Question

The figure below shows a human arm that weighs 39.3 N. The arm is extended outward and is motionless. The gravitational force

F_g

on the arm acts at point A, a distance of 0.290 m from the shoulder joint, which is represented by point O. The shoulder pushes down and to the right on the humerus bone of the arm with a force

F_s

at point O, at an angle θ as shown. The deltoid muscle pulls back on the arm toward the shoulder with a tension force

F_t.

This tension force acts at a point 0.080 m to the right of point O, and it is directed up and to the left, at a 12.0° angle with respect to the horizontal.

An 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 F_s acts down and right on point O at an acute angle θ below the horizontal.
• A force vector F_t acts up and left on the unlabeled point at an angle 12.0°above the horizontal. Vector F_t is longer than vector F_s.
• A force vector F_g acts vertically down on point A. Vector F_g is shorter than vector F_s.

Find the magnitudes (in N) of the forces

F_t

and

F_s.

(Enter your answers to the nearest whole number. Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)

F_t

What is the equilibrium condition for the net torque on the arm? Which pivot point should you choose to have only one unknown force? What is the magnitude and direction of the torque due to

F_t?

Of the torque due to

F_g?

N

F_s

= N

Answer 1

Solution of above question is given below-

IK 7 = 39. 3N since For Arm is Equilibrium motin less & It condition for is in Equilibriun. the net torque is choose either point o u t to have single unknown force. Torque about point o, 1 -0 (A sin 12") x 0. 080 - Fg x 0.290 =0 given Fg = 39,3N soling, F = 685 2067N Now Torque about point B. (F. sino) x 0.080 - 39.3(0.290 -0.080) = Now, in x dirn net force will be zero. became of eruilibriun. cos 12° = Fs coso 685 2067 2012° - } cosa soling es 0 & ③.

F = 678.126 2N] 10 = 8.759 Torque of about point to To = E sin 12 ) x 0.080 = 11.397 Nm Canticlock wise) Torque of about point o. ZF = F x 0.290 - 11.397 Nm (close wise)