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Mean anomaly - function of mean longitude

In celestial mechanics, the mean anomaly is an angle used in calculating the position of a body in an elliptical orbit in the classical two-body problem. ... more

Chebychev–Grübler–Kutzbach criterion

The Chebychev–Grübler–Kutzbach criterion determines the degree of freedom of a kinematic chain, that is, a coupling of rigid bodies by means of mechanical ... more

Radiation Pressure by Absorption (using classical electromagnetism: waves)

Radiation pressure is the pressure exerted upon any surface exposed to electromagnetic radiation. Radiation pressure implies an interaction between ... more

Allowable Strength Design Load combination (eq7)

In structural engineering, a structure is a body or combination of pieces of rigid bodies in space to form a fitness system for supporting loads. ... 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/

Lift coefficient (for an airfoil section)

The lift coefficient is a dimensionless coefficient that relates the lift generated by a lifting body to the density of the fluid around the body, its ... more

Load and Resistance Factor Design (LRFD) - Load combinations (eq1)

In structural engineering, a structure is a body or combination of pieces of rigid bodies in space to form a fitness system for supporting loads. ... more

Radau parameter - related to Darwin / Radau equation

In astrophysics, the Darwin / Radau equation gives an approximate relation between the moment of inertia factor of a planetary body and its rotational ... more

Moment of inertia factor

In planetary sciences, the moment of inertia factor or normalized polar moment of inertia is a dimensionless quantity that characterizes the radial ... more

Mean anomaly at epoch

In celestial mechanics, the mean anomaly is an angle used in calculating the position of a body in an elliptical orbit in the classical two-body problem. ... more

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