Answered: The upper leg muscle (quadriceps)…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

The upper leg muscle (quadriceps) exerts a force of 1300 N, which is carried by a tendon over the kneecap (the patella) at the angles shown in Figure 5.43. Find the magnitude and direction of the force exerted by the kneecap on the upper leg bone (the femur).

Expert Solution

Step 1

Given,

$F_{O} = 1300 N$

Net force along the x-axis:

$\begin{array}{rcl} F_{X} & = & F_{O} \cos 55 ° + F_{O} \cos 75 ° \\ = & (1300 N) \cos 55 ° + (1300 N) \cos 75 ° \\ = & 1082 N \end{array}$

Net force along the y-axis:

$\begin{array}{rcl} F_{Y} & = & F_{O} \sin 55 ° - F_{O} \sin 75 ° \\ = & (1300 N) \sin 55 ° - (1300 N) \sin 75 ° \\ = & - 190.8 N \end{array}$

The magnitude of net-force F:

$\begin{array}{rcl} F & = & \sqrt{{F_{X}}^{2} + {F_{Y}}^{2}} \\ = & \sqrt{{(1082 N)}^{2} + {(- 190.8 N)}^{2}} \\ = & 1098.7 N \end{array}$

The direction of net-force F which it makes with the x-axis:

$\begin{array}{rcl} θ & = & \tan^{- 1} (\frac{F_{Y}}{F_{X}}) \\ = & \tan^{- 1} (\frac{- 190.8 N}{1082 N}) \\ = & - 10 ° \end{array}$

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