The upper leg muscle (quadriceps) exerts a force of\({\rm{1250 N}}\), which is carried by a tendon over the kneecap (the patella) at the angles shown inFigure\({\rm{9}}{\rm{.38}}\). Find the direction and magnitude of the force exerted by the kneecap on the upper leg bone (the femur).

A force is an external factor that can change the rest or motion of a body. It has a size and a general direction.

The diagram for the leg muscles is shown below:

The net horizontal force is,

\(\begin{align}{F_x} &= - 1250\cos 55^\circ + 1250\cos 75^\circ \\ &= - 1040\;N\end{align}\)

The component of the force along Y-direction is,

\(\begin{align}{F_y} &= 1250\sin 55^\circ - 1250\sin 75^\circ \\ &= - 183\;N\end{align}\)

The magnitude of the force is,

\(\begin{align}F &= \sqrt {{F_x}^2 + {F_y}^2} \\ &= \sqrt {{{\left( { - 1040} \right)}^2} + {{\left( { - 183} \right)}^2}} \\ &= 1.1 \times {10^3}\;N\end{align}\)

The direction of the force is,

\(\begin{align}\theta &= {\tan ^{ - 1}}\left( {\frac{{{F_y}}}{{{F_x}}}} \right)\\ &= {\tan ^{ - 1}}\left( {\frac{{ - 183}}{{ - 1040}}} \right)\\ &= 190^\circ \end{align}\)

Hence, the force is \({\rm{1}}{\rm{.1 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{N}}\) and the direction with horizontal is \({\rm{190^\circ }}\).