The moment arm of the biceps is approximately 2 cm from the axis of rotation of the elbow and the force vector representing the biceps is approximately 115 degrees from the positive x-axis. if the weight of the forearm and hand is 21 n and the center of mass of the segment is 13 cm from the elbow joint:
calculate the muscle force needed to hold the arm in the static position
calculate the vertical component of the joint reaction force
calculate the horizontal component of the joint reaction force
calculate the orientation of the vector representing the joint reaction force.

F = 150.6 N

Rx = 63.65 N

Ry = - 115.5 N

orientation is - 61.14 degree  with horizontal

Explanation:

given data

moment arm m  = 2 cm

angel = 115 degree

distance = 13 cm

force = 21 N

to find out

muscle force, vertical component force , horizontal component force , orientation

solution

we consider here Ra and Rb is reaction upside  and

we know moment formula that is

moment at joint  = F × sin(θ) × m - force × distance

0 = F  × sin(115) × 2 - 21 × 13

F = 150.6 N

so

for horizontal component

sum of horizontal force = Rx + Fcosθ

0 =  Rx + (150.6) cos115

Rx = 63.65 N

and

for vertical component

sum of vertical force = Ry + Fsinθ

0 =  Ry + (150.6) sin(115)

Ry = - 115.5 N

and

reaction direction are

tan∅ = Ry / Rx

tan∅ = -115.5 / 63.65

∅  = - 61.14 degree

so orientation is - 61.14 degree  with horizontal

asthenosphere and convection

nah.   i'm just sleepy.