Question

Most of the motion generated by joints in the human body are examples of levers. An example of a third class lever is the forearm; the force which generates the motion is located in‑between the pivot point and the resistance force. The elbow acts as the pivot point, the biceps muscle pulls up on the forearm to move the mass which is held in the hand. Imagine someone is holding an object in her hand with her forearm horizontal and her upper arm vertical so her elbow creates a 90∘ angle. If her biceps attach to her forearm 0.0597 m from the elbow and make an angle of 29.0∘ with the vertical while a 2.00 kg mass is held in her hand 29.03 cm from the elbow, what force ????F is exerted by the biceps muscle? Ignore the mass of the forearm.

Answer 1

Answer:

force due to Biceps is 109.1 N

Explanation:

As we know that the system of mass with the arm is at rest

So here net force and torque must be zero about the elbow must be zero

So we will have

Torque due to force of Biceps = Torque due to weight held

$F d sin(90 - \theta) = mg (L)$

now by force balance

$F = \frac{mgL}{dcos\theta}$

$F = \frac{2 \times 9.81 \times 29.03}{5.97 cos29}$

$F = 109.1 N$

So force due to Biceps is 109.1 N