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.

#### Solution:

Show me the final answerâ†“

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 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)}

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

\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)}

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)}

Hi! Where does the 0.04 in part A come from? Also where does the 0.1584 come from in part B? Thank you!

Hi, that’s 40 mm converted to m. Same with the other value. ðŸ™‚

The axis aren’t perpendicular so how could we analyze the force into its components using sin and cos angle and say that we can why didn’t we use cos15 to calculate Ma why did we use sin5 are n’t the both give the same outcome

What components are you referring to?