Calculating Torque by Muscle: Bicep Flexor Force & Upper Arm Bone Force

[mattmannmf]

A man holds a 196 N ball in his hand with the forearm horizontal as shown on the drawing to the left. He can support the ball in this position because of force, Fb from his bicep flexor muscule which is applied perpendicular to the forearm. This force applies a torque about the elbow joint.
Let's assume the forearm weighs 22 N and that the distance between the elbow joint and the center of the ball is L= 0.326 m. The distance between the point where the tendon connects the bicep to the forearm and the center of gravity of the arm is x1 = 0.089 m. The distance between the point of connection of the tendon and the elbow joint is x2 = 0.05 m.

a) What is the magnitude of Fb?

b) What is the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint?

not sure what to do for this one

 

[mattmannmf]

i found out that the magnitude of Fb is 1339.08 N

 

[nlightner]

I would approach this problem by using the principles of torque and equilibrium. Torque is the rotational force that is applied to an object, and it is calculated by multiplying the force by the distance from the pivot point. In this case, the pivot point is the elbow joint.

a) To find the magnitude of Fb, we can use the equation for torque:
T = Fd
Where T is the torque, F is the force, and d is the distance from the pivot point. In this case, the torque is caused by the weight of the ball, so we can set T = mg, where m is the mass of the ball and g is the acceleration due to gravity. We can also find the distance from the pivot point by subtracting x1 from L: d = L - x1. Plugging in the values given, we get:
mg = Fb(L-x1)
Solving for Fb, we get:
Fb = mg/(L-x1)
Substituting in the values given, we get:
Fb = (196 N)/(0.326 m - 0.089 m) = 761.54 N.
Therefore, the magnitude of Fb is 761.54 N.

b) To find the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint, we need to use the principle of equilibrium. This means that the sum of all the forces acting on the forearm must equal zero, since the forearm is not moving. We can write this as:
ΣF = 0
Where ΣF is the sum of all the forces. In this case, we have two forces acting on the forearm: Fb and the force applied by the upper arm bone. We can write this as:
ΣF = Fb + Fbone = 0
Solving for Fbone, we get:
Fbone = -Fb
Substituting in the value for Fb that we calculated in part a, we get:
Fbone = -761.54 N
Therefore, the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint is 761.54 N, in the opposite direction of Fb.