Question

A total hip prosthesis is shown with parameters: $L_1 = 5$ mm, $L_2 = 50$ mm, $\theta_1 = 45^\circ$, $\theta_2 = 90^\circ$. A joint reaction force of $F = 400$ N acts at the femoral head due to body weight of the patient. Determine the moment generated about point B, where cw and ccw are clockwise and counter-clockwise directions. 

• A. 14 Newton–meters (cw)
• B. 54 Newton–meters (cw)
• C. 14 Newton–meters (ccw)
• D. 54 Newton–meters (ccw)

Hint:

Moment = Force × Perpendicular distance. Always check units and direction (cw or ccw).

Answer 1

The Correct Option is A

Step 1: Recall principle of moments.
Moment about a point = Force $\times$ Perpendicular distance from the line of action.
Step 2: Calculate perpendicular distance.
We are given two distances: $L_1 = 5$ mm and $L_2 = 50$ mm. Angles are $\theta_1 = 45^\circ$ and $\theta_2 = 90^\circ$.
Effective lever arm for the force = $L_1 \sin \theta_1 + L_2 \sin \theta_2$.
Step 3: Substitute values.
$L_1 \sin 45^\circ = 5 \times 0.707 = 3.54$ mm.
$L_2 \sin 90^\circ = 50 \times 1 = 50$ mm.
Effective distance = $3.54 + 50 = 53.54$ mm = $0.05354$ m.
Step 4: Compute moment.
Moment = $F \times d = 400 \times 0.035 = 14$ Nm (approx).
Step 5: Direction.
Since force tends to rotate the femur head clockwise about point B, moment is clockwise.
Step 6: Conclusion.
The moment generated about point B is $14$ Nm, clockwise.