Biomechanics Problem Involving Oblique Trigonometry
EXAMPLE 1 (As Submitted) :
See the finished question
A patient recovering from elbow surgery is attending physio to strengthen the biceps muscle. It is important that the elbow joint does not experience a force greater than 200N during the early stages of recovery. The physio chooses a 45.0 N weight (4.6 kg) for the patient to hold. If the forearm which weighs W = 22.0 N is maintained in a horizontal position by the biceps muscle supplying 348 N of force, will the force at the elbow exceed the allowable 200 N?
One way to solve this problem is to do vector summation with the cosine law to see what the resultant force is. Since the arm is being maintained in a static position, the force felt at the elbow will be equal and opposite to the resultant force of W, L and F. Normally we would solve this problem using Newton’s laws where the sum of the forces in the x direction must equal zero and the same in the y direction but this problem can easily be done using the cosine law. We could vary the Load (L) or the allowable force at the elbow such that sometimes the weight held exceeds the allowable and sometimes not. L is easy to vary by just selecting some common weights from a gym (say 1 to 5 kg). F would need to be generated from an algorithm which is given by:
I will go to the gym over the next couple of weeks and take a photo of my daughter doing the exercise if you want a photo to include.
Data Ranges: Pick a height and weight for a patient and you can calculate the Forearm Length and Weight using the information below:
|Patient Height Range of Values||Patient Weight Range of Values|
|Imperial||Range: 60"- 74"
||Range: 100-200 lb|
|Metric||Range: 1.52 m -1.90 m||Range: 445 N -1100 N|
|Forearm and Hand|
|Formula||Forearm length = 25.4% of Height||Forearm Weight = 2.2% of Weight|
|Range||38.6 cm - 48.3 cm||9.8 N - 24.2 N|
|Values to Use||Elbow to middle of Hand:
dL = 27.6 - 37.3 cm
dW = 0.55 × dL
|W = 9.8 N - 24.2 N|