When you bend your knee, the quadriceps muscle is stretched. This increases the tension in the quadriceps tendon attached to your kneecap (patella), which, in turn, increases the tension in the patella tendon that attaches your kneecap to your lower leg bone (tibia). Simultaneously, the end of your upper leg bone (femur) pushes outward on the patella. The figure below shows how these parts of a knee joint are arranged. What size *normal* force does the femur exert on the kneecap if the tendons are oriented as in the figure with 𝜃 = 24º and the tension in each tendon is 102 N? Answer to 1 decimal place.

To solve this, we need to analyze the forces acting on the kneecap in terms of equilibrium.

### Step-by-Step Breakdown:

1. **Free-Body Diagram:**
   The forces acting on the kneecap include:
   - The tension \( T \) in the quadriceps tendon (attached to the top of the patella).
   - The tension \( T \) in the patellar tendon (attached to the tibia, pulling downward).
   - The normal force \( F \) exerted by the femur on the patella (pushing outward).

   Since the tendons pull at an angle of \( \theta = 24^\circ \) relative to the horizontal, we can resolve these tensions into horizontal and vertical components.

2. **Horizontal Force Equilibrium:**
   The horizontal components of the two tensions contribute to the net horizontal force on the kneecap, and this is balanced by the normal force exerted by the femur. Thus:

   \[
   F = 2T \cos(\theta)
   \]

   Here, \( F \) is the normal force exerted by the femur on the kneecap, and \( T \) is the tension in each tendon.

3. **Substituting the values:**
   We know that:
   - \( T = 102 \, \text{N} \)
   - \( \theta = 24^\circ \)

   Now, substitute these values into the equation:

   \[
   F = 2 \times 102 \times \cos(24^\circ)
   \]

4. **Calculation:**

   \[
   F = 204 \times \cos(24^\circ)
   \]

   Using \( \cos(24^\circ) \approx 0.9135 \):

   \[
   F = 204 \times 0.9135 = 186.354 \, \text{N}
   \]

5. **Final Answer:**
   The normal force exerted by the femur on the kneecap is approximately \( 186.4 \, \text{N} \).

Would you like to dive deeper into the steps, or do you have any follow-up questions?  

Here are 5 questions related to this problem:
1. How would the result change if the angle \( \theta \) increased?
2. What happens to the normal force if the tension in the tendons is doubled?
3. Can you explain how the equilibrium of forces helps in analyzing this problem?
4. How would you incorporate friction if it existed between the femur and patella?
5. What role does the patella play in knee biomechanics, beyond force transmission?

**Tip:** Understanding free-body diagrams is essential in analyzing mechanical systems in equilibrium, especially in biomechanics!

What size *normal* force does the femur exert on the kneecap if the tendons are oriented as in the figure with 𝜃 = 24º and the tension in each tendon is 102 N? Answer to 1 decimal place.

Find out how to calculate the normal force exerted by the femur on the kneecap using trigonometry. This biomechanical problem involves resolving forces into horizontal components and applying force equilibrium principles. With an angle of 24º and a tendon tension of 102 N, this step-by-step guide walks you through the solution to find the normal force at 186.4 N. Suitable for undergraduate students studying physics or biomechanics.

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