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Strategy

The force is equal to the weight supported:

and the cross-sectional area of the upper leg bone(femur) is:

To find the change in length we use the Young’s modulus formula. The Young’s modulus reference value for a bone under compression is known to be **9×10 ^{9} N/m^{2}**. Now,all quantities except

**ΔL**are known. Thus:

Discussion

This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or muscle, several of the substances listed in Table 5.3(*see reference below*) have larger values of Young’s modulus Y . In other words, they are more rigid.

**Reference:**

This worksheet is a modified version of Example 5.4 page 188 found in :

OpenStax College,College Physics. OpenStax College. 21 June 2012.

http://openstaxcollege.org/textbooks/college-physics

Creative Commons License : http://creativecommons.org/licenses/by/3.0/

Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in the figure below, and compare this force with the weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures.

**(a)** The figure shows the forearm of a person holding a book. The biceps exert a force **F _{B}** to support the weight of the forearm and the book. The triceps are assumed to be relaxed.

**(b)**Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint

Strategy

There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is **F _{B}**, that of the elbow joint is

**F**, that of the weights of the forearm is

_{E}**w**, and its load is

_{a}**w**. Two of these are unknown

_{b}**F**, so that the first condition for equilibrium cannot by itself yield

_{B}**F**. But if we use the second condition and choose the pivot to be at the elbow, then the torque due to

_{B}**F**is zero, and the only unknown becomes

_{E}**F**.

_{B}Solution

The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the second condition for equilibrium (net **τ = 0**) becomes

Note that **sin θ = 1** for all forces, since **θ = 90º** for all forces. This equation can easily be solved for **F _{B}** in terms of known quantities,yielding. Entering the known values gives

which yields

Now, the combined weight of the arm and its load is known, so that the ratio of the force exerted by the biceps to the total weight is

Discussion

This means that the biceps muscle is exerting a force **7.38** times the weight supported.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.

http://openstaxcollege.org/textbooks/college-physics

Creative Commons License : http://creativecommons.org/licenses/by/3.0/

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Calculate the change in length of the upper leg bone (the femur) when a

70.0 kgman supports62.0 kgof his mass on it, assuming the bone to be equivalent to a uniform rod that is45.0 cmlong and2.00 cmin radius.