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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Textbook Question
Chapter 13.3, Problem 13.129P

The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track. The brakes are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the time required to bring the train to a stop, (b) the force in each coupling.

Chapter 13.3, Problem 13.129P, The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the

(a)

Expert Solution
Check Mark
To determine

Find the time required to bring the train (t) to a stop.

Answer to Problem 13.129P

The time required to bring the train (t) to a stop is 5.64sec_.

Explanation of Solution

Given information:

The initial speed of the train (v1) is 30mi/h.

The coefficient of kinetic friction (μk) is 0.35.

The weight of the rail car A (WA) is 40tons.

The weight of the rail car B (WB) is 50tons.

The weight of the rail car C (WC) is 40tons.

The acceleration due to gravity (g) is 32.2ft/s2.

Calculation:

Show the impulse momentum diagram for the entire train as Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 13.3, Problem 13.129P , additional homework tip  1

Convert the initial speed of the train (v1) from mi/h to ft/s:

v1=(v1)mi/h×(5280ftmi)×(1hr3600sec)

Here, (v1)mi/h is the initial speed of the train in mi/h.

Substitute 30mi/h for (v1)mi/h.

v1=30×(5280ftmi)×(1hr3600sec)=44ft/s

Calculate the masses of the rail cars A (mA) using the relation:

mA=WAg

Substitute 40tons for WA and 32.2ft/s2 for g.

mA=40tons32.2ft/s2=4032.2×(2000lbs1ton)=2484lb.s2/ft

Calculate the mass of the rail car B (mB) using the relation:

mB=WBg

Substitute 50tons for WB and 32.2ft/s2 for g.

mB=50tons32.2ft/s2=5032.2×(2000lbs1ton)=3106lb.s2/ft

Calculate the mass of the rail car C (mC) using the relation:

mC=WCg

Substitute 40tons for WC and 32.2ft/s2 for g.

mC=40tons32.2ft/s2=4032.2×(2000lbs1ton)=2484lb.s2/ft

Calculate the frictional force acting on the car B after application of brakes [(Ff)B] using the relation:

(Ff)B=μkmBg

Substitute 3106lb.s2/ft for mB, 0.35 for μk, and 32.2ft/s2 for g.

(Ff)B=(0.35)(3106lb.s2/ft)(32.2ft/s2)=35000lb

Calculate the frictional force acting on the car C after application of brakes [(Ff)C] using the relation:

(Ff)C=μkmCg

Substitute 2484lb.s2/ft for mC, 0.35 for μk, and 32.2ft/s2 for g.

(Ff)C=(0.35)(2484lb.s2/ft)(32.2ft/s2)=28000lb

The brakes are not applied on the wheels of car A [(Ff)A] so there is no frictional force acting on the car A that is zero.

Calculate the total mass of the train (m) using the relation:

m=mA+mB+mC

Substitute 2484lb.s2/ft for mA, 2484lb.s2/ft for mC, and 3106lb.s2/ft for mB.

m=(2484lb.s2/ft)+(3106lb.s2/ft)+(2484lb.s2/ft)=8074lb.s2/ft

Calculate the total frictional force acting on the whole train (Ff) using the relation:

Ff=[(Ff)A+(Ff)B+(Ff)C]

Substitute 0 for (Ff)A, 28000lb for (Ff)C and 35000lb for (Ff)C.

Ff=[0+(35000lb)+(28000lb)]=63000lb

The expression for the impulse acting on the train due to frictional force (ImpT) as follows:

ImpT=Fft

Here, t is the time taken by the train to come to rest.

Use the principle of impulse-momentum to the entire train to find the time taken by the train to stop by application of brakes.

The expression or the principle of impulse-momentum as follows:

mv1+ImpT=mv2

Substitute Fft for ImpT.

mv1+Fft=mv2Fft=m(v2v1)t=m(v2v1)Ff

Substitute 8074lb.s2/ft for m, 44ft/s for v1, 0 for v2, and 63000lb for Ff.

t=(8074lb.s2/ft)(044ft/s)63000lb=35525663000=5.64sec

Therefore, the time required to bring the train (t) to a stop is 5.64sec_.

(b)

Expert Solution
Check Mark
To determine

Find the force in each coupling.

Answer to Problem 13.129P

The force in AB (FAB) and BC (FBC) are 19380lb_ and 8620lb_ respectively.

Explanation of Solution

Given information:

The initial speed of the train (v1) is 30mi/h.

The coefficient of kinetic friction (μk) is 0.35.

The weight of the rail car A (WA) is 40tons.

The weight of the rail car B (WB) is 50tons.

The weight of the rail car C (WC) is 40tons.

The acceleration due to gravity (g) is 32.2ft/s2.

Calculation:

Show the impulse-momentum diagram of rail car A as in Figure (2).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 13.3, Problem 13.129P , additional homework tip  2

The expression for the impulse acting on the rail car A (ImpA) as follows:

ImpA=[(Ff)A+FAB]t

Here, FAB is the coupling force acting between the rail car A and B.

The expression for the principle of impulse-momentum to rail car A alone as follows:

mAv1+ImpA=mAv2

Substitute [(Ff)A+FAB]t for ImpA.

mAv1[(Ff)A+FAB]t=mAv2

Substitute 2484lb.s2/ft for mA, 44ft/s for v1, 0 for (Ff)A, 0 for v2, and 5.64sec for t.

(2484lb.s2/ft)(44ft/s)[0+FAB](5.64sec)=0(5.64)FAB=109,296FAB=109,2965.64FAB=19,379lbFAB19,380lb

Show the impulse-momentum diagram of rail car C as in Figure (3).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 13.3, Problem 13.129P , additional homework tip  3

The expression for the impulse acting on the rail car C ,(ImpC) as follows:

ImpC=[FBC(Ff)C]t

Here, FBC is the coupling force acting between the rail car B and C.

The expression for principle of impulse-momentum to car C alone as follows:

mCv1+ImpC=mCv2

Substitute [FBC(Ff)C]t for ImpC.

mCv1+[FBC(Ff)C]t=mCv2

Substitute 2484lb.s2/ft for mC, 44ft/s for v1, 28000lb for (Ff)C, 0 for v2, and 5.64sec for t.

(2,484lb.s2/ft)(44ft/s)+[FBC28,000lb](5.64s)=0(5.64)FBC157,920=109,296(5.64)FBC=48,624FBC=48,6245.64FBC=8,620lb

Therefore, the force in AB (FAB) and BC (FBC) are 19,380lb_ and 8,620lb_ respectively.

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Chapter 13 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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