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.

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Answer 1

Textbook Question

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

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 5.727 ft/s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 1

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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=120 ton32.2 ft/s2(2000 lb1 ton)=7453.42 lb.s2/ft

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 2

Apply Impulse momentum principle

0+FΔt=mBvB FΔt=(6211.18 lb.s2/ft)vB→(1)

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 3

Apply impulse momentum principle

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

The relative velocity vT/B

vT/B=vT−vB=5.4 ft/s

According to above equations

FΔt=(6211.18 lb.s2/ft)vB(7453.42 lb.s2/ft)(6 ft/s)−(7453.42 lb.s2/ft)vT=(6211.18 lb.s2/ft)[vT−5.4 ft/s]

Solve

vT=5.727 ft/s

Conclusion:

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

Expert Solution

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 2031 lb.s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 4

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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 5.727 ft/s.

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

The velocity vB of barge when cable becomes taut

vT−vB=5.4 ft/s vB=5.727 ft/s−5.4 ft/s vB =0.327 ft/s

The impulse FΔt exerted on the barge

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

Conclusion:

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

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Answer 2

Textbook Question

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

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 5.727 ft/s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 1

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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=120 ton32.2 ft/s2(2000 lb1 ton)=7453.42 lb.s2/ft

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 2

Apply Impulse momentum principle

0+FΔt=mBvB FΔt=(6211.18 lb.s2/ft)vB→(1)

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 3

Apply impulse momentum principle

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

The relative velocity vT/B

vT/B=vT−vB=5.4 ft/s

According to above equations

FΔt=(6211.18 lb.s2/ft)vB(7453.42 lb.s2/ft)(6 ft/s)−(7453.42 lb.s2/ft)vT=(6211.18 lb.s2/ft)[vT−5.4 ft/s]

Solve

vT=5.727 ft/s

Conclusion:

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

Expert Solution

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 2031 lb.s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 4

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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 5.727 ft/s.

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

The velocity vB of barge when cable becomes taut

vT−vB=5.4 ft/s vB=5.727 ft/s−5.4 ft/s vB =0.327 ft/s

The impulse FΔt exerted on the barge

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

Conclusion:

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

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Answer 3

Textbook Question

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

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 5.727 ft/s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 1

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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=120 ton32.2 ft/s2(2000 lb1 ton)=7453.42 lb.s2/ft

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 2

Apply Impulse momentum principle

0+FΔt=mBvB FΔt=(6211.18 lb.s2/ft)vB→(1)

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 3

Apply impulse momentum principle

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

The relative velocity vT/B

vT/B=vT−vB=5.4 ft/s

According to above equations

FΔt=(6211.18 lb.s2/ft)vB(7453.42 lb.s2/ft)(6 ft/s)−(7453.42 lb.s2/ft)vT=(6211.18 lb.s2/ft)[vT−5.4 ft/s]

Solve

vT=5.727 ft/s

Conclusion:

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

Expert Solution

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 2031 lb.s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 4

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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 5.727 ft/s.

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

The velocity vB of barge when cable becomes taut

vT−vB=5.4 ft/s vB=5.727 ft/s−5.4 ft/s vB =0.327 ft/s

The impulse FΔt exerted on the barge

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

Conclusion:

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

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Answer 4

Textbook Question

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

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 5.727 ft/s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 1

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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=120 ton32.2 ft/s2(2000 lb1 ton)=7453.42 lb.s2/ft

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 2

Apply Impulse momentum principle

0+FΔt=mBvB FΔt=(6211.18 lb.s2/ft)vB→(1)

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 3

Apply impulse momentum principle

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

The relative velocity vT/B

vT/B=vT−vB=5.4 ft/s

According to above equations

FΔt=(6211.18 lb.s2/ft)vB(7453.42 lb.s2/ft)(6 ft/s)−(7453.42 lb.s2/ft)vT=(6211.18 lb.s2/ft)[vT−5.4 ft/s]

Solve

vT=5.727 ft/s

Conclusion:

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

Expert Solution

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 2031 lb.s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 4

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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 5.727 ft/s.

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

The velocity vB of barge when cable becomes taut

vT−vB=5.4 ft/s vB=5.727 ft/s−5.4 ft/s vB =0.327 ft/s

The impulse FΔt exerted on the barge

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

Conclusion:

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

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

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 5.727 ft/s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 1

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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=120 ton32.2 ft/s2(2000 lb1 ton)=7453.42 lb.s2/ft

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 2

Apply Impulse momentum principle

0+FΔt=mBvB FΔt=(6211.18 lb.s2/ft)vB→(1)

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 3

Apply impulse momentum principle

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

The relative velocity vT/B

vT/B=vT−vB=5.4 ft/s

According to above equations

FΔt=(6211.18 lb.s2/ft)vB(7453.42 lb.s2/ft)(6 ft/s)−(7453.42 lb.s2/ft)vT=(6211.18 lb.s2/ft)[vT−5.4 ft/s]

Solve

vT=5.727 ft/s

Conclusion:

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

Expert Solution

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 2031 lb.s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 4

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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 5.727 ft/s.

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

The velocity vB of barge when cable becomes taut

vT−vB=5.4 ft/s vB=5.727 ft/s−5.4 ft/s vB =0.327 ft/s

The impulse FΔt exerted on the barge

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

Conclusion:

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

Answer 8

Expert Solution

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 5.727 ft/s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 1

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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=120 ton32.2 ft/s2(2000 lb1 ton)=7453.42 lb.s2/ft

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 2

Apply Impulse momentum principle

0+FΔt=mBvB FΔt=(6211.18 lb.s2/ft)vB→(1)

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 3

Apply impulse momentum principle

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

The relative velocity vT/B

vT/B=vT−vB=5.4 ft/s

According to above equations

FΔt=(6211.18 lb.s2/ft)vB(7453.42 lb.s2/ft)(6 ft/s)−(7453.42 lb.s2/ft)vT=(6211.18 lb.s2/ft)[vT−5.4 ft/s]

Solve

vT=5.727 ft/s

Conclusion:

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

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

Velocity of the tugboat after the cable becomes taut

Answer 13

Answer to Problem 13.145P

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

Answer 14

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 1

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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=120 ton32.2 ft/s2(2000 lb1 ton)=7453.42 lb.s2/ft

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

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

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 2

Apply Impulse momentum principle

0+FΔt=mBvB FΔt=(6211.18 lb.s2/ft)vB→(1)

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 3

Apply impulse momentum principle

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

The relative velocity vT/B

vT/B=vT−vB=5.4 ft/s

According to above equations

FΔt=(6211.18 lb.s2/ft)vB(7453.42 lb.s2/ft)(6 ft/s)−(7453.42 lb.s2/ft)vT=(6211.18 lb.s2/ft)[vT−5.4 ft/s]

Solve

vT=5.727 ft/s

Conclusion:

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

Answer 15

Expert Solution

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 2031 lb.s.

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 4

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

Cable is being unwound at a constant rate of 5.4 ft/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 5.727 ft/s.

Mass mB of barge

mB=100 ton32.2 ft/s2(2000 lb1 ton)=6211.18 lb.s2/ft

The velocity vB of barge when cable becomes taut

vT−vB=5.4 ft/s vB=5.727 ft/s−5.4 ft/s vB =0.327 ft/s

The impulse FΔt exerted on the barge

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

Conclusion:

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

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

Impulse exerted on the barge as the cable becomes taut

Answer 20

Answer to Problem 13.145P

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

Answer 21

Explanation of Solution

Given information:

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS, Chapter 13.3, Problem 13.145P , additional homework tip 4

Weight of tugboat is 120 ton.

Weight of barge is 100 ton.

The velocity of tugboat is 6 ft/s.

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