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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

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

Stokes' law (Excess force due to the difference of the weight of the sphere and the buoyancy on the sphere)

The weight of an object is the force on the object due to gravity. Buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed ... 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

Arithmetic mean size - 1st moment

Calculates the arithmetic mean size (arithmetic method of moments) of the particles’ size distribution of a soil, in metric scale. In statistics, the ... more

Buoyant force (Archimedes' principle)

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

Knudsen diffusivity for diffusing species

Knudsen diffusion is a means of diffusion that occurs when the scale length of a system is comparable to or smaller than the mean free path of the ... more

Critical Buckling Stress of a Column with Buckling Coefficient

Column or pillar in architecture and structural engineering is a structural element that transmits, through compression, the weight of the structure above ... more

Propellant mass fraction

In aerospace engineering, the propellant mass fraction is the portion of a vehicle’s mass which does not reach the destination, usually used as a ... more

Specific gravity of solids

Silts, sands and gravels are classified by their size, and hence they may consist of a variety of minerals. Owing to the stability of quartz compared to ... more

Gravity Acceleration by Altitude

The gravity of Earth, which is denoted by g, refers to the acceleration that the Earth imparts to objects on or near its surface due to gravity. In SI ... more

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