Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9780073398242
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Textbook Question
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Chapter 16.2, Problem 16.141P

Two rotating rods in the vertical plane are connected by a slider block P of negligible mass. The rod attached at A has a weight of 1.6 lb and a length of 8 in. Rod BP weighs 2 lb and is 10 in. long and the friction between block P and AE is negligible. The motion of the system is controlled by a couple M applied to rod BP. Knowing that rod BP has a constant angular velocity of 20 rad/s clockwise, determine (a) the couple M, (b) the components of the force exerted on AE by block P.

Chapter 16.2, Problem 16.141P, Two rotating rods in the vertical plane are connected by a slider block P of negligible mass. The

Fig. P16.141 and Fig. P16.142

(a)

Expert Solution
Check Mark
To determine

Find the value of couple M.

Answer to Problem 16.141P

The value of couple M is 5.82lbft_.

Explanation of Solution

Given information:

The weight of the rod AE is WAE=1.6lb.

The weight of the rod BP is WBP=2lb.

The length of the rod AE is LAE=8in.

The length of the rod BP is LBP=10in.

The angular velocity is ω=20rad/s.

Calculation:

Consider the acceleration due to gravity as g=32.2ft/s2.

Calculate the position vector (r) as shown below.

The position of P with respect to A.

rP/A=(10in.×1ft12in.sin30°)j=0.4167j

The position of P with respect to B.

rP/B=(10in.×1ft12in.cos30°)i(10in.×1ft12in.sin30°)j=0.7217i0.4167j

The position of P with respect to E.

rP/E=(8in.×1ft12in.)j=0.6667j

The angular velocity of rod BP in vector form is ωBP=(20rad/s)k.

Calculate the velocity of rod BP (vP) as shown below.

vP=ωBP×rP/B

Substitute (20rad/s)k for ωBP, and 0.7217i0.4167j for rP/B.

vP=((20rad/s)k)×(0.7217i0.4167j)=14.434j8.334i=8.334i+14.434j

Consider the relative angular velocity of rod AE as ωAE=ωAEk and the collar P slides on the rod with relative velocity vP/AE=uj.

Calculate the velocity of point P (vP) as shown below.

vP=ωAE×rP/A+vP/AE

Substitute ωAEk for ωAE, 0.4167j for rP/A, 8.334i+14.434j for vp, and uj for vP/AE.

8.334i+14.434j=ωAEk×(0.4167j)+uj=0.4167ωAEi+uj

Resolving i and j components as shown below.

8.334=0.4167ωAEωAE=20rad/s14.434=uu=14.434ft/s

Calculate the acceleration of rod BP (aP) as shown below.

aP=αBP×rP/BωBP2rP/B

Substitute 0 for αBP, 0.7217i0.4167j for rP/B, and 20rad/s for ωBP.

aP=0(20)2(0.7217i0.4167j)=288.68i+166.68j

Calculate the acceleration of point P with respect to point E (aP) as shown below.

aP=αAE×rP/AωAE2rP/A

Substitute αAEk for αAE, 0.4167j for rP/A, (20rad/s)k for ωAE.

aP=(αAEk)×(0.4167j)(20)2(0.4167j)=0.4166αAEi(400)(0.4167j)=0.4167αAEi+166.68j

Calculate the acceleration of point P (aP) as shown below.

aP=aP+aP/AE+2ωBP×vP/AE

Substitute 288.68i+166.68j for aP, 0.4167αAEi+166.68j for aP, 0 for aP/AE, (20rad/s)k, for ωBP, and 14.434j for vP/AE.

288.68i+166.68j=0.4167αAEi+166.68j+0+2(20k)×(14.434j)=0.4167αAEi+166.68j+577.36i=(0.4167αAE+577.36)i+166.68j

Resolving i and j components as shown below.

288.68=0.4167αAE+577.360.4167αAE=288.68αAE=692.8rad/s2

Calculate the mass of (m) as shown below.

m=Wg (1)

For rod AE.

Substitute 1.6 lb for W and 32.2ft/s2 for g.

mAE=1.632.2=0.049689lbs2/ft

For rod BP.

Substitute 2 lb for W and 32.2ft/s2 for g.

mBP=232.2=0.062112lbs2/ft

Calculate the mass moment of inertia (I¯) as shown below.

I¯=mL212 (2)

For rod AE.

Substitute 0.049689lbs2/ft for m and 8 in. for L in Equation (2).

IAE=(0.049689lbs2/ft)×(8in.×1ft12in.)212=0.00184lbs2ft

For rod BP.

Substitute 0.062112lbs2/ft for m and 10 in. for L in Equation (2).

IBP=(0.062112lbs2/ft)×(10in.×1ft12in.)212=0.00359lbs2ft

Calculate the position vector (r) as shown below.

The position of mass center G with respect to the rod AE.

rG/A=(4in.×1ft12in.)j=0.3333j

The position of mass center H with respect to the rod BP.

rH/B=(10in.×1ft12in.cos30°)i(10in.×1ft12in.sin30°)j=0.7217i0.4167j

Calculate the acceleration of point G (aG) as shown below.

aG=αAErG/AωAE2rG/A

Substitute (692.8rad/s2)k for αAE, 0.3333j for rG/A, and (20rad/s)k for ωAE.

aG=(692.8k)×(0.3333j)(20)2(0.3333j)=230.91i+133.32j

Calculate the acceleration of point H (aH) as shown below.

aH=αBPrH/BωBP2rH/B

Substitute 0 for αBP, 0.7217i0.4167j for rH/B, and (20rad/s)k for ωBP.

aH=0(20)2(0.7217i0.4167j)=288.68i+166.68j

Calculate the inertial terms of the mass center (ma) as shown below.

For rod AE.

mAEaG=(0.049689)(230.91i+133.32j)=11.474i+6.625j

For rod BP.

mBPaH=(0.062112)(288.68i+166.68j)=17.93i+10.353j

Calculate the effective couples at mass center (I¯α) as shown below.

For rod AE.

IAEαAE=(0.00184)(692.8k)=1.275k

For rod BP.

IBPαBP=(0.00359)(0)=0

Sketch the Free Body Diagram of rod AE as shown in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.141P , additional homework tip  1

Refer to Figure 1.

Apply the Equilibrium of moment about A as shown below.

MA=IGα+madrP/A×(Pi)=IAEαAE+rG/A×(mAEaG)

Substitute 0.4167j for rP/A, 1.275k for IAEαAE, 0.3333j for rG/A, and 11.474i+6.625j for mAEaG.

(0.4167j)×(Pi)=1.275k+(0.3333j)×(11.474i+6.625j)0.4167Pk=1.275k3.8243k0.4167P=5.0993P=12.24lb

Sketch the Free Body Diagram of rod BP as shown in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.141P , additional homework tip  2

Refer to Figure 2.

Apply the Equilibrium of moment about A as shown below.

MB=IGα+madrP/B×Pi+rH/B2×(WBPj)+Mk=rH/B×(mBPaH)+IBPαBP

Substitute 0.7217i0.4167j for rP/B, 12.24lb for P, 0.7217i0.4167j for rH/B, 2lb for WBP, 17.93i+10.353j for mBPaH, and 0 for IBPαBP.

[(0.7217i0.4167j)×12.24i+12(0.7217i0.4167j)×(2j)+Mk]=[(0.7217i0.4167j)×(17.93i+10.353j)+0]5.1004k+0.7217k+Mk=7.4718k+7.4714kMk=5.8225kM=5.82lbft

Hence, the couple M is 5.82lbft_.

(b)

Expert Solution
Check Mark
To determine

The components of the force exerted on AE by block.

Answer to Problem 16.141P

The components of the force exerted on AE by block is P=12.24lb_.

Explanation of Solution

Given information:

The weight of the rod AE is WAE=1.6lb.

The weight of the rod BP is WBP=2lb.

The length of the rod AE is LAE=8in.

The length of the rod BP is LBP=10in.

The angular velocity is ω=20rad/s.

Calculation:

Refer to part (a).

The components of the force exerted on AE by block P=12.24lb.

Therefore, the components of the force exerted on AE by block is P=12.24lb_.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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