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A MAN HOLDS A 178 N BALL IN HIS HAND, WITH THE FOREARM HORIZONTAL. HE CAN SUPPORT HE BALL IN THIS POSITION BECAUSE OF THE FLEXOR MUSCLE FORCE, M, WHICH IS APPLIED PERPENDICULAR TO THE FOREARM. THE FOREARM WEIGHS 22 N AND HAS A CENTER GRAVITY. FIND THE MAGNITUDE OF M AND B)THE MAGNITUDE AND DIRECTION OF THE FORCE APPLIED BY THE UPPER ARM BONE TO THE FOREARM AT THE ELBOW JOINT.

A MAN HOLDS A 178 N BALL IN HIS HAND, WITH THE FOREARM HORIZONTAL. HE CAN SUPPORT HE BALL IN THIS POSITION BECAUSE OF THE FLEXOR MUSCLE FORCE, M, WHICH IS APPLIED PERPENDICULAR TO THE FOREARM. THE FOREARM WEIGHS 22 N AND HAS A CENTER GRAVITY. FIND THE MAGNITUDE OF M AND B)THE MAGNITUDE AND DIRECTION OF THE FORCE APPLIED BY THE UPPER ARM BONE TO THE FOREARM AT THE ELBOW JOINT.

Answer:The Muscle Force (M) is 1212.16 N.The Upper-arm bone force is 1012.10 N directed downward.Explanation:This problem is about torques and rotational dynamics. First, we analyse the torques in the elbow joint, as it's shown in the image. As you can see, the only present torques are: muscle torque, fore-arm torque and ball torque. Now, using the distances we can calculate each of them, because the definition of a torque is force multiplied by distance, being the distance from the application point and the support point (elbow joint).The problem specify that the flexor muscle has a M force, so its torque would be:(Counterclockwise, CCW).The fore-arm torque would be:(Clockwise, CW).The ball torque would be:(CW).Now, we know that those three torques are on the same support point, and applying Newton's Law, we solve:Therefore the Muscle Force is 1212.16 N, which means that the muscle is doing a great effort.Now, to calculate the magnitude and direction of the upper arm bone, we used forces with rotations, because that bone doesn't rotate. Again, applying Newton's Law, we analyse the sum of forces:Therefore, the magnitude of the force applied by the upper-arm is 1012.10 N. The sign indicates that this force is downwards, that's the direction....

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