Chapter 13, Problem 3AP

A 35-N hand and forearm are held at a 45° angle to the vertically oriented humerus. The CG of the forearm and hand is located at a distance of 15 cm from the joint center at the elbow, and the elbow flexor muscles attach at an average distance of 3 cm from the joint center. (Assume that the muscles attach at an angle of 45° to the bones.)

a. How much force must be exerted by the forearm flexors to maintain this position?

b. How much force must the forearm flexors exert if a 50-N weight is held in the hand at a distance along the arm of 25 cm? (Answers: a. 175 N; b. 591.7 N)

Summary Introduction

To determine: The force exerted by the forearm.

Answer to Problem 3AP

The force exerted by the forearm is $\underset{\_}{175\text{N}}$.

Explanation of Solution

Calculation:

Consider torque acts on the counter clockwise direction for forearm of the hand will be taken as positive and the torque acts on the clockwise direction for joint center of the elbow will be taken as negative.

Express the magnitude of the net torque acts on the joints.

${\displaystyle \sum M_{\text{tri}}=0}$ (I)

Here, $M_{\text{tri}}$ is the moment of torque acts on the joints.

Rewrite the equation (I) for the net forces acts on the joints of the hand.

$F_{\text{m}}d\cos \theta -F_{\text{h and f}}x\cos \theta =0$ (II)

Here, $F_{\text{m}}$ is the muscle force exerted by the forearm flexors, $F_{\text{h and f}}$ is the force exerted by the hand and forearm vertically to the humerus, $d$ is the distance of the elbow flexor muscles from the joint center, $x$ is the distance of the forearm and hand from joint center at the elbow, and $\theta$ is the angle of the muscles attached to the bones.

Substitute $3\text{cm}$ for $d$, $15\text{cm}$ for $x$, $35\text{N}$ for $F_{\text{h and f}}$, and $45°$ for $\theta$ to find $F_{\text{m}}$ in equation (II).

$F_{\text{m}}\left(3\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\cos \left(45°\right)-\left(35\text{N}\right)\left(15\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\cos \left(45°\right)=0$

$\begin{array}{c}{F}_{\text{m}}\left(0.0212\text{m}\right)-3.71\text{N}\cdot \text{m}=0\\ F_{\text{m}}\left(0.0212\text{m}\right)=3.71\text{N}\cdot \text{m}\\ F_{\text{m}}=175\text{N}\end{array}$

Conclusion

Therefore, the force exerted by the forearm is $\underset{\_}{175\text{N}}$.

Summary Introduction

To determine: The force on the forearm flexors exerts on the weight hold in the hand.

Answer to Problem 3AP

The force on the forearm flexors exerts on the weight hold in the hand is $\underset{\_}{592\text{N}}$.

Explanation of Solution

Calculation:

Rewrite the equation (I) for the net forces acts on the weight holds in the hand.

$F_{\text{m}}d\cos \theta -F_{\text{h and f}}x\cos \theta -F_{\text{w}}l\cos \theta =0$ (III)

Here, $F_{\text{w}}$ is the force exerted on the weight held in the hand, $l$ is the distance of the arm, and $\theta$ is the angle of the muscles attached to the bones.

Substitute $3\text{cm}$ for $d$, $15\text{cm}$ for $x$, $35\text{N}$ for $F_{\text{h and f}}$, $50\text{N}$ for $F_{\text{w}}$, $25\text{cm}$ for $l$, and $45°$ for $\theta$ to find $F_{\text{m}}$ in equation (III).

$\begin{array}{l}{F}_{\text{m}}\left(3\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\cos \left(45°\right)\\ -\left(35\text{N}\right)\left(15\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\cos \left(45°\right)\\ -\left(50\text{N}\right)\left(25\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\cos \left(45°\right)\end{array}\}=0$

$\begin{array}{c}{F}_{\text{m}}\left(0.0212\text{m}\right)-3.71\text{N}\cdot \text{m}-8.84\text{N}\cdot \text{m}=0\\ F_{\text{m}}\left(0.0212\text{m}\right)-12.55\text{N}\cdot \text{m}=0\end{array}$

Solve the relation for $F_{\text{m}}$.

$\begin{array}{c}{F}_{\text{m}}\left(0.0212\text{m}\right)=12.55\text{N}\cdot \text{m}\\ F_{\text{m}}=\frac{12.55\text{N}\cdot \text{m}}{0.0212\text{m}}\\ =592\text{N}\end{array}$

Conclusion

Therefore, the force on the forearm flexors exerts on the weight hold in the hand is $\underset{\_}{592\text{N}}$.