GR9277 #8


Problem


\prob{8}
A solid cone hangs from a frictionless pivot at the origin O, as shown above. If , , are unit vectors, and a, b, and c are positive constants, which of the following forces applied to the rim of the cone at a point P results in a torque on the cone with a negative component ?
 , P is (0,b,c)
 , P is (0,b,c)
 , P is (b,0,c)
 , P is (b,0,c)
 , P is (b,0,c)

Mechanics}Torque
The problem wants a negative component for . Recall that whenever and are parallel (or antiparallel). Thus, choices (A), (B), (E) are immediately eliminated. One can work out the crossproduct to find that (D) yields a positive , thus (C) must be it.


Alternate Solutions 
BerkeleyEric 20100914 15:29:40  The zcomponent of the torque (if you write out the cross product) is . A, B, and E can be immediately thrown out since there is no or component, and then you can just check to see that C gives the write sign.  

Comments 
MallSaint 20161022 17:08:05  For a torque with a negative z component, the base of cylinder must be rotating clockwise as viewed from above (by the right hand rule). This can be given by a force applied in the y direction to a point on the far side of the base. Hence (C).   hooverbm 20121005 08:21:01  For those who are getting confused, for some reason I looked at this problem as if the zdirection were out of the page (force of habit). Clearly z is up/down in the plane of the page.
Torque = r X F and when you determine the direction, point your fingers along the direction of "r" and curl through the smallest of the two angles toward "F". Your thumb is the direction of the torque.
You can go through each option fairly quickly. You soon discover that if F is directed along "y", the only way to get a torque that is in the negative "z" is if the "r" has components into the page (x direction). Thus C.   Setareh 20111108 14:15:47  I really don't understand how you say r and F are parallel in A, B and E...!   alimerzairan 20111014 16:31:48  this is just a test of the righthand rule.
r = thumb
F = index finger
torque = middle finger
we know that the (A), (B), and (E) can't be true, otherwise the cross product will be 0.
And they want the torque to be in the z direction. So point your middle finger to the z direction and you find that the r vector must be in the i direction. This means a negative xvalue which is present in (C).   jondiced 20101103 14:15:18  Remember that , not ! Oops.   BerkeleyEric 20100914 15:29:40  The zcomponent of the torque (if you write out the cross product) is . A, B, and E can be immediately thrown out since there is no or component, and then you can just check to see that C gives the write sign.
QuantumCat 20140901 12:06:50 
That's exactly how I solved this problem. I noted that to get , you have to have a negative crossed with or crossed with negative . Since C & D are the only answer choices with a force not in the direction, and the force in C & D is the in the direction, you must have a negative component, C.
Tl;dr: to get you have to have for the given force choices.

QuantumCat 20140901 12:08:42 
Crap, accidentally put a minus sign in the cross product. It should read:

  yosun 20070601 23:07:02  test   RebeccaJK42 20070322 11:16:55  Does the R vector point from O to P or P to O?
yosun 20070601 23:06:19 
R would point from O to P, of course. Recall the right hand rule, and the two axe vectors ( and ) that spawn from the same origin.

 

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