The size of force exerted by the femur on the kneecap is 48.81 Newton.
Given the following data:
First of all, we would determine the horizontal component of the force acting on the kneecap.
\(F_y = Fcos(\theta)\\\\F_y = 60cos(42)\\\\F_y = 60 \times 0.7432\\\\F_y = 44.59 \; Newton\)
For the vertical component of the force:
\(F_x = Fsin(\theta)\\\\ F_x = 60sin(42)\\\\ F_x = 60 \times 0.6691\\\\ F_x = 40.15\; Newton\)
Next, we would find the vertical component of the net force:
\(\sum F_x = F - F_xsin(\theta)\\\\\sum F_x = 60 - 40.15\\\\\sum F_x = 19.85\; Newton\)
Finally, the resultant force acting on the kneecap:
\(F_R^2 = F_y^2 + F_x^2\\\\F_R = \sqrt{F_y^2 + F_x^2} \\\\F_R = \sqrt{44.59^2 + 19.85^2}\\\\F_R = \sqrt{1988.27 + 394.02}\\\\F_R = \sqrt{2382.29}\)
Resultant force = 48.81 Newton
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