The moment arm of the biceps is approximately 2 cm from the axis of rotation of...

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.

Solutions

Expert Solution

The forces are as shown above. muscle force act at 115 deg with the horizontal.

Only the vertical comp. produce torque about the joint. . Horizontal comp. passes through the joint and no torque. The force makes (115-90) = 25 deg with the vertical.

For the joint to be in static equilibrium

moments about the joint

M = F_m Cos(25) * 2cm - 21* 13 cm = 0

F_m = 21*13/(2 *Cos(25)) = 150.6 N

Let R_x and R_y be the joint reaction in x and y (up) directions

F_x = R_x + F_m Cos(115) =0 ; sum of forces in x-direction

= R_x + 150.6 * Cos(115) =0

R_x = 63.65 N - horizontal reaction

F_y = R_y + F_m Sin(115) =0 ; sum of forces in y-direction

= R_y + 150.6 * Sin(115) - 21=0

R_y = -115.5 N - vertical reaction

direction of the reaction

tan(r) = R_y /R_x = -115.5/63.65

r = -61.14 ⁰ ,

The joint reaction is oriented -61.14 deg. with the horizontal. i.e. it is below the horizontal.

genius_generous answered 1 year ago

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