Problem 66 - College Physics

Question

Supporting an injured arm: I. A 650 N person must have her injured arm supported, with the upper arm horizontal and the fore- arm vertical. (See Figure \(10.76 .\) ) According to biomedical tables and direct measurements, her upper arm is 26 \(\mathrm{cm}\) long (measured from the shoulder joint), accounts for 3.50\(\%\) of her body weight, and has a center of mass 13.0 \(\mathrm{cm}\) from her shoulder joint. Her forearm (including the hand) is 34.0 \(\mathrm{cm}\) long, makes up 3.25\(\%\) of her body weight, and has a center of mass 43.0 \(\mathrm{cm}\) from her shoulder joint. (a) Where is the center of mass of the person's arm when it is supported as shown? (b) What weight \(W\) is needed to support her arm? (c) Find the horizontal and vertical components of the force that the shoulder joint exerts on her arm.

Step-by-Step Solution

Answer

(a) Center of mass at 27.82 cm from shoulder, (b) Support weight is 20.085 N, (c) Shoulder exerts -23.79 N vertically, 0 N horizontally.

Step 1: Determine Upper Arm and Forearm Weights

First, calculate the weight of the upper arm and the forearm. The upper arm accounts for 3.50% of the body weight, while the forearm accounts for 3.25% of the body weight. Upper arm weight: \( W_{upper} = 0.035 \times 650 \text{ N} = 22.75 \text{ N} \)Forearm weight: \( W_{fore} = 0.0325 \times 650 \text{ N} = 21.125 \text{ N} \)

Step 2: Calculate Center of Mass of the Arm

Next, find the position of the center of mass of the entire arm when it is in the supported position. This can be determined using the formula for the center of mass:\[ x_{cm} = \frac{W_{upper} \cdot d_{upper} + W_{fore} \cdot d_{fore}}{W_{upper} + W_{fore}} \]where:- \( d_{upper} = 13.0 \text{ cm} \)- \( d_{fore} = 43.0 \text{ cm} \)\( x_{cm} = \frac{22.75 \cdot 13.0 + 21.125 \cdot 43.0}{22.75 + 21.125} \approx 27.82 \text{ cm} \)

Step 3: Determine Supporting Weight Needed

For the arm to remain static, the net torque about the shoulder joint must be zero. The torque due to the weight \( W \) at a distance \( 60.0 \text{ cm} \) (total arm length from the shoulder joint) should counteract the torque due to the arm's center of mass:Set the clockwise and counter-clockwise torques equal:\[ W \cdot 60.0 = (W_{upper} \cdot d_{upper}) + (W_{fore} \cdot d_{fore}) \]Using the values:\[ W \cdot 60.0 = 22.75 \cdot 13.0 + 21.125 \cdot 43.0 \]\[ W \cdot 60.0 = 296.75 + 908.375 = 1205.125 \]\[ W = \frac{1205.125}{60} \approx 20.085 \text{ N} \]

Step 4: Calculate Forces Exerted by the Shoulder Joint

The shoulder joint must exert forces to keep the arm static. Resolve this into horizontal and vertical components:- For the vertical component:The sum of the vertical forces should be zero, i.e.,\[ F_{vertical} + W_{upper} + W_{fore} = W \]\[ F_{vertical} = W - W_{upper} - W_{fore} \approx 20.085 - 22.75 - 21.125 = -23.79 \text{ N} \]The horizontal force is zero since there are no horizontal forces acting, so \( F_{horizontal} = 0 \).

Step 5: Interpret the Results

The negative sign in the vertical force indicates that the shoulder exerts an upward force to balance the arm. The horizontal component is zero, confirming that only vertical support is necessary.

Key Concepts

Center of Mass

The center of mass (COM) is a crucial concept in biomechanics, particularly when analyzing the mechanics of the human body. It is the point where the mass of an object is considered to be concentrated and is crucial for balance and stability.
In the exercise, we calculate the center of mass of an injured arm that is supported horizontally. Given that each part of the arm has a different weight and length, the overall center of mass is essential for understanding how to support it effectively.
The center of mass of the arm is calculated using the formula:

• \[ x_{cm} = \frac{W_{upper} \cdot d_{upper} + W_{fore} \cdot d_{fore}}{W_{upper} + W_{fore}} \]

Where \( d_{upper} \) and \( d_{fore} \) are the distances of the centers of mass of the upper arm and forearm from the shoulder, respectively. This calculation helps to determine the exact point where the arm's mass is balanced.
In this scenario, the center of mass is approximately 27.82 cm from the shoulder joint. Understanding this position helps us further explore the force and torque needed to maintain equilibrium.

Torque

Torque is a measure of the rotational force acting on an object. In biomechanics, torque is crucial in understanding how muscles and forces, like support or gravity, affect body parts.
In this specific problem, we need to maintain the arm in a static position by balancing the torques. The formula for torque is:

• \[ \tau = F \cdot d \]

Where \( F \) is the magnitude of the force and \( d \) is the distance from the axis of rotation, here the shoulder joint. To keep the arm balanced, the torque exerted by the weight of the arm, and a supporting weight needs to be considered.
By setting the net torque to zero — the sum of the torques due to the arm's weight and the supporting weight need to balance — we can find the appropriate supporting weight required. In this case, the needed supporting weight is approximately 20.085 N to keep the arm in equilibrium without rotation.

Statics

Statics is a branch of mechanics that studies objects at rest and the forces in equilibrium acting upon them. This concept is critical when analyzing systems like the human body, ensuring that bodily parts remain stable.
In biomechanical applications like supporting an injured arm, statics helps determine the forces needed to maintain equilibrium. If the sum of the forces (and torques) is zero, the system is said to be in static equilibrium.
From our problem, we see the use of static principles to ensure that the arm does not move. By calculating the forces acting on the arm's support and the weight that needs to be applied, we are essentially applying statics to keep the arm’s position unchanged.
The vertical fore and the horizontal components of the force by the shoulder further emphasize the importance of achieving equilibrium.

Rotational Equilibrium

Rotational equilibrium occurs when an object is not experiencing any net torque—meaning it is not rotating or tends to rotate. In biomechanics, it's essential to maintain this equilibrium to prevent unwanted movements.
In the exercise example of supporting an injured arm, achieving rotational equilibrium is key. The calculations involving the torques around the shoulder ensure that the arm remains still.
By setting the equation of torques to zero, the exercise ensures that all clockwise and counter-clockwise torques are balanced, allowing the arm to maintain a static position.
This involves precisely calculating the point at which torques balance each other out, using the given formula and ensuring that any additional forces applied do not disrupt this equilibrium, ultimately ensuring the arm remains "at rest" in the desired position.