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The speed of sound is the distance travelled per unit time by a sound wave propagating through an elastic medium. The SI unit of the speed of sound is the ... 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/
Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational.
(a) Calculate the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child).
(b) Calculate the force on the baby due to Jupiter if it is at its closest distance to Earth, some 6.29e+11 m away. How does the force of Jupiter on the baby compare to the force of the father on the baby?
Father’s gravitational force on the baby is:
Jupiter’s gravitational force on the baby is:
(c) What should be the father’s weight, so that he exerts the same force on the baby as that of Jupiter? **
**this section is not included in the Reference material
Discussion
Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)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/
Dedicated to little Konstantinos
A horn antenna or microwave horn is an antenna that consists of a flaring metal waveguide shaped like a horn to direct radio waves in a beam. Horns are ... more
A horn antenna or microwave horn is an antenna that consists of a flaring metal waveguide shaped like a horn to direct radio waves in a beam. Horns are ... more
A horn antenna or microwave horn is an antenna that consists of a flaring metal waveguide shaped like a horn to direct radio waves in a beam. Horns are ... 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.