Class Work Equilibrium Name An outstretched arm that is parallel to the floor is
ID: 2033031 • Letter: C
Question
Class Work Equilibrium Name An outstretched arm that is parallel to the floor is pulling downward on a ring in order to hold a 98 N weight stationary. The arm is able to pull on the ring because the latissimus dorsi mu applies a force Fw to the arm that is at 0.069 m from the shoulder joint and at an angle of 2 weight of the arm W is 47 N and its center of gravity (cg) is 0.28 m from the shoulder joint. The shoulder joint exerts a reaction force FR on the arm that is at an angle ? to the horizontal. a) With what force Fp does the ring pull up 9°. The 061 m 0.28 m- 0 069 m uis at on the arm? (Hint: What is the tension in the rope connected to the ring?) Calculate the magnitudes of FM and FR and the angle Hin b) FM 0Explanation / Answer
given
W = 98 N
m = 0.069 m
thetam = 29 deg
W' = 47 N
d = 0.28 m
Fr = ?
Fm = ?
theta = ?
a. lte tension in the string be T
T = W = 98 N
hence
Fp = T = 98 N
b, from moment balance
W'*0.28 + Fm*sin(thetam)*0.069 = Fp*0.61
hence
Fm = 1393.6443 N
From force balance
Fp + Frsin(theta) = W' + Fm*sin(thetam)
Fr*cos(theta) = Fm*cos9thetam)
hence
Fr^2 = (W' + Fm*sin(thetam) - Fp)^2 + (Fm*cos(thetam))^2
Fr = 1369.64554327412 N
theta = 27.1337033 deg