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).

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).

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter5: Displacement And Force In Two Dimensions

Section: Chapter Questions

Problem 6STP

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Concept explainers

Rotational Equilibrium And Rotational Dynamics

In physics, the state of balance between the forces and the dynamics of motion is called the equilibrium state. The balance between various forces acting on a system in a rotational motion is called rotational equilibrium or rotational dynamics.

Equilibrium of Forces

The tension created on one body during push or pull is known as force.

Question

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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