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 13.3, Problem 13.145P

A 120-ton tugboat is moving at 6 ft/s with a slack towing cable attached to a 100-ton barge that is at rest. The cable is being unwound from a drum on the tugboat at a constant rate of 5.4 ft/s and that rate is maintained after the cable becomes taut. Neglecting the resistance of the water, determine (a) the velocity of the tugboat after the cable becomes taut, (b) the impulse exerted on the barge as the cable becomes taut.

Chapter 13.3, Problem 13.145P, A 120-ton tugboat is moving at 6 ft/s with a slack towing cable attached to a 100-ton barge that is

Expert Solution
Check Mark
To determine

(a)

Velocity of the tugboat after the cable becomes taut

Answer to Problem 13.145P

The velocity of the tugboat after the cable becomes taut is equal to 5.727ft/s.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.3, Problem 13.145P , additional homework tip  1

Weight of tugboat is 120ton.

Weight of barge is 100ton.

The velocity of tugboat is 6ft/s.

Cable is being unwound at a constant rate of 5.4ft/s.

“A force acting on a particle during a very short time interval but large enough to produce a definite change in momentum is called an impulsive force.”

Impulse momentum principle for impulsive motion is defined as,

mv1+FavgΔt=mv2

Calculation:

Mass mT of tugboat

mT=120ton32.2ft/s2(2000lb1ton)=7453.42lb.s2/ft

Mass mB of barge

mB=100ton32.2ft/s2(2000lb1ton)=6211.18lb.s2/ft

Assume vT,vB as end velocities of tug boat and barge respectively.

Vector Mechanics For Engineers, Chapter 13.3, Problem 13.145P , additional homework tip  2

Apply Impulse momentum principle

0+FΔt=mBvBFΔt=(6211.18lb.s2/ft)vB(1)

Vector Mechanics For Engineers, Chapter 13.3, Problem 13.145P , additional homework tip  3

Apply impulse momentum principle

(7453.42lb.s2/ft)(6ft/s)FΔt=(7453.42lb.s2/ft)vT(2)

The relative velocity vT/B

vT/B=vTvB=5.4ft/s

According to above equations

FΔt=(6211.18lb.s2/ft)vB(7453.42lb.s2/ft)(6ft/s)(7453.42lb.s2/ft)vT=(6211.18lb.s2/ft)[vT5.4ft/s]

Solve

vT=5.727ft/s

Conclusion:

The velocity of the tugboat after the cable becomes taut is equal to 5.727ft/s.

Expert Solution
Check Mark
To determine

(b)

Impulse exerted on the barge as the cable becomes taut

Answer to Problem 13.145P

The impulse FΔt exerted on the barge is equal to 2031lb.s.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.3, Problem 13.145P , additional homework tip  4

Weight of tugboat is 120ton.

Weight of barge is 100ton.

The velocity of tugboat is 6ft/s.

Cable is being unwound at a constant rate of 5.4ft/s.

“A force acting on a particle during a very short time interval but large enough to produce a definite change in momentum is called an impulsive force.”

Impulse momentum principle for impulsive motion is defined as

mv1+FavgΔt=mv2

Calculation:

According to sub part A

Velocity vT of the barge when cable becomes taut is 5.727ft/s.

Mass mB of barge

mB=100ton32.2ft/s2(2000lb1ton)=6211.18lb.s2/ft

The velocity vB of barge when cable becomes taut

vTvB=5.4ft/svB=5.727ft/s5.4ft/svB=0.327ft/s

The impulse FΔt exerted on the barge

FΔt=(6211.18lb.s2/ft)vBFΔt=(6211.18lb.s2/ft)(0.327ft/s)=2031lb.s

Conclusion:

The impulse FΔt exerted on the barge is equal to 2031lb.s.

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Chapter 13 Solutions

Vector Mechanics For Engineers

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