Work Done on Femur: Calculating Force Applied on 435mm Femur

[as76]

How much work is done by an external force of 151 N when it is applied to a 435 mm femur? Model the bone as a hollow cylinder with inner radius R1 = 4.20 mm and outer radius R2 = 12.60 mm. Young's modulus for the bone is Y = 1.67×1010 Pa.

Could someone please help me approach this problem?
Thanks

 

[Mapes]

Hi as76, welcome to PF.

(1) How is the force applied (e.g., axial, torsion, shear?)

(2) Do you know how to calculate work done by an applied force on a deformable object?

 

[kerimek]

for reaching out! I would be happy to assist you in approaching this problem.

Firstly, we need to understand the concept of work in physics. Work is defined as the product of force and displacement in the direction of the force. In this case, the external force of 151 N is applied to the femur, which has a displacement of 435 mm (0.435 m).

Now, to calculate the work done on the femur, we need to use the formula W = F*d*cosθ, where W is the work done, F is the force applied, d is the displacement, and θ is the angle between the force and the displacement.

Since the force is applied in the same direction as the displacement, the angle θ is 0 degrees. Therefore, the equation becomes W = F*d*cos0 = F*d.

Next, we need to model the femur as a hollow cylinder. The formula for the volume of a hollow cylinder is V = π*h*(R2^2 - R1^2), where h is the height of the cylinder, R1 is the inner radius, and R2 is the outer radius.

In this case, the height of the femur (h) is equal to the displacement (d) of 0.435 m. Substituting the given values, we get V = π*0.435*(0.0126^2 - 0.0042^2) = 0.0018405 m^3.

Now, we can use the formula for work to calculate the force applied on the femur. W = F*d = (0.0018405 m^3)*(1.67×10^10 Pa) = 3.07×10^7 Nm.

Therefore, the work done by the external force of 151 N on the femur is approximately 3.07×10^7 Nm.

I hope this helps in approaching the problem. Let me know if you have any further questions.