As a swimmer pulls his arm through the water, various muscles exert forces on the upper arm. The figure below shows a force F exerted on the humerus (upper arm bone) by the pectoral muscle. The muscle is connected to the bone d = 7.00 cm from the center point O of the shoulder joint. Find the magnitude of F, if this force's torque on the arm provides half of the total torque balancing the torque produced by the water pushing against the hand. (Let Fhand = 107 and D = 30.5 cm. )
To solve this problem, we can use the principle of torque equilibrium. Torque is the measure of a force's tendency to rotate an object around a pivot point. In this case, the pivot point is the shoulder joint.
According to the problem statement, the torque provided by the pectoral muscle is half of the torque produced by the water pushing against the hand. Let's denote the torque provided by the pectoral muscle as T_p and the torque produced by the water as T_w.
To find the magnitude of the force F exerted by the pectoral muscle, we can use the equation:
T_p = F * d
where F is the magnitude of the force and d is the distance from the center point O of the shoulder joint.
First, let's find the torque produced by the water pushing against the hand, T_w:
T_w = F_hand * D
where F_hand is given as 107 N and D is given as 30.5 cm. We need to convert D from centimeters to meters:
D = 30.5 cm = 0.305 m
Substituting the values:
T_w = 107 N * 0.305 m = 32.735 N·m
Since T_p is half of T_w, we have:
T_p = 0.5 * T_w = 0.5 * 32.735 N·m = 16.3675 N·m
Now, we can solve for the magnitude of the force F:
16.3675 N·m = F * 0.07 m
Rearranging the equation to solve for F:
F = 16.3675 N·m / 0.07 m = 233.8214 N
Therefore, the magnitude of the force F exerted by the pectoral muscle is approximately 233.82 N.
To find the magnitude of force F exerted on the humerus by the pectoral muscle, we first need to understand the concept of torque.
Torque is the measure of the ability of a force to rotate an object around an axis. It depends on the magnitude of the force, the distance from the axis of rotation, and the angle between the force and the lever arm.
In this case, the torque provided by the water pushing against the hand is provided by the force Fhand acting on the hand. Let's call this torque Twater.
The total torque balancing the torque produced by the water pushing against the hand is twice the torque provided by the pectoral muscle, so we can call this torque Tmuscle.
To find the magnitude of F, we need to find the torque provided by the pectoral muscle. The equation for torque is:
Torque = Force x Distance x sin(θ)
Where:
- Torque is the rotational force or moment (in N·m or N·cm)
- Force is the applied force (in N or dynes)
- Distance is the distance between the axis of rotation and the point where the force is applied (in meters or centimeters)
- θ is the angle between the force and the lever arm (in radians or degrees)
Since the problem provides the values for Fhand and D, we can set up the equation as follows:
Twater = Twater/2
Twater = Fhand x D x sin(θ)
Twater/2 = F x d x sin(θ)
107 x 30.5 x sin(θ) = F x 7 x sin(θ)
We can now solve for the magnitude of F by rearranging the equation:
F = (107 x 30.5 x sin(θ)) / (7 x sin(θ))
The sin(θ) terms on both sides of the equation cancel out, leaving:
F = (107 x 30.5) / 7
Calculating this expression will give us the magnitude of force F exerted on the humerus by the pectoral muscle.