'

Search results

Found 921 matches
Moment of Inertia - I-Beam (Ideal cross section)

An I-beam, also known as H-beam, W-beam (for “wide flange”), Universal Beam (UB), Rolled Steel Joist (RSJ), or ... more

Depth of the contact region between a rigid conical indenter and an elastic half-space

Contact mechanics is the study of the deformation of solids that touch each other at one or more points. Hertzian contact stress refers to the localized ... more

Maximum axial load that a long, slender, ideal column can carry without buckling

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

Second moment of area - Hollow Rectangular profile

The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a ... more

Curved Area of a Frustum of a Right Circulacircular Cone

The curved area of the frustum cone, can be calculated from the radius of the lower base and the radius of the upper base and the height of the cone (the ... more

Static balance

Rotating unbalance is the uneven distribution of mass around an axis of rotation. A rotating mass, or rotor, is said to be out of balance when its center ... more

Couple balance

Rotating unbalance is the uneven distribution of mass around an axis of rotation. A rotating mass, or rotor, is said to be out of balance when its center ... more

Second moment of area - I-Beam (W-section)

An I-beam, also known as H-beam, W-beam (for “wide flange”), Universal Beam (UB), Rolled Steel Joist (RSJ), or ... more

Slant height of a right regular pyramid

Slant height is the height of any slant triangle of the regular pyramid, or the distance from the apex, down the side, to a point on the base ... 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/

...can't find what you're looking for?

Create a new formula