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The Gibbs free energy is a thermodynamic potential that measures the “usefulness” or process-initiating work obtainable from a thermodynamic system, at a ... more
The capstan equation or belt friction equation, also known as Eytelwein’s formula, relates the hold-force to the load-force if a flexible line is ... 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/
Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T , for the four-rocket propulsion system shown in the Figure below. The sled’s initial acceleration is 49 m/s 2, the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.
A sled experiences a rocket thrust that accelerates it to the right.Each rocket creates an identical thrust T . As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight,w.The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger than scale.
Assumptions: The mass of the Sled remains steady throughout the operation
Strategy
Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with
Fnet is the net force along the horizontal direction, m is the rocket’s mass and a the acceleration.
We can see from the Figure at the top, that the engine thrusts add, while friction opposes the thrust.
Tt is the total thrust from the 4 rockets, Fnet the net force along the horizontal direction and Ff the force of friction.
Finally, since there are 4 rockets, we calculate the thrust that each one provides:
T is the individual Thrust of each engine, b is the number of rocket engines
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/
Electromagnetic induction is the production of an electromotive force across a conductor when it is exposed to a varying magnetic field. The induced ... more
(a) Calculate the buoyant force on 10,000 metric tons (1.00×10 7 kg) of solid steel completely submerged in water, and compare this with the steel’s weight.
(b) What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace 1.00×10 5 m 3 of water?
Strategy for (a)
To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in Table [insert table #] We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight
First, we use the definition of density to find the steel’s volume, and then we substitute values for mass and density. This gives :
Because the steel is completely submerged, this is also the volume of water displaced, Vw. We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives:
By Archimedes’ principle, the weight of water displaced is m w g , so the buoyant force is:
The steel’s weight is 9.80×10 7 N , which is much greater than the buoyant force, so the steel will remain submerged.
Strategy for (b)
Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water.
The mass of water displaced is found from its relationship to density and volume, both of which are known. That is:
The maximum buoyant force is the weight of this much water, or
Discussion
The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a load nine times its own weight without sinking.
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/
Buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. Buoyant force equivalent to the weight of the fluid that ... more
The hardness number is determined by the load over the surface area of the indentation and not the area normal to the force, and is therefore not ... more
A cantilever is a beam anchored at only one end. The beam carries the load to the support where it is forced against by a moment and shear stress. A ... more
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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.