Question

The upper leg muscle (quadriceps) exerts a force of
FQ = 1320 N,
which is carried by a tendon over the kneecap (the patella) at the angles 55 and 75. Find the magnitude in newtons and direction in degrees counter-clockwise from an axis directed to the left of the force exerted by the kneecap on the upper leg bone (the femur).

Answer 1

Interpretation/assumption
- The tendon on one side pulls up-right at 55° from the vertical, the other pulls up-left at 75° from the vertical; each tension = 1320 N.
- Sum the two tendon forces to get the resultant force on the patella; the force the kneecap exerts on the femur is the equal-and-opposite of that resultant.

Compute components (x right, y up)
- F1 (1320 N, 55° right-of-vertical): Fx1 = 1320 sin55 = 1320(0.81915) = 1081.3 N; Fy1 = 1320 cos55 = 1320(0.57358) = 757.3 N.
- F2 (1320 N, 75° left-of-vertical): Fx2 = −1320 sin75 = −1320(0.96593) = −1274.0 N; Fy2 = 1320 cos75 = 1320(0.25882) = 341.6 N.

Resultant on patella: R = F1 + F2
- Rx = 1081.3 − 1274.0 = −192.7 N (left)
- Ry = 757.3 + 341.6 = 1098.9 N (up)
- |R| = sqrt((−192.7)^2 + 1098.9^2) = 1116.6 N ≈ 1.12 × 10^3 N.

Force on femur (kneecap on femur) = −R = (192.7 N right, −1098.9 N down).
Direction: measured counterclockwise from an axis pointing to the left:
- The left-pointing axis is 180° from +x; the force on the femur lies at ≈ 279.98° from +x, so measured CCW from the left axis the angle = 279.98° − 180° ≈ 100.0°.

Answer
- Magnitude ≈ 1.12 × 10^3 N.
- Direction ≈ 100.0° counterclockwise from an axis directed to the left.