### Question 6 of 19The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of the forearm about 2.15 cm beyond the joint. Assume the forearm has a mass of 2.25 kg and a length of 0.425 m.**Illustration Description:**The diagram shows a side view of a human arm. The arm is bent at the elbow, with the forearm extended horizontally.- The **humerus** is labeled as the upper arm bone.- The **elbow** marks the hinge point between the humerus and forearm.- The **forearm** consists of two bones: the **radius** and the **ulna**.- The **biceps muscle** is shown with an arrow indicating the force it exerts upwards.- A block labeled **M** represents a mass being held in the hand.**Question:**When the humerus and the biceps are nearly vertical and the forearm is horizontal, if a person wishes to hold an object of mass 4.55 kg so that their forearm remains motionless, what is the force \( F \) exerted by the biceps muscle?**Formula Input Area:**\[ F = \ \ \ \ \text{N} \]**Tools:**An option to input scientific notation is provided with \( \times 10^y \). This setup explores the concepts of torque and equilibrium.

Given: Mass of Forearm, mF=2.25kgMass of Biceps, mB=4.55kgLength of the forearm, l=0.425mDistance, d=2.15cm=2.15×10-2m The torque produced by the biceps is τ=0. So, the clockwise torque is equal to the anticlockwise torque, i.e., FB×d=mBgl+mFgl2FB=mBgl+mFgl2d ..... (1)Substitute the value in equation (1). FB=4.55×9.8×0.425+2.25×9.8×0.42522.15×10-2FB=18.9507+9.371222.15×10-2FB=18.9507+4.68562.15×10-2FB=23.63632.15×10-2FB=10.9936×102N

Answered: The bones of the forearm (radius and…