VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)
VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)
12th Edition
ISBN: 9781260916942
Author: BEER
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
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_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
The 2-mass system shown below depicts a disk which rotates about its center and has rotational moment of inertia Jo and radius r. The angular displacement of the disk is given by 0. The spring with constant k₂ is attached to the disk at a distance from the center. The mass m has linear displacement & and is subject to an external force u. When the system is at equilibrium, the spring forces due to k₁ and k₂ are zero. Neglect gravity and aerodynamic drag in this problem. You may assume the small angle approximation which implies (i) that the springs and dampers remain in their horizontal / vertical configurations and (ii) that the linear displacement d of a point on the edge of the disk can be approximated by d≈re. Ө K2 www m 4 Cz 777777 Jo Make the following assumptions when analyzing the forces and torques: тв 2 0>0, 0>0, x> > 0, >0 Derive the differential equations of motion for this dynamic system. Start by sketching LARGE and carefully drawn free-body-diagrams for the disk and the…
A linear system is one that satisfies the principle of superposition. In other words, if an input u₁ yields the output y₁, and an input u2 yields the output y2, the system is said to be linear if a com- bination of the inputs u = u₁ + u2 yield the sum of the outputs y = y1 + y2. Using this fact, determine the output y(t) of the following linear system: given the input: P(s) = = Y(s) U(s) = s+1 s+10 u(t) = e−2+ sin(t) =e
The manometer fluid in the figure given below is mercury where D = 3 in and h = 1 in. Estimate the volume flow in the tube (ft3/s) if the flowing fluid is gasoline at 20°C and 1 atm. The density of mercury and gasoline are 26.34 slug/ft3 and 1.32 slug/ft3 respectively. The gravitational force is 32.2 ft/s2.

Chapter 14 Solutions

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

Ch. 14.1 - A system consists of three identical 19.32-lb...Ch. 14.1 - A system consists of three identical 19.32-lb...Ch. 14.1 - A system consists of three particles A, B, and C....Ch. 14.1 - For the system of particles of Prob. 14.13,...Ch. 14.1 - A 13-kg projectile is passing through the origin O...Ch. 14.1 - Prob. 14.16PCh. 14.1 - A 2-kg model rocket is launched vertically and...Ch. 14.1 - An 18-kg cannonball and a 12-kg cannonball are...Ch. 14.1 - 14.19 and 14.20 Cruiser A was traveling east at 60...Ch. 14.1 - 14.19 and 14.20 Cruiser A was traveling east at 60...Ch. 14.1 - Prob. 14.21PCh. 14.1 - Two spheres, each of mass m, can slide freely on a...Ch. 14.1 - In a game of pool, ball A is moving with a...Ch. 14.1 - Prob. 14.24PCh. 14.1 - Prob. 14.25PCh. 14.1 - In a scattering experiment, an alpha particle A is...Ch. 14.1 - Derive the relation HO=rmv+HG between the angular...Ch. 14.1 - Prob. 14.28PCh. 14.1 - Prob. 14.29PCh. 14.1 - Show that the relation MA=HA, where HA is defined...Ch. 14.2 - Determine the energy lost due to friction and the...Ch. 14.2 - In Prob. 14.3, determine the energy lost (a) when...Ch. 14.2 - Prob. 14.33PCh. 14.2 - Determine the energy lost as a result of the...Ch. 14.2 - Prob. 14.35PCh. 14.2 - Prob. 14.36PCh. 14.2 - Prob. 14.37PCh. 14.2 - Ball B is suspended from a cord of length l...Ch. 14.2 - A 15-lb block B starts from rest and slides on the...Ch. 14.2 - A 40-lb block B is suspended from a 6-ft cord...Ch. 14.2 - Prob. 14.41PCh. 14.2 - 14.41 and 14.42 In a game of pool, ball A is...Ch. 14.2 - Prob. 14.43PCh. 14.2 - In a game of pool, ball A is moving with the...Ch. 14.2 - Prob. 14.45PCh. 14.2 - Prob. 14.46PCh. 14.2 - Four small disks A, B, C, and D can slide freely...Ch. 14.2 - In the scattering experiment of Prob. 14.26, it is...Ch. 14.2 - Three identical small spheres, each weighing 2 lb,...Ch. 14.2 - Three small spheres A, B, and C, each of mass m,...Ch. 14.2 - Prob. 14.51PCh. 14.2 - Prob. 14.52PCh. 14.2 - Two small disks A and B of mass 3 kg and 1.5 kg,...Ch. 14.2 - Two small disks A and B of mass 2 kg and 1 kg,...Ch. 14.2 - Three small identical spheres A, B, and C, which...Ch. 14.2 - Prob. 14.56PCh. 14.3 - A stream of water with a density of = 1000 kg/m3...Ch. 14.3 - A jet ski is placed in a channel and is tethered...Ch. 14.3 - Tree limbs and branches are being fed at A at the...Ch. 14.3 - Prob. 14.60PCh. 14.3 - Prob. 14.61PCh. 14.3 - Prob. 14.62PCh. 14.3 - Prob. 14.63PCh. 14.3 - Prob. 14.64PCh. 14.3 - Prob. 14.65PCh. 14.3 - Prob. 14.66PCh. 14.3 - Prob. 14.67PCh. 14.3 - Prob. 14.68PCh. 14.3 - Prob. 14.69PCh. 14.3 - Prob. 14.70PCh. 14.3 - Prob. 14.71PCh. 14.3 - Prob. 14.72PCh. 14.3 - Prob. 14.73PCh. 14.3 - Prob. 14.74PCh. 14.3 - Prob. 14.75PCh. 14.3 - Prob. 14.76PCh. 14.3 - The propeller of a small airplane has a...Ch. 14.3 - Prob. 14.78PCh. 14.3 - Prob. 14.79PCh. 14.3 - Prob. 14.80PCh. 14.3 - Prob. 14.81PCh. 14.3 - Prob. 14.82PCh. 14.3 - Prob. 14.83PCh. 14.3 - Prob. 14.84PCh. 14.3 - Prob. 14.85PCh. 14.3 - Prob. 14.86PCh. 14.3 - Solve Prob. 14.86, assuming that the chain is...Ch. 14.3 - Prob. 14.88PCh. 14.3 - Prob. 14.89PCh. 14.3 - Prob. 14.90PCh. 14.3 - Prob. 14.91PCh. 14.3 - Prob. 14.92PCh. 14.3 - A rocket sled burns fuel at the constant rate of...Ch. 14.3 - Prob. 14.94PCh. 14.3 - Prob. 14.95PCh. 14.3 - Prob. 14.96PCh. 14.3 - Prob. 14.97PCh. 14.3 - Prob. 14.98PCh. 14.3 - Determine the distance traveled by the spacecraft...Ch. 14.3 - A rocket weighs 2600 lb, including 2200 lb of...Ch. 14.3 - Determine the altitude reached by the spacecraft...Ch. 14.3 - Prob. 14.102PCh. 14.3 - Prob. 14.103PCh. 14.3 - Prob. 14.104PCh. 14 - Three identical cars are being unloaded from an...Ch. 14 - A 50-kg mother and her 26-kg son are sledding down...Ch. 14 - An 80-Mg railroad engine A coasting at 6.5 km/h...Ch. 14 - Prob. 14.108RPCh. 14 - Mass C, which has a mass of 4 kg, is suspended...Ch. 14 - Prob. 14.110RPCh. 14 - A 6000-kg dump truck has a 1500-kg stone block...Ch. 14 - For the ceiling-mounted fan shown, determine the...Ch. 14 - Prob. 14.113RPCh. 14 - Prob. 14.114RPCh. 14 - Prob. 14.115RPCh. 14 - A chain of length l and mass m falls through a...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Dynamics - Lesson 1: Introduction and Constant Acceleration Equations; Author: Jeff Hanson;https://www.youtube.com/watch?v=7aMiZ3b0Ieg;License: Standard YouTube License, CC-BY