Consider the total hip joint prosthesis as shown in the figure. The geometric parameters of the prosthesis are such that \(L_1=40\) mm, \(L_2=60\) mm, \(\theta_1=45^\circ\), \(\theta_2=90^\circ\). Assume that, when standing symmetrically on both feet, a joint reaction force of \(400~\text{N}\) is acting vertically at the femoral head (point A) due to the body weight of the subject. Calculate the magnitude of the moment (in Nm) about point C. (Round off the answer to one decimal place.)
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- When a force is vertical (or horizontal), the moment about a point reduces to the force times the horizontal (or vertical) offset of the point of application from that point.
- Keep track of units: convert mm to m before computing moments in N·m.
Step 1: The force at \(A\) is vertical; hence the moment about \(C\) equals the force magnitude times the perpendicular (horizontal) distance from \(C\) to the line of action of the force through \(A\).
From the geometry, \(B\) is directly above \(C\) by \(L_2\) (\(\theta_2=90^\circ\)), and \(A\) is located from \(B\) by \(L_1\) at an angle \(\theta_1\) to the horizontal. Therefore the horizontal distance of \(A\) from \(C\) is
\[
x_A=L_1\cos\theta_1.
\]
Step 2: Compute the moment magnitude:
\[
M_C=F\,x_A
=400\;\text{N}\times (0.040\;\text{m})\cos45^\circ
=400\times 0.02828
\approx 11.3~\text{N}\cdot\text{m}.
\]
