4. (30pts) Given the forearm of a person holding an object (mass: M): (a) (5pts)

Question

4. (30pts) Given the forearm of a person holding an object (mass: M): (a) (5pts) In addition to the muscle forces drawn in the figure, identify all the other forces (i.e., joint reaction forces, and weights) applied to the forearm and draw them on the free-body diagram below (i.e., complete this free-body diagram). For the muscles crossing elbow joints, we will consider only two muscle forces, biceps brachii (flexor; Fmi) and triceps brachii (extensor; Fm2). Assume that the direction of these muscle forces is vertical (i.e., no x component), as the orientation of the upper arm segment is vertical. FT m1 Forearm mass: m (kg) Object mass: M (kg) (b) (1Opts) Assuming that the triceps brachii muscle is not activated (i.e., Fm2-0), determine the magnitude of the muscle force (Fm) and the joint reaction force at elbow joint as functions of the elbow angle . Here, all the weight and length information is given: a = 2cm, b = 10cm, L = 25cm m 10kg, M-20kg (c) (15pts) When both biceps and triceps are activated, determine the magnitude of the muscle forces (Fml and Fm), as well as the elbow joint reaction force as functions of elbow angle . Here, assume that the biceps force is always three times larger than the triceps force (Fm,-3x Fm) 2cm, b = 10cm, c = 25cm m = 10kg, M 20kg a

Explanation / Answer

4. a. the following is the complete fbd of the scenario

the additional forces are

W = weight of the fore arm

Mg = weight of the mass M

R = elbow reaction R at angle theta' with the fore arm

b. assuming Fm2 = 0

a = 2cm

b = 10 cm

L = 25 cm

m = 10 kg

M = 20 kg

from force balance

Fm1 = mg + Mg + Rsin(theta' + theta)

Rcos(theta' + theta) = 0

hence

theta' + theta = 90 deg

theta' = 90 - theta

hence

Fm1 = mg + Mg + R

also from moment balance

Fm1*a*cos(theta) = mg*cos(theta) b + Mg(cos(theta))*L

hence

Fm1 = 9.81/a + 49.05/a

Fm1 = 2943 N

R = 2648.7 N

c. assuming Fm2= Fm1/3

from force balance

Fm1 + Fm2 = mg + Mg + Rsin(theta' + theta) = 4Fm1/3

Rcos(theta' + theta) = 0

hence

theta' + theta = 90 deg

theta' = 90 - theta

hence

4Fm1/3 = mg + Mg + R

also from moment balance

Fm1*a*cos(theta) = mg*cos(theta) b + Mg(cos(theta))*L + Fm2*a*cos(theta)

hence

Fm1 = 3(9.81/a + 49.05/a)/2

Fm1 = 4414.5 N

R = 5591.7 N

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