The total hip replacement is subjected 2


The total hip replacement is subjected to a force of F = 120 N. Determine the moment of this force about the neck at A and the stem at B.

The total hip replacement is subjected

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

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We will first draw a diagram showing the angle at which the force is applied, the angles inside the triangle that is formed and the lengths.

The total hip replacement is subjected

The moment at A can be calculated by noticing that there is a 5^0 angle between the force and the line of action though point A. Thus, we can write:

\circlearrowright M_A=(-120)(\sin5^0)(0.04)

M_A=-0.42\,\text{N}\cdot\text{m}

M_A=0.42\,\text{N}\cdot\text{m}\circlearrowleft\,\text{(Counterclockwise)}

 

To figure out the moment at B, we must first calculate the lengths of the triangle. This can be easily found using sine law. Pay close attention to the diagram to see how the angles and lengths were found.

\dfrac{\text{length}}{\sin150^0}=\dfrac{55}{\sin10^0}

\text{length}=158.4 mm

 

We can now write a moment equation at point B.

\circlearrowright M_B=(120)(\sin15^0)(0.1584)

M_B=4.92\,\text{N}\cdot\text{m}\circlearrowright\,\text{(Clockwise)}

 

Final Answers:

M_A=0.42\,\text{N}\cdot\text{m}\circlearrowleft\,\text{(Counterclockwise)}

M_B=4.92\,\text{N}\cdot\text{m}\circlearrowright\,\text{(Clockwise)}

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 4, question 4-17.

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