Confusion regarding sin and cos

  • Thread starter ness9660
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Generally Im confused about the use of sin and cos in physics problems.

http://img188.imageshack.us/img188/3162/eg2gu.gif [Broken]


The torque about the beam's attachment to the wall is:

T * 8 * sin(53)


Where T is the tension of the wire.

Why is sin the choice and not cos?

The best correlation Ive come up with so far was in two dimensional collisions, where motion in the y axis is always associated with sin, while the x axis with cos.


Can anyone give any insight?
 
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Answers and Replies

  • #2
robphy
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ness9660 said:
Generally Im confused about the use of sin and cos in physics problems.

[snip]
The best correlation Ive come up with so far was in two dimensional collisions, where motion in the y axis is always associated with sin, while the x axis with cos.

Can anyone give any insight?

better:
cos goes with ADJACENT side
sin goes with OPPOSITE side
(from the definitions of sin() and cos(), of course).
 
  • #3
Curious3141
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No, don't use "blind" methods of association to learn stuff like that - you will make mistakes later on (and they are not always applicable). (edit : this is in reference to the orig. post, not robphy's reply)

The magnitude of the moment (torque) of a force about a point is the product of the force and the perpendicular distance from the point to the line of force (this is called the "moment arm"). Draw a perpendicular from the point of attachment at the wall to the wire (which corresponds to the direction of the tensional force) and calculate the length of the perpendicular segment with trig.

More properly, the definition of torque is [tex]\tau[/tex] = r X F, meaning the cross product of the position vector of the point of application of force (taking the fulcrum to be the origin) and the force itself.

By the definition of the cross product, the magnitude of the torque will always come out to the product of the magnitudes of the distance and the force times the sine of the angle between them, i.e.

[tex]|\tau| = |r||F|\sin \theta[/tex]

which you can verify is the case in this problem too (though in this case, [tex]\theta[/tex] is actually (180 - 53) = 127 degrees, which has the same sine as 53 degrees). The only thing is that torque (properly defined) is a vector quantity, and its direction is at right angles to the other two vectors, in this case, the torque vector will be pointing out of the page at you.
 
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Good info already stated. What i'll say is that perhaps it's better to view the result as 8*(T*sin(53)). Now look at the term in parenthesis...

T*sin(53). what would happen to the tension in the rope if the 53 degrees went to 0? Since the rope and the beam would then be the same length the tension would be 0 also. So ask the question "if theta went to 0, what trig function sin or cos would also give me 0?" You have to have the problem set up right with axes and all that, but this helps when trying to decide between sin and cos...
 

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