Chapter 6, Problem 9AP

a)

Summary Introduction

To determine: The torque produced at $30°$.

Answer to Problem 9AP

The torque created at $30°$ is $1.5\text{N-m}$.

Explanation of Solution

Calculation:

Write an expression to find the torque at $30°$.

$T_{30}=F\sin \theta \cdot d$

Here, $T_{30}$ is the torque created at $30°$, $F$ is the force exerted by the joint center, $\theta$ is the angle, and d is the distance.

Substitute 100 N for $F$, $30°$ for $\theta$, and 3 cm for $d$ to find $T_{30}$.

$\begin{array}{c}{T}_{30}=\left(100\text{N}\text{sin}\text{30}°\right)\left(3\text{cm}\right)\\ =\left(100\text{N}\text{sin}\text{30}°\right)\left(3\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =1.5\text{N-m}\end{array}$

Therefore, the torque created at $30°$ is $1.5\text{N-m}$.

Conclusion

Therefore, the torque created at $30°$ is $1.5\text{N-m}$.

b)

Summary Introduction

To determine: The torque produced at $60°$.

Answer to Problem 9AP

The torque created at $60°$ is $2.6\text{N-m}$.

Explanation of Solution

Calculation:

Write an expression to find the torque at $60°$.

$T_{60}=F\sin \theta \cdot d$

Here, $T_{60}$ is the torque created at $60°$, $F$ is the force exerted by the joint center, $\theta$ is the angle, and d is the distance.

Substitute 100 N for $F$, $60°$ for $\theta$, and 3 cm for $d$ to find $T_{60}$.

$\begin{array}{c}{T}_{60}=\left(100\text{N}\text{sin}\text{60}°\right)\left(3\text{cm}\right)\\ =\left(100\text{N}\text{sin}\text{60}°\right)\left(3\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =2.59\text{N-m}\\ \approx 2.6\text{N-m}\end{array}$

Therefore, the torque created at $60°$ is $2.6\text{N-m}$.

Conclusion

Therefore, the torque created at $60°$ is $2.6\text{N-m}$.

c)

Summary Introduction

To determine: The torque produced at $90°$.

Answer to Problem 9AP

The torque created at $90°$ is $3\text{N-m}$.

Explanation of Solution

Calculation:

Write an expression to find the torque at $90°$.

$T_{90}=F\sin \theta \cdot d$

Here, $T_{90}$ is the torque created at $90°$, $F$ is the force exerted by the joint center, $\theta$ is the angle, and d is the distance.

Substitute 100 N for $F$, $90°$ for $\theta$, and 3 cm for $d$ to find $T_{90}$.

$\begin{array}{c}{T}_{60}=\left(100\text{N}\text{sin}\text{90}°\right)\left(3\text{cm}\right)\\ =\left(100\text{N}\text{sin}\text{90}°\right)\left(3\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =3\text{N-m}\end{array}$

Therefore, the torque created at $90°$ is $3\text{N-m}$.

Conclusion

Therefore, the torque created at $90°$ is $3\text{N-m}$.

d)

Summary Introduction

To determine: The torque produced at $120°$.

Answer to Problem 9AP

The torque created at $120°$ is $2.6\text{N-m}$.

Explanation of Solution

Calculation:

Write an expression to find the torque at $120°$.

$T_{120}=F\sin \theta \cdot d$

Here, $T_{120}$ is the torque created at $120°$, $F$ is the force exerted by the joint center, $\theta$ is the angle, and d is the distance.

Substitute 100 N for $F$, $120°$ for $\theta$, and 3 cm for $d$ to find $T_{60}$.

$\begin{array}{c}{T}_{60}=\left(100\text{N}\text{sin}\text{120}°\right)\left(3\text{cm}\right)\\ =\left(100\text{N}\text{sin}\text{120}°\right)\left(3\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =2.59\text{N-m}\\ \approx 2.6\text{N-m}\end{array}$

Therefore, the torque created at $120°$ is $2.6\text{N-m}$.

Conclusion

Therefore, the torque created at $120°$ is $2.6\text{N-m}$.

e)

Summary Introduction

To determine: The torque produced at $150°$.

Answer to Problem 9AP

The torque created at $150°$ is $1.5\text{N-m}$.

Explanation of Solution

Calculation:

Write an expression to find the torque at $150°$.

$T_{150}=F\sin \theta \cdot d$

Here, $T_{150}$ is the torque created at $150°$, $F$ is the force exerted by the joint center, $\theta$ is the angle, and d is the distance.

Substitute 100 N for $F$, $150°$ for $\theta$, and 3 cm for $d$ to find $T_{60}$.

$\begin{array}{c}{T}_{60}=\left(100\text{N}\text{sin}\text{150}°\right)\left(3\text{cm}\right)\\ =\left(100\text{N}\text{sin}\text{150}°\right)\left(3\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =1.49\text{N-m}\\ \approx 1.5\text{N-m}\end{array}$

Therefore, the torque created at $150°$ is $1.5\text{N-m}$.

Conclusion

Therefore, the torque created at $150°$ is $1.5\text{N-m}$.