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Auger electron spectroscopy is a common analytical technique used specifically in the study of surfaces and, more generally, in the area of materials ... more
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 9×109 N/m2. 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/
In chemistry, a reaction quotient: Qr is a function of the activities or concentrations of the chemical species involved in a chemical reaction. In the ... more
Low-density lipoprotein (LDL) is one of the five major groups of lipoproteins. These groups, from least dense to most dense, are: ... more
Auger electron spectroscopy is a common analytical technique used specifically in the study of surfaces and, more generally, in the area of materials ... more
Low-density lipoprotein (LDL) is one of the five major groups of lipoproteins. These groups, from least dense to most dense, are: ... more
Auger electron spectroscopy is a common analytical technique used specifically in the study of surfaces and, more generally, in the area of materials ... more
Optimum HDTV viewing distance is the distance that provides the viewer with the optimum immersive visual ... more
In 1954, Miles developed his version of this equation for GRMS as he was researching fatigue failure of aircraft structural ... more
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 FB 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 FB, that of the elbow joint is FE, that of the weights of the forearm is wa , and its load is wb. Two of these are unknown FB, so that the first condition for equilibrium cannot by itself yield FB . But if we use the second condition and choose the pivot to be at the elbow, then the torque due to FE is zero, and the only unknown becomes FB .
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 FB 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 kg man supports 62.0 kg of his mass on it, assuming the bone to be equivalent to a uniform rod that is 45.0 cm long and 2.00 cm in radius.