Basic Biomechanics
Basic Biomechanics
7th Edition
ISBN: 9780073522760
Author: Susan J Hall
Publisher: McGraw-Hill Education
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Chapter 6, Problem 9AP

a)

Summary Introduction

To determine: The torque produced at 30°.

a)

Expert Solution
Check Mark

Answer to Problem 9AP

The torque created at 30° is 1.5N-m.

Explanation of Solution

Calculation:

Write an expression to find the torque at 30°.

T30=Fsinθd

Here, T30 is the torque created at 30°, F is the force exerted by the joint center, θ is the angle, and d is the distance.

Substitute 100 N for F, 30° for θ, and 3 cm for d to find T30.

T30=(100Nsin30°)(3cm)=(100Nsin30°)(3cm(1m100cm))=1.5N-m

Therefore, the torque created at 30° is 1.5N-m.

Conclusion

Therefore, the torque created at 30° is 1.5N-m.

b)

Summary Introduction

To determine: The torque produced at 60°.

b)

Expert Solution
Check Mark

Answer to Problem 9AP

The torque created at 60° is 2.6N-m.

Explanation of Solution

Calculation:

Write an expression to find the torque at 60°.

T60=Fsinθd

Here, T60 is the torque created at 60°, F is the force exerted by the joint center, θ is the angle, and d is the distance.

Substitute 100 N for F, 60° for θ, and 3 cm for d to find T60.

T60=(100Nsin60°)(3cm)=(100Nsin60°)(3cm(1m100cm))=2.59N-m2.6N-m

Therefore, the torque created at 60° is 2.6N-m.

Conclusion

Therefore, the torque created at 60° is 2.6N-m.

c)

Summary Introduction

To determine: The torque produced at 90°.

c)

Expert Solution
Check Mark

Answer to Problem 9AP

The torque created at 90° is 3N-m.

Explanation of Solution

Calculation:

Write an expression to find the torque at 90°.

T90=Fsinθd

Here, T90 is the torque created at 90°, F is the force exerted by the joint center, θ is the angle, and d is the distance.

Substitute 100 N for F, 90° for θ, and 3 cm for d to find T90.

T60=(100Nsin90°)(3cm)=(100Nsin90°)(3cm(1m100cm))=3N-m

Therefore, the torque created at 90° is 3N-m.

Conclusion

Therefore, the torque created at 90° is 3N-m.

d)

Summary Introduction

To determine: The torque produced at 120°.

d)

Expert Solution
Check Mark

Answer to Problem 9AP

The torque created at 120° is 2.6N-m.

Explanation of Solution

Calculation:

Write an expression to find the torque at 120°.

T120=Fsinθd

Here, T120 is the torque created at 120°, F is the force exerted by the joint center, θ is the angle, and d is the distance.

Substitute 100 N for F, 120° for θ, and 3 cm for d to find T60.

T60=(100Nsin120°)(3cm)=(100Nsin120°)(3cm(1m100cm))=2.59N-m2.6N-m

Therefore, the torque created at 120° is 2.6N-m.

Conclusion

Therefore, the torque created at 120° is 2.6N-m.

e)

Summary Introduction

To determine: The torque produced at 150°.

e)

Expert Solution
Check Mark

Answer to Problem 9AP

The torque created at 150° is 1.5N-m.

Explanation of Solution

Calculation:

Write an expression to find the torque at 150°.

T150=Fsinθd

Here, T150 is the torque created at 150°, F is the force exerted by the joint center, θ is the angle, and d is the distance.

Substitute 100 N for F, 150° for θ, and 3 cm for d to find T60.

T60=(100Nsin150°)(3cm)=(100Nsin150°)(3cm(1m100cm))=1.49N-m1.5N-m

Therefore, the torque created at 150° is 1.5N-m.

Conclusion

Therefore, the torque created at 150° is 1.5N-m.

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