Question

Three muscles are the major contributors to flexing the forearm: the biceps, the brachioradialis and the brachialis. Determine the force in the biceps muscle, the joint reaction force, and the joint moment at the elbow for an individual holding a 10 kg mass (98.1 N) with its forearm at 90° of flexion. The force in the brachialis muscle is 5 N and in the brachioradialis is 3.6 N.

The following anatomical data is given:

Length of forearm and hand: 0.5 m

Mass of forearm and hand: 2 kg

Distance from muscle insertion to elbow:

Biceps: 4 cm

Brachialis: 3.5 cm

Brachioradialis: 20 cm

Angle of muscle at 90° flexion:

Biceps: 80

Brachialis: 68

Brachioradialis: 23

Answer 1

SOLUTION:

∑F = 0

∑Fy = -W – W (forearm and hand) + FBRD sin 23° + FBIC sin 80.3° + FBRA sin 68.7° + Ry = 0

                                                  -98 – 19.6  + 0.39 (3.6 ) + 0.98 FBIC + 0.93 (5 ) + Ry = 0 | 1 |

                                                                           

                          ∑Fx = -FBRD cos 23° – FBIC cos 80.3° – FBRAcos 68.7° + Rx = 0                                 

                                                                       -0.92 (3.6 ) – 0.17 FBIC – 0.36 (5 ) = 0 | 2 |

FBIC = - 30.1 N

Then      

Ry = 98 N +19.6 N -1.40 N + 29.4 N - 4.65 N

Ry = 141 N

∑M= 0

∑M = -W(0.5 ) – W_FH (0.5  / 2) + FBRA sin 68.7° (0.035 ) + FBIC sin 80.3° (0.04 ) + FBRD sin 23° (0.2 ) = 0

                                              -49  – 4.9  + 0.033 FBRA + 0.039 FBIC + 0.078 FBRD + M = 0 | 3 |

     

M = 49 N-m +4.9 N-m - 0.033 (5 N) + 0.039 (-30.1 N) - 0.078 (3.6 N)

                       M = 52.3 N-m, clockwise