The arm in the figure below weighs 44.1 N. The force of gravity acting on the arm acts through point A. Determine the magnitudes of the tension force Ft in the deltoid muscle and the force Fs exerted by the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown. (Enter your answers to at least the nearest newton.)Ft  =  Enter a number. NFs  =  Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. NAn arm is extending horizontally to the right with a cutout that shows muscles in the upper arm. A horizontal dashed line extends along the bottom of the muscles and another along the top. Point O is labeled at the left end of the arm on the lower horizontal line. An unlabeled point lies 0.080 m horizontally to the right of point O. Point A lies 0.290 m horizontally to the right of point O. Three forces are acting on the arm.A force vector Fs acts down and right on point O at an acute angle 𝜃 below the horizontal.A force vector Ft acts up and left on the unlabeled point at an angle 12° above the horizontal. Vector Ft is longer than vector Fs.A force vector Fg acts vertically down on point A. Vector Fg is shorter than vector Fs

The deltoids exert a force of 100 N at an angle of 45 degrees to the humerus.The tendon inserts 7.5 cm from the shoulder joint axis of rotation.The arm is abducted 45 degrees from the vertical or anatomical position.The combined weight of the upper arm, forearm, and hand is 75 N acts at the center of gravity of the upper extremity which is 15 cm from the shoulder joint axis.What is the torque about the shoulder developed by the deltoids in Nm?

Hold your upper arm vertical and your lower arm horizontal with your hand holding a 191 N dumb-bell, as shown in the Figure. You’ll feel that your biceps muscle has contracted and is trying to pivot your lower arm about the elbow joint (black dot in the figure). For a person with the arm dimensions shown, what force, in N, must the biceps muscle provide? You can ignore the mass of the arm and hand in your calculation.

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

Order the diagrams to show the relationship between the length of the effort arm and the force needed to lift the weight.RESET

The large quadriceps muscle in the upper leg terminates at its lower end in a tendon attached to the upper end of the tibia (see Figure a). The forces on the lower leg when the leg is extended are modeled as in Figure b, where T is the force of tension in the tendon, w is the force of gravity acting on the lower leg, and F is the force of gravity acting on the foot. Find T when the tendon is at an angle of 25.0° with the tibia, assuming that w = 31.0 N, F = 10.8 N, and the leg is extended at an angle 𝜃 of 40.0° with the vertical. Assume that the center of gravity of the lower leg is at its center and that the tendon attaches to the lower leg at a point one-fifth of the way down the leg. (Give the magnitude.)Figure (a) depicts the leg of a skeleton. The upper leg is horizontal and the lower leg extends down and right. Quadriceps run along the top of the upper leg and connect to the tibia on the lower leg via a tendon. The foot is shown extending to the right at the base of the leg.Figure (b) is a force diagram of figure (a) where the leg is represented by two rods. The right end of a horizontal rod connects to a second rod that extends down and right at an acute angle 𝜃 from the vertical. Three forces act on the second rod.A force vector T acts up and left on the second rod a short distance down from the connection point at an acute angle 25.0° above the line of the rod.A force vector w acts vertically downwards on the center of the second rod.A force vector F acts vertically downwards on the bottom end of the second rod. N