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Stiffness is the rigidity of an object — the extent to which it resists deformation in response to an applied force. In general, stiffness is not the same ... more
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/
In magnetostatics, the force of attraction or repulsion between two current-carrying wires is often called Ampère’s force law. The ... more
Capital market line (CML) is the tangent line drawn from the point of the risk-free asset to the feasible region for risky ... more
The constant K_M (motor size constant) is a value used to describe characteristics of electrical motors.
The motor constant is winding independent ... more
A spheroid, or ellipsoid of revolution is a quadric surface obtained by rotating an ellipse about one of its principal axes; in other words, an ellipsoid ... more
A spheroid, or ellipsoid of revolution is a quadric surface obtained by rotating an ellipse about one of its principal axes; in other words, an ellipsoid ... more
In astrodynamics, the vis viva equation, also referred to as orbital energy conservation equation, is one of the fundamental equations that govern the ... more
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.
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/
The second law states that the net force on an object is equal to the rate of change of its linear momentum in an inertial reference frame. The second law ... more
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(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?