Biomechanics Biceps Problem - Classical Mechanics (Moment Arms/Forces)

[sKyHigh]

Please let me know if I did this wrong or right, and if I did it wrong, please correct me :)

1. Homework Statement
The biceps brachii, a muscle in the arm, connects the radius, a bone in the forearm, to the scapula in the shoulder (see below). The muscle attaches at two places on the scapula but at only one on the radius. To move or hold the arm in place, the biceps muscle balances the weight of the arm and the force at the elbow joint. The centre of mass of the arm is at 15 cm from the elbow joint. The horizontal force of the elbow joint is 6.5 N when the forearm is held parallel to the ground and the forearm weighs 15.3 N. If the biceps supports the entire weight of the forearm, calculate the necessary force from each branch of the biceps to hold the forearm parallel to the ground and the vertical force at the elbow.

2. Homework Equations
net F = 0; net F_x = 0, net F_y = 0
net torque = 0

The Attempt at a Solution

Choose E as point of rotation (i.e., Moment Arm of E, MA_E = 0)

MA_arm = (15.3 N)(15 cm) = 229.5 N cm

Determine angles of insertions of A and B:

Firstly, I assume that, based on the diagram, points E and B (the point where B attaches to the scapula) line up vertically - is this assumption valid?

Thus angle of insertion of A = theta_A = tan^-1(30/2) = 86.19 deg
Similarly, theta_B = tan^-1(30/5) = 80.54 deg

Torque balance:

MA_arm = MA_A + MA_B
229.5 N cm = F_A * (5 cm)sin(86.19 deg) + F_B * (5 cm)sin(80.54 deg)

Rearranging for F_A,
F_A = [229.5 N cm - F_B * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) ... Eq. 1

Note: The "(5 cm)sin(86.19 deg)" and "(5 cm)sin(80.54 deg)" represent, respectively, the lengths of the lines that originate at E and intersect the F_A and F_B vectors such that the lines form right angles with said vectors.

Force balance:

x-direction:
F_Ex = F_Ax + F_Bx
6.5 N = F_Acos(86.19 deg) + F_Bcos(80.54 deg)
F_A = [6.5 N - F_Bcos(80.54 deg)] / cos(86.19 deg) ... Eq. 2

Substituting Eq. 1 into Eq. 2 yields:

[229.5 N cm - F_B * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) = [6.5 N - F_Bcos(80.54 deg)] / cos(86.19 deg)
46 N - 0.989F_B = 97.82 N - 2.473F_B
1.484F_B = 51.82 N
F_B = 34.92 N

Substituting this result into either of Eqs. 1 or 2 yields F_A = 11.47 N

y-dir'n force balance:
15.3 N = F_Ay + F_By + F_Ey
F_Ey = 15.3 N - (11.47 N)sin(86.19 deg) - (34.92 N)sin(80.54 deg)
F_Ey = -30.6 N - am I supposed to get a negative answer here?!

 

[sKyHigh]

Buuuuump! xD

 

[sKyHigh]

It's a fun problem for anyone who enjoys classical mechanics. Can anyone afford the time to work through the problem and see if they get the same numbers as me? Thanks!

 

[haruspex]

Please let me know if I did this wrong or right, and if I did it wrong, please correct me :)

1. Homework Statement
The biceps brachii, a muscle in the arm, connects the radius, a bone in the forearm, to the scapula in the shoulder (see below). The muscle attaches at two places on the scapula but at only one on the radius. To move or hold the arm in place, the biceps muscle balances the weight of the arm and the force at the elbow joint. The centre of mass of the arm is at 15 cm from the elbow joint. The horizontal force of the elbow joint is 6.5 N when the forearm is held parallel to the ground and the forearm weighs 15.3 N. If the biceps supports the entire weight of the forearm, calculate the necessary force from each branch of the biceps to hold the forearm parallel to the ground and the vertical force at the elbow.

View attachment 73651

2. Homework Equations
net F = 0; net F_x = 0, net F_y = 0
net torque = 0

The Attempt at a Solution

Choose E as point of rotation (i.e., Moment Arm of E, MA_E = 0)

MA_arm = (15.3 N)(15 cm) = 229.5 N cm

Determine angles of insertions of A and B:

Firstly, I assume that, based on the diagram, points E and B (the point where B attaches to the scapula) line up vertically - is this assumption valid?

Thus angle of insertion of A = theta_A = tan^-1(30/2) = 86.19 deg
Similarly, theta_B = tan^-1(30/5) = 80.54 deg

Torque balance:

MA_arm = MA_A + MA_B
229.5 N cm = F_A * (5 cm)sin(86.19 deg) + F_B * (5 cm)sin(80.54 deg)

Rearranging for F_A,
F_A = [229.5 N cm - F_B * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) ... Eq. 1

Note: The "(5 cm)sin(86.19 deg)" and "(5 cm)sin(80.54 deg)" represent, respectively, the lengths of the lines that originate at E and intersect the F_A and F_B vectors such that the lines form right angles with said vectors.

Force balance:

x-direction:
F_Ex = F_Ax + F_Bx
6.5 N = F_Acos(86.19 deg) + F_Bcos(80.54 deg)
F_A = [6.5 N - F_Bcos(80.54 deg)] / cos(86.19 deg) ... Eq. 2

Substituting Eq. 1 into Eq. 2 yields:

[229.5 N cm - F_B * (5 cm)sin(80.54 deg)] / (5 cm)sin(86.19 deg) = [6.5 N - F_Bcos(80.54 deg)] / cos(86.19 deg)
46 N - 0.989F_B = 97.82 N - 2.473F_B
1.484F_B = 51.82 N
F_B = 34.92 N

Substituting this result into either of Eqs. 1 or 2 yields F_A = 11.47 N

y-dir'n force balance:
15.3 N = F_Ay + F_By + F_Ey
F_Ey = 15.3 N - (11.47 N)sin(86.19 deg) - (34.92 N)sin(80.54 deg)
F_Ey = -30.6 N - am I supposed to get a negative answer here?!

That all looks right. To see whether the vertical force should be up or down, consider moments about the point where the muscle attaches to the radius.

 

[sKyHigh]

That all looks right. To see whether the vertical force should be up or down, consider moments about the point where the muscle attaches to the radius.

Thanks for responding! Was my use of trigonometry correct, particularly when calculating the moment arms? I neglected the 6 cm width of the forearm, not sure if it was okay to do so. That's where I had a bit of uncertainty. Also, in retrospect, F_Ey should be negative; the elbow ALWAYS exerts a net downward force in this configuration.

 

[haruspex]

Thanks for responding! Was my use of trigonometry correct, particularly when calculating the moment arms? I neglected the 6 cm width of the forearm, not sure if it was okay to do so. That's where I had a bit of uncertainty. Also, in retrospect, F_Ey should be negative; the elbow ALWAYS exerts a net downward force in this configuration.

The 6cm width is not in itself relevant, but your working is not quite accurate because the point of attachment of the muscle to the radius is slightly above the joint. The height difference is unclear - it looks to be less than 3cm.

 

[sKyHigh]

The 6cm width is not in itself relevant, but your working is not quite accurate because the point of attachment of the muscle to the radius is slightly above the joint. The height difference is unclear - it looks to be less than 3cm.

Thank you, that is precisely where my uncertainty lies in my attempt to solve the problem. Do you know of a more accurate way to solve the problem than assuming a height differential of, say, 2.5 cm? That is, I would instead use "(5.6 cm)sin(86.19 deg)" and "(5.6 cm)sin(80.54 deg)", respectively, when calculating MA_A and MA_B, where the "5.6 cm" is obtained via

c² = a² + b²
c² = 5² + 2.5²
c = 5.6 cm

The other uncertainty is the assumption that points B and E are in vertical alignment. What do you think of this assumption? Is it necessary to solve the problem? Valid?

Many thanks for helping me out!

 

[sKyHigh]

Bump! :p

 

[sKyHigh]

By assuming a height differential of 2.5 cm between E and the point of insertion of the biceps into the radius (i.e., a distance of 5.6 cm between E and said point of insertion), I get F_A = 3.26 N, F_B = 38.24 N, and F_Ey = 25.67 N [down]. Can anyone verify this? Thanks!