If the biceps brachii, attaching to the radius 2.5 cm from the elbow joint produces 250 N of tension perpendicular to the bone, and the triceps brachii attaching 3 cm away from the elbow joint exerts 200 N of tension perpendicular to the bone how much net torque is present at the joint?Question 23Select one:a.0.25 Nm flexionb.0.25 Nm extensionc.No movement at the jointd.2.5 Nm flexione.2.5 Nm extension
Question
If the biceps brachii, attaching to the radius 2.5 cm from the elbow joint produces 250 N of tension perpendicular to the bone, and the triceps brachii attaching 3 cm away from the elbow joint exerts 200 N of tension perpendicular to the bone how much net torque is present at the joint?Question 23Select one:a.0.25 Nm flexionb.0.25 Nm extensionc.No movement at the jointd.2.5 Nm flexione.2.5 Nm extension
Solution
To solve this problem, we need to calculate the torque produced by each muscle and then subtract them to find the net torque.
Torque (τ) is calculated by multiplying the force (F) by the distance (d) from the pivot point, so:
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For the biceps brachii: τ = F * d τ = 250 N * 2.5 cm = 625 Ncm Since 1 m = 100 cm, we convert the torque to Nm: τ = 625 Ncm * (1 m/100 cm) = 6.25 Nm
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For the triceps brachii: τ = F * d τ = 200 N * 3 cm = 600 Ncm Again, convert the torque to Nm: τ = 600 Ncm * (1 m/100 cm) = 6 Nm
The net torque is the difference between the two torques. Since the biceps brachii is producing a larger torque, the net torque will be a flexion:
Net Torque = τ_biceps - τ_triceps Net Torque = 6.25 Nm - 6 Nm = 0.25 N*m flexion
So, the correct answer is a. 0.25 Nm flexion.
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