An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as to extend inward and outward. It can be used to simulate driving, fixed-wing aircraft flying, and helicopter maneuvering. In one training scenario, the trainer rotates and translates in the horizontal plane, where the location of the pilot is defined by the relationships r = 10 + 2 cos ( π 3 t ) and θ = 0.1 ( 2 t 2 − t ) , where r , θ , and t are expressed in feet, radians: and seconds, respectively. Knowing that the pilot has a weight of 175 Ibs, ( a ) determine the magnitude of the resulting force acting on the pilot at t = 5 s, ( b ) plot the magnitudes of the radial and transverse components of the force exerted on the pilot from 0 to 10 seconds.

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Answer 1

Textbook Question

Chapter 12.1, Problem 12.66P

An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as to extend inward and outward. It can be used to simulate driving, fixed-wing aircraft flying, and helicopter maneuvering. In one training scenario, the trainer rotates and translates in the horizontal plane, where the location of the pilot is defined by the relationships r = 10 + 2 cos ( π 3 t ) and θ = 0.1 ( 2 t 2 − t ) , where r , θ , and t are expressed in feet, radians: and seconds, respectively. Knowing that the pilot has a weight of 175 Ibs, (a) determine the magnitude of the resulting force acting on the pilot at t = 5 s, (b) plot the magnitudes of the radial and transverse components of the force exerted on the pilot from 0 to 10 seconds.

Chapter 12.1, Problem 12.66P, An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as

Expert Solution

To determine

(a)

The magnitude of the resulting force on pilot.

Answer to Problem 12.66P

We got force F=230.113 lb

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]Fr=175[−2π29cos(π3×5)−(10+2cos(π3×5))×(0.1(4×5−1))2]Fr=−7141.2087 lb-ft/s2

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]Fθ=175[0.4(10+2cos(π3×5))+2(−2π3sin(π3×5))(0.1(4×5−1))]Fθ=1976.333 lb-ft/s2

Resultant force,

F=Fr2+Fθ2F=(−7141.2087)2+1976.3332F=7409.64 lb-ft/s2F=7409.6432.2 lbF=230.113 lb

Conclusion:

We got force F=230.113 lb

Expert Solution

To determine

(b)

Plot the radial and transverse components of force.

Answer to Problem 12.66P

Plot is in explanation part.

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]

Table of Fr, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 1

Table of Fθ, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 2

Conclusion:

Plots are mentioned the explanation part.

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P3: A differential band brake shown in the figure below uses a woven lining having a design value of the friction coefficient f=0.20. Dimensions are b=80 mm, r=250 mm, c=700 mm, a = 150 mm, s=35 mm, and 0=240°. Find 1) the brake torque if the maximum lining pressure is 0.5 MPa, 2) the corresponding actuating force F, and 3) the values of dimensions that would cause the brake to be self-locking. (25%) -240° F-250 mm Band width, b-80 mm Rotation Friction coefficient, -0.20 Maximum lining pressure, P-0.5 MPa 3-35 mm la-150 mm e-700 mm-

Answer 2

Textbook Question

Chapter 12.1, Problem 12.66P

An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as to extend inward and outward. It can be used to simulate driving, fixed-wing aircraft flying, and helicopter maneuvering. In one training scenario, the trainer rotates and translates in the horizontal plane, where the location of the pilot is defined by the relationships r = 10 + 2 cos ( π 3 t ) and θ = 0.1 ( 2 t 2 − t ) , where r , θ , and t are expressed in feet, radians: and seconds, respectively. Knowing that the pilot has a weight of 175 Ibs, (a) determine the magnitude of the resulting force acting on the pilot at t = 5 s, (b) plot the magnitudes of the radial and transverse components of the force exerted on the pilot from 0 to 10 seconds.

Chapter 12.1, Problem 12.66P, An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as

Expert Solution

To determine

(a)

The magnitude of the resulting force on pilot.

Answer to Problem 12.66P

We got force F=230.113 lb

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]Fr=175[−2π29cos(π3×5)−(10+2cos(π3×5))×(0.1(4×5−1))2]Fr=−7141.2087 lb-ft/s2

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]Fθ=175[0.4(10+2cos(π3×5))+2(−2π3sin(π3×5))(0.1(4×5−1))]Fθ=1976.333 lb-ft/s2

Resultant force,

F=Fr2+Fθ2F=(−7141.2087)2+1976.3332F=7409.64 lb-ft/s2F=7409.6432.2 lbF=230.113 lb

Conclusion:

We got force F=230.113 lb

Expert Solution

To determine

(b)

Plot the radial and transverse components of force.

Answer to Problem 12.66P

Plot is in explanation part.

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]

Table of Fr, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 1

Table of Fθ, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 2

Conclusion:

Plots are mentioned the explanation part.

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Answer 3

Textbook Question

Chapter 12.1, Problem 12.66P

An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as to extend inward and outward. It can be used to simulate driving, fixed-wing aircraft flying, and helicopter maneuvering. In one training scenario, the trainer rotates and translates in the horizontal plane, where the location of the pilot is defined by the relationships r = 10 + 2 cos ( π 3 t ) and θ = 0.1 ( 2 t 2 − t ) , where r , θ , and t are expressed in feet, radians: and seconds, respectively. Knowing that the pilot has a weight of 175 Ibs, (a) determine the magnitude of the resulting force acting on the pilot at t = 5 s, (b) plot the magnitudes of the radial and transverse components of the force exerted on the pilot from 0 to 10 seconds.

Chapter 12.1, Problem 12.66P, An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as

Expert Solution

To determine

(a)

The magnitude of the resulting force on pilot.

Answer to Problem 12.66P

We got force F=230.113 lb

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]Fr=175[−2π29cos(π3×5)−(10+2cos(π3×5))×(0.1(4×5−1))2]Fr=−7141.2087 lb-ft/s2

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]Fθ=175[0.4(10+2cos(π3×5))+2(−2π3sin(π3×5))(0.1(4×5−1))]Fθ=1976.333 lb-ft/s2

Resultant force,

F=Fr2+Fθ2F=(−7141.2087)2+1976.3332F=7409.64 lb-ft/s2F=7409.6432.2 lbF=230.113 lb

Conclusion:

We got force F=230.113 lb

Expert Solution

To determine

(b)

Plot the radial and transverse components of force.

Answer to Problem 12.66P

Plot is in explanation part.

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]

Table of Fr, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 1

Table of Fθ, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 2

Conclusion:

Plots are mentioned the explanation part.

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Answer 4

Textbook Question

Chapter 12.1, Problem 12.66P

An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as to extend inward and outward. It can be used to simulate driving, fixed-wing aircraft flying, and helicopter maneuvering. In one training scenario, the trainer rotates and translates in the horizontal plane, where the location of the pilot is defined by the relationships r = 10 + 2 cos ( π 3 t ) and θ = 0.1 ( 2 t 2 − t ) , where r , θ , and t are expressed in feet, radians: and seconds, respectively. Knowing that the pilot has a weight of 175 Ibs, (a) determine the magnitude of the resulting force acting on the pilot at t = 5 s, (b) plot the magnitudes of the radial and transverse components of the force exerted on the pilot from 0 to 10 seconds.

Chapter 12.1, Problem 12.66P, An advanced spatial disorientation trainer allows the cab to rotate around multiple axes, as well as

Expert Solution

To determine

(a)

The magnitude of the resulting force on pilot.

Answer to Problem 12.66P

We got force F=230.113 lb

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]Fr=175[−2π29cos(π3×5)−(10+2cos(π3×5))×(0.1(4×5−1))2]Fr=−7141.2087 lb-ft/s2

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]Fθ=175[0.4(10+2cos(π3×5))+2(−2π3sin(π3×5))(0.1(4×5−1))]Fθ=1976.333 lb-ft/s2

Resultant force,

F=Fr2+Fθ2F=(−7141.2087)2+1976.3332F=7409.64 lb-ft/s2F=7409.6432.2 lbF=230.113 lb

Conclusion:

We got force F=230.113 lb

Expert Solution

To determine

(b)

Plot the radial and transverse components of force.

Answer to Problem 12.66P

Plot is in explanation part.

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]

Table of Fr, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 1

Table of Fθ, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 2

Conclusion:

Plots are mentioned the explanation part.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The magnitude of the resulting force on pilot.

Answer to Problem 12.66P

We got force F=230.113 lb

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]Fr=175[−2π29cos(π3×5)−(10+2cos(π3×5))×(0.1(4×5−1))2]Fr=−7141.2087 lb-ft/s2

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]Fθ=175[0.4(10+2cos(π3×5))+2(−2π3sin(π3×5))(0.1(4×5−1))]Fθ=1976.333 lb-ft/s2

Resultant force,

F=Fr2+Fθ2F=(−7141.2087)2+1976.3332F=7409.64 lb-ft/s2F=7409.6432.2 lbF=230.113 lb

Conclusion:

We got force F=230.113 lb

Expert Solution

To determine

(b)

Plot the radial and transverse components of force.

Answer to Problem 12.66P

Plot is in explanation part.

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]

Table of Fr, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 1

Table of Fθ, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 2

Conclusion:

Plots are mentioned the explanation part.

Answer 8

Expert Solution

To determine

(a)

The magnitude of the resulting force on pilot.

Answer to Problem 12.66P

We got force F=230.113 lb

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]Fr=175[−2π29cos(π3×5)−(10+2cos(π3×5))×(0.1(4×5−1))2]Fr=−7141.2087 lb-ft/s2

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]Fθ=175[0.4(10+2cos(π3×5))+2(−2π3sin(π3×5))(0.1(4×5−1))]Fθ=1976.333 lb-ft/s2

Resultant force,

F=Fr2+Fθ2F=(−7141.2087)2+1976.3332F=7409.64 lb-ft/s2F=7409.6432.2 lbF=230.113 lb

Conclusion:

We got force F=230.113 lb

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The magnitude of the resulting force on pilot.

Answer 13

Answer to Problem 12.66P

We got force F=230.113 lb

Answer 14

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]Fr=175[−2π29cos(π3×5)−(10+2cos(π3×5))×(0.1(4×5−1))2]Fr=−7141.2087 lb-ft/s2

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]Fθ=175[0.4(10+2cos(π3×5))+2(−2π3sin(π3×5))(0.1(4×5−1))]Fθ=1976.333 lb-ft/s2

Resultant force,

F=Fr2+Fθ2F=(−7141.2087)2+1976.3332F=7409.64 lb-ft/s2F=7409.6432.2 lbF=230.113 lb

Conclusion:

We got force F=230.113 lb

Answer 15

Expert Solution

To determine

(b)

Plot the radial and transverse components of force.

Answer to Problem 12.66P

Plot is in explanation part.

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]

Table of Fr, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 1

Table of Fθ, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 2

Conclusion:

Plots are mentioned the explanation part.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

Plot the radial and transverse components of force.

Answer 20

Answer to Problem 12.66P

Plot is in explanation part.

Answer 21

Explanation of Solution

Given information:

Time t=5 sec

r=10+2cos(π3t)

θ=0.1(2t2−t)

Concept used:

Fr=m(r¨−rθ˙2)

Fθ=m(rθ¨+2r˙θ˙)

Calculation:

Derivatives,

r˙=ddt(r)r˙=ddt(10+2cos(π3t))r˙=−2π3sin(π3t)

r¨=ddt(r˙)r¨=ddt(−2π3sin(π3t))r¨=−2π29cos(π3t)

θ˙=dθdtθ˙=ddt(0.1(2t2−t))θ˙=0.1(4t−1)

θ¨=dθ˙dtθ¨=ddt(0.1(4t−1))θ¨=0.4

Force components,

Fr=m(r¨−rθ˙2)Fr=175[−2π29cos(π3t)−(10+2cos(π3t))×(0.1(4t−1))2]

Fθ=m(rθ¨+2r˙θ˙)Fθ=175[0.4(10+2cos(π3t))+2(−2π3sin(π3t))(0.1(4t−1))]

Table of Fr, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
Fr (lb)	-12.6	-11.0	-11.3	-13.6	-18.0	-26.1	-40.7	-65.6	-104.1	-157.1	-221.8	-291.5

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
Fr (lb)	-356.9	-408.8	-441.7	-457.0	-464.0	-478.9	-520.7	-604.9	-738.2

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 1

Table of Fθ, with respect to t,

t (sec)	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5
F? (lb)	26.1	24.4	18.0	10.4	5.8	7.7	17.4	32.8	49.2	60.4	61.4	49.4

t (sec)	6.0	6.5	7.0	7.5	8.0	8.5	9.0	9.5	10.0
F? (lb)	26.0	-3.0	-29.4	-44.3	-41.5	-19.5	17.5	60.2	96.5

Plot,

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.66P , additional homework tip 2

Conclusion:

Plots are mentioned the explanation part.

Answer 22

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