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Time to reach a required delta-v - rocket propulsion

Spacecraft propulsion is any method used to accelerate spacecraft and artificial satellites. Space propulsion or in-space propulsion exclusively deals with ... more

Worksheet 296

(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 :

Density

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:

Density

By Archimedes’ principle, the weight of water displaced is m w g , so the buoyant force is:

Force (Newton's second law)

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:

Density

The maximum buoyant force is the weight of this much water, or

Force (Newton's second law)

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/

Acceleration of a particle in an electric field

The electric field is a component of the electromagnetic field. It is a vector field, and it is generated by electric charges or time-varying magnetic ... more

Specific Impulse

Specific impulse (usually abbreviated Isp) is a way to describe the efficiency of rocket and jet engines. It represents the force with respect to the ... more

Lorentz force

Lorentz force is the force applied on a charged particle, moving with velocity v, vertically into a magnetic field. A positively charged particle will be ... more

Angular velocity

In physics, the angular velocity is defined as the rate of change of angular displacement and is a vector quantity (more precisely, a pseudovector) which ... more

Worksheet 306

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

Force (Newton's second law)
Torque
Force (Newton's second law)
Torque

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

Mechanical equilibrium - 3=3 Torque example

which yields

Torque
Addition

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

Division

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/

Flux (as a single scalar)

Flux is two separate simple and ubiquitous concepts throughout physics and applied mathematics. Within a discipline, the term is generally used ... more

Volumetric flow rate

The volumetric flow rate is the volume of fluid which passes through a given surface per unit time. Fow velocity in fluid dynamics or drift velocity in ... more

Volumetric flow rate (parallel to the unit normal)

volumetric flow rate, (also known as volume flow rate, rate of fluid flow or volume velocity) is the volume of fluid which passes per unit time. The only ... more

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