Question

A therapist applies a lateral force of 80 N to the forearm at a distance of 25 cm from the elbow joint center. The biceps attaches to the radius at a 90 ∘ angle at a distance of 3 cm from the elbow. a. How much force is required of the biceps to stabilize the arm in position? b. What is the magnitude of the reaction force exerted by the humerus on the elbow?

Answer 1

a) The force required of the biceps to stabilize the arm in position is approximately 666.67 N. b) The magnitude of the reaction force exerted by the humerus on the elbow is 80 N.

Step-by-step explanation:

a. To stabilize the arm in position, the biceps muscle needs to exert a force equal in magnitude and opposite in direction to the lateral force applied by the therapist. The force required of the biceps can be determined using the principle of torque, which states that the torque (force x distance) exerted by the therapist is equal to the torque exerted by the biceps. Therefore, the force required of the biceps can be calculated as follows:

The torque exerted by the therapist = Torque exerted by the biceps

(Force applied by the therapist) x (Distance from therapist's force to the elbow) = (Force required of the biceps) x (Distance from biceps to the elbow)

(80 N) x (25 cm) = (Force required of the biceps) x (3 cm)

Force required of the biceps = (80 N x 25 cm) / 3 cm = 666.67 N

Therefore, the force required of the biceps to stabilize the arm in position is approximately 666.67 N.

b. The magnitude of the reaction force exerted by the humerus on the elbow can be determined using Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. Since the therapist is applying a lateral force of 80 N to the forearm, the reaction force exerted by the humerus on the elbow will be 80 N.

Answer 2

The force required by the biceps to stabilize the arm when an 80 N lateral force is applied is approximately 666.67 N. The magnitude of the reaction force exerted by the humerus on the elbow, in this case, is approximately 669.05 N.

Step-by-step explanation:

To answer the student's question regarding the amount of force required of the biceps to stabilize the arm in position when a lateral force of 80 N is applied, we refer to an understanding of torque and equilibrium in the context of biomechanics. We know that torque (τ) is the product of force (F) and the perpendicular distance from the pivot point (lever arm, r), or τ = r x F. The biceps muscle generates a force at a 90° angle; thus, the applied force is perpendicular, and the entire force contributes to the torque.

To find the required force of the biceps (Fb), we set the torque created by the applied lateral force (torqueexternal) equal to the torque created by the biceps (torquebiceps) since the system is in equilibrium. Thus, Fb x rb = 80 N x 0.25 m, where rb is 0.03 m. This yields Fb = (80 N x 0.25 m) / 0.03 m = 666.67 N.

To find the magnitude of the reaction force exerted by the humerus on the elbow (Fe), we use the equilibrium in vertical forces. Since we have established that the biceps muscle exerts a force of 666.67 N to maintain equilibrium, and the applied external lateral force is 80 N, the elbow joint must provide a supporting force equal to the vector sum of these forces, to maintain static equilibrium. If these forces are represented in a free body diagram with the biceps and external force at right angles, we can calculate using the Pythagorean theorem: Fe = √(Fb² + 80⁲) N. Substituting the determined value for Fb, we find, Fe = √(666.67⁲ + 80⁲) N = approximately 669.05 N.