9).

H˙A=12mr2×ω1ω2i+14mr2×ω˙2j+12mr2×ω˙1k .........(12)

Write the expression for the velocity of mass centre of the disk.

v¯=ω2j×ci .........(13)

Here, velocity of mass centre is v¯ and distance between B and A is c.

Write the expression for the matrix multiplication of the vector product for Equation (13).

v¯=cω2k .........(14)

Write the expression for the acceleration of the mass centre of the disk.

a¯=ω˙2j×ci+ω˙2j×v¯ .........(15)

Write the expression for the matrix multiplication of the vector product for Equation (15).

a¯=cω˙2kcω22i .........(16)

Write the expression for the the sum of the forces acting on the system.

F=Cxi+Czk+Dxi+Dzk .........(17)

Write the expression for the force in terms of mass and acceleration.

F=ma¯ .........(18)

Substitute Cxi+Czk+Dxi+Dzk for F in Equation (18).

Cxi+Czk+Dxi+Dzk=ma¯ .........(19)

Here, force at C in x-direction is Cx, force at C in z-direction is Cz. force at D in x-direction is Dx, and force at D in z-direction is Dz.

Substitute cω˙2kcω22i for a¯ in Equation (19)

Cxi+Czk+Dxi+Dzk=m×(cω˙2kcω22i)Cxi+Czk+Dxi+Dzk=mcω˙2kmcω22i .........(20)

Compare the coefficients of the unit vector of i on both side of Equation (20).

Cx+Dx=mcω22 .........(21)

Compare the coefficients of the unit vector of i on both side of Equation.

Cz+Dz=mcω˙2 .........(22)

Write the expression for the rate of angular momentum about D.

H˙D=H˙A+rA/D×ma¯ .........(23)

Here, distance between A and D is rA/D.

Write the expression for rA/D in vector form.

rA/D=cibj

Here, distance from the centre of disk to point B is c and distance of BD is b.

Substitute (12mr2×ω1ω2i+14mr2×ω˙2j+12mr2×ω˙1k) for H˙A, (cibj) for rA/D and ω˙2j×ci+ω˙2j×v¯ for a¯ in Equation(23).

H˙D=(12mr2×ω1ω2i+14mr2×ω˙2j+12mr2×ω˙1k)+(cibj)×m(ω˙2j×ci+ω˙2j×v¯) .........(24)

Write the expression for the matrix multiplication for vector product for equation (24).

H˙D=[(12mr2×ω1ω2i+14mr2×ω˙2j+12mr2×ω˙1k)mbω˙2i+mc2ω˙2j+mbcω22k]H˙D=[m(12r2ω1ω2bcω˙2)i+m(14r2+c2)ω˙2j+m(12r2ω˙1+bcω22)k] .........(25)

Write the expression for the moment about D

MD=M0j+2bj×(Cxi+Czk) .........(26)

Here, length of BC member is b and moment couple when system is at rest is M0.

Write the expression for the matrix multiplication for the vector product for equation (26).

MD=2bCzi+M0j2bCxk .........(27)

Here 2b is the length of DC.

The sum of the moment at D is equal to the rate of change of angular momentum at D.

MD=H˙D .........(29)

Substitute 2bCzi+M0j2bCxk for MD in equation (25).

[2bCzi+M0j2bCxk]=[m(12r2ω1ω2bcω˙2)i+m(14r2+c2)ω˙2j+m(12r2ω˙1+bcω22)k] .........(30)

Compare the coefficients of the unit vector of j on both side of Equation (30).

M0=m(14r2+c2)ω˙2 .........(31)

Compare the coefficients of the unit vector of k on both side of Equation (30).

2bCx=m(12r2ω˙1+bcω22)Cx=m2b(12r2ω˙1+bcω22) .........(32).

Compare the coefficients of the unit vector of i on both side of Equation (30).

2bCz=m(12r2ω1ω2bcω˙2)Cz=m2b(12r2ω1ω2bcω˙2) .........(33)

Substitute m2b(12r2ω˙1+bcω22) for Cx in Equation (21).

[m2b(12r2ω˙1+bcω22)+Dx]=mcω22Dx=m2b(12r2ω˙1+bcω22)mcω22Dx=mcω222+m2b12r2ω˙1Dx=(m2b)(12r2ω˙1+bcω22) .........(34)

Substitute m2b(12r2ω1ω2bcω˙2) for Cz in Equation (22).

[m2b(12r2ω1ω2bcω˙2)+Dz]=mcω˙2Dz=mcω˙2m2b(12r2ω1ω2bcω˙2)Dz=mcω˙22(m2b)(12r2ω1ω2)Dz=(m2b)(12r2ω1ω2+bcω˙2) .........(35)

Calculation:

Substitute 6lb for W and 32.2ft/s2 for g in Equation (1).

m=6lb32.2ft/s2=0.1863lbs2ft

Substitute values of 0.1863lbs2ft for m, (3in) for r, (5in) for c and 6rad/s2 for ω˙2 in Equation (31).

M0=0.1863lbs2ft(14(3in)2+(5in)2)×6rad/s2M0=0.1863lbs2ft×(14(3in(1ft12in))2+(5in(1ft12in))2)×6rad/s2M0=0.1863lbs2ft×((0.25ft)2+(0.4166ft)2)×6rad/s2M0=0.212ftlb

Thus value of couple M0 is 0.212ftlb.

Conclusion:

The value of couple is 0.212ftlb.

To determine

(b)

The dynamic reaction at C.

The dynamic reaction at D.

Expert Solution
Check Mark

Answer to Problem 18.102P

The dynamic reaction at C .is (12.58lb)i(9.43lb)k .

The dynamic reaction at D .is (12.58lb)i(9.43lb)k .

Explanation of Solution

Given information:

Write the expression for the angular velocity in terms of time in y-direction.

ω2=(ω2)0+ω˙2t .........(36)

Here time is t.

Calculation:

Substitute 0 for (ω2)0, 6rad/s2 for ω˙2 and 3s for t in Equation (36).

ω2=0+7.089rad/s2×2s=14.179rad/s

Substitute values of 0.1863lbs2ft for m, (3in) for r, (4in) for b, (5in) for c, 0 for ω˙1, and 18rad/s for ω2 in Equation (32).

Cx=[0.1863lbs2ft2(4in)×(12(3in)2(0)+(4in)(5in)(18rad/s2))]Cx=[0.1863lbs2ft2(4in(1ft12in))×(12(3in(1ft12in))2(0)+(4in(1ft12in))(5in(1ft12in))(18rad/s2))]=0.2795(0+44.999)=12.58lb

Substitute values of 0.1863lbs2ft for m, (3in) for r, (4in) for b, (5in) for c, 60rad/s for ω1, 18rad/s for ω2 and 0 for ω˙2 in Equation (33).

Cz=[0.1863lbs2ft2(4in)×(12(3in)2(60rad/s)(18rad/s)(4in)(5in)×0)]Cz=[0.1863lbs2ft2(4in(1ft12in))×(12(3in(1ft12in))2(60rad/s)(18rad/s)(4in(1ft12in))(5in(1ft12in))×0)]=0.2795(33.75+0)=9.43lb

Hence, dynamic reaction at C is (12.58lb)i(9.43lb)k.

Substitute values of 0.1863lbs2ft for m, (3in) for r, (4in) for b, (5in) for c, 0 for ω˙1, and 18rad/s for ω2, Equation (34).

Dx=(0.1863lbs2ft2(4in))(12(3in)2×0+(4in)(5in)(18rad/s)2)Dx=(0.1863lbs2ft2(4in(1ft12in)))(12(3in(1ft12in))2×0+(4in(1ft12in))(5in(1ft12in))(18rad/s)2)Dx=0.2795(0+44.999)lbDx=12.58lb

Substitute values of 0.1863lbs2ft for m, (3in) for r, (4in) for b, (5in) for c, 60rad/s for ω1, 18rad/s for ω2 and 0 for ω˙2 in Equation (35).

Dz=(0.1863lbs2ft2(4in))(12(3in)2×(60rad/s)×(18rad/s)+(4in)(5in)×0)Dz=(0.1863lbs2ft2(4in(1ft12in)))(12(3in(1ft12in))2×(60rad/s)×(18rad/s)+(4in(1ft12in))(5in(1ft12in))×0)Dz=9.43lb

Hence, dynamic reaction at D is (12.58lb)i(9.43lb)k

Conclusion:

The dynamic reactions at C is (12.58lb)i(9.43lb)k.

The dynamic reactions at D is (12.58lb)i(9.43lb)k.

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Chapter 18 Solutions

Vector Mechanics For Engineers

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