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 16.142P

Two rotating rods in the vertical plane are connected by a slider block P of negligible mass. The rod attached at A has a mass of 0.8 kg and a length of 160 mm. Rod BP has a mass of 1 kg and is 200 mm long and the friction between block P and AE is negligible. The motion of the system is controlled by a couple M applied to bar BP. Knowing that at the instant shown rod BP has an angular velocity of 20 rad/s clockwise and an angular acceleration of 80 rad/s2 clockwise, determine (a) the couple M, (b) the components of the force exerted on AE by block P.

Chapter 16.2, Problem 16.142P, 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 16.142P

The value of couple M is 7.19Nm_.

Explanation of Solution

Given information:

The mass of the rod AE is mAE=0.8kg.

The mass of the rod BP is mBP=1kg.

The length of the rod AE is LAE=160mm

The length of the rod BP is LBP=200mm

The angular velocity is ω=20rad/s.

The angular acceleration is α=80rad/s2.

Calculation:

Consider the acceleration due to gravity as g=9.81m/s2.

Calculate the position vector (r) as shown below.

The position of P with respect to A.

rP/A=(200mm×1m1,000mmsin30°)=(0.1m)j

The position of P with respect to B.

rP/B=(200mm×1m1,000mmcos30°)(200mm×1m1,000mmsin30°)=(0.1732i0.1j)m

The position of P with respect to E.

rP/E=(160mm×1m1,000mm)j=(0.16m)j

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

The angular acceleration of rod BP in vector form is αBP=(80rad/s2)k.

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

vP=ωBP×rP/B

Substitute (20rad/s)k for ωBP, and (0.1732i0.1j)m for rP/B.

vP=(20k)×(0.1732i0.1j)=3.464j2i=2i+3.464j

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, (0.1m)j for rP/A, 2i+3.464j for vp, and uj for vP/AE.

2i+3.464j=ωAEk×(0.1j)+uj=0.1ωAEi+uj

Resolving i and j components as shown below.

2=0.1ωAEωAE=20rad/s3.464=uu=3.464m/s

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

aP=αBP×rP/BωBP2rP/B

Substitute (80rad/s2)k for αBP, (0.1732i0.1j)m for rP/B, and 20rad/s for ωBP.

aP=80k×(0.1732i0.1j)(20)2(0.1732i0.1j)=13.855j8i+69.28i+40j=61.28i+53.855j

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

aP=αAE×rP/AωAE2rP/A

Substitute αAEk for αAE, (0.1m)j for rP/A, 20rad/s for ωAE.

aP=(αAEk)×(0.1j)(20)2(0.1j)=0.1αAEi+40j

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

aP=aP+aP/AE+2ωBP×vP/AE

Substitute 61.28i+53.855j for aP, 0.1αAEi+40j for aP, 0 for aP/AE, (20rad/s)k, for ωBP, and 3.464j for vP/AE.

61.28i+53.855j=0.1αAEi+40j+0+2(20k)×(3.464j)=0.1αAEi+40j+138.56i=(0.1αAE+138.56)i+40j

Resolving i and j components as shown below.

61.28=0.1αAE+138.560.1αAE=77.28αAE=772.8rad/s2

Calculate the weight of (m) as shown below.

W=mg (1)

For rod AE.

Substitute 0.8 kg for W and 9.81m/s2 for g.

WAE=0.8×9.81=7.848kgm/s2×1N1kgm/s2=7.848N

For rod BP.

Substitute 1 kg for W and 9.81m/s2 for g.

WBP=1×9.81=9.81kgm/s2×1N1kgm/s2=9.81N

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

I¯=mL212 (2)

For rod AE.

Substitute 0.8kg for m and 160 mm for L in Equation (2).

IAE=(0.8kg)×(160mm×1m1,000mm)212=0.001707kgm2

For rod BP.

Substitute 1kg for m and 200 mm for L in Equation (2).

IBP=(1kg)×(200mm×1m1,000mm)212=0.003333kgm2

Calculate the position vector (r) as shown below.

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

rG/A=(80mm×1m1,000mm)j=(0.08m)j

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

rH/B=(100mm×1m1,000mmcos30°)i(100mm×1m1,000mmsin30°)j=0.0866i0.05j

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

aG=αAErG/AωAE2rG/A

Substitute (772.8rad/s2)k for αAE, (0.08m)j for rG/A, and (20rad/s) for ωAE.

aG=(772.8k)×(0.08j)(20)2(0.08j)=61.824i+32j

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

aH=αBPrH/BωBP2rH/B

Substitute (80rad/s2)k for αBP, 0.0866i0.05j for rH/B, and (20rad/s)k for ωBP.

aH=(80k)×(0.0866i0.05j)(20)2(0.0866i0.05j)=6.928j4j+34.64i+20j=34.64i+22.928j

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

For rod AE.

mAEaG=(0.8)(61.824i+32j)=49.46i+25.6j

For rod BP.

mBPaH=(1)(34.64i+22.928j)=34.64i+22.928j

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

For rod AE.

IAEαAE=(0.001707)(772.8k)=1.319k

For rod BP.

IBPαBP=(0.003333)(80k)=0.2667

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

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.142P , additional homework tip  1

Refer to Figure 1.

Apply the Equilibrium of moment about A as shown below.

MA=IGα+mad0.1×P=0.08×49.461.3190.1P=5.2758P=52.758N

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

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16.2, Problem 16.142P , additional homework tip  2

Refer to Figure 2.

Apply the Equilibrium of moment about A as shown below.

MB=IGα+mad52.758×0.1+9.81×0.0866+M=8×0.10.26676.1253+M=1.0667M=7.192Nm

Hence, the couple M is 7.19Nm_.

(b)

Expert Solution
Check Mark
To determine

Find the components of the force exerted on AE by block.

Answer to Problem 16.142P

The components of the force exerted on AE by block is P=52.8N_.

Explanation of Solution

Given information:

The mass of the rod AE is mAE=0.8kg.

The mass of the rod BP is mBP=1kg.

The length of the rod AE is LAE=160mm

The length of the rod BP is LBP=200mm

The angular velocity is ω=20rad/s.

The angular acceleration is α=80rad/s2.

Calculation:

Refer to part (a).

The components of the force exerted on AE by block P=52.758N.

Therefore, the components of the force exerted on AE by block is P=52.8N_.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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