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Q5.1-3PE

Expert-verifiedFound in: Page 188

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) What is the maximum frictional force in the knee joint of a person who supports 66.0 kg**** of her mass on that knee? **

**(b) During strenuous exercise it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain**.

- The maximum frictional force in the knee joint of a person who supports 66 kg is ${\mathit{\text{F}}}_{\text{max}}\text{= 10.35 N}$.
- The maximum friction force when the weight supported by the knee is 10 times its initial weight is 97.02 N

**Friction is the force that prevents motion when one body's surface comes in contact with another's surface.**

(a)The maximum friction force is equal to the static friction. Its value can be calculated by using the following formula:

$F\text{}=\text{}{\mu}_{s}N$

Here, *N* is a normal reaction and ${\mu}_{s}$_{ }is the coefficient of static friction.

A mass of is supported by the knee. Therefore, the force exerted by the mass on the knee is

localid="1655798162636" $\begin{array}{c}W\text{}=\text{}mg\\ W\text{}=\text{}66.0\times 9.8\text{}\\ =\text{}646.8\text{\hspace{0.33em}N}\end{array}$

Although a force of 646.8 N is acting on the knee, it is still at rest. We know from Newton’s second law that if there is an external force on a system, there must be an acceleration. However, the knee does not have any acceleration, which means that there must be an equal and opposite force that is balancing the weight. (Newton’s third law)

So, the normal force is the force that balances the weight of the mass that acts downward. Therefore,

localid="1655798166804" $\text{N = W = 646.8\hspace{0.33em}N}$

Now, the value of the maximum friction force can be calculated by the following formula:

${\mathit{\text{F}}}_{\mathit{\text{max}}}{\mathit{\text{= \mu}}}_{\mathit{\text{s}}}\mathit{\text{N}}$

The coefficient of static friction (from Table ) for bones lubricated by the synovial fluid ${\mu}_{s}=0.016$ is:

$\begin{array}{c}{F}_{\mathrm{max}}=\text{}0.016\times 646.8\\ {F}_{\mathrm{max}}=\text{}10.35\text{\hspace{0.33em}N}\end{array}$

(b)

During the strenuous exercise, the weight supported by the knee is W’= 10W = 10 mg.

So, the normal force is $\text{N\u2019 = W\u2019 = 10 mg}$.

The following formula can now be used to compute the maximum friction force:

$F{\u2019}_{\mathrm{max}}=\text{}{\mu}_{s}N\u2019$

The coefficient of static friction (from Table 5.1) for bones lubricated by the synovial fluid ${\mathit{\text{\mu}}}_{\mathit{\text{s}}}=\text{0.015}$ is

$\begin{array}{l}F{\u2019}_{\mathrm{max}}=\text{}0.015\times 10\times 646.8\\ F{\u2019}_{\mathrm{max}}=\text{97.02\hspace{0.33em}N}\end{array}$

The maximum friction force when the weight supported by the knee is 10 times its initial weight is 97.02 N

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