Question

Dr. Alexander is holding a briefcase with straight arm. Assume weight of the briefcase FB = 40 N, weight of the forearm FFA = 15 N, weight of the upper arm FUA = 25 N, moment arm of FB = 0.50 m, moment arm of FFA = 0.35 m, moment arm of FUA = 0.12 m, and moment arm of Fm = 0.025 m. Also,

FM is 42 degrees above the horizontal. Calculate (a) muscle force FM, and (2) joint reaction forces Fj,x and Fj,y. (20 pts

Answer 1

Given data of forces and moment lengths:

Weight of the briefcase: F_B = 40 N                                   moment arm of the briefcase: R_B = 0.50 m

weight of the Forearm: F_FA = 15 N                                       moment arm of the Forearm: R_FA = 0.35 m

weight of the Upper arm: F_UA = 25 N                                   moment arm of the Upper arm: R_UA = 0.12 m

                          moment arm of the muscle: F_FA = 0.025 m

also the angle of F_M with horizontal: 42^o

a) the torque or moment created by the forces applied in their specified directions:

Choosing the elbow as the pivot,

Torque produced by Briefcase:

Torque produced by Forearm:

Torque produced by Upperarm: (since the upper arm is considered as the pivot)

Torque produced by Muscle Force:

Balancing the torques:

(ANS a)

b) Joint Reaction Force: It is defined as the force produced within a joint as a result of the force exerted on the joint.

Thus the parallel and perpendicular components of joint reaction forces are: F_{j x} and F_{j y} respectively.

From Free Body diagram,

(as there are no resultant force in horizontal)

Note: -ve sign infers the joint reaction force acts on +ve y axis.