Component of weight perpendicular to bar

  • Thread starter theone
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  • #1
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Homework Statement


http://postimg.org/image/tlvhadljz/
I am trying to find the component of weight perpendicular to the bar.

Homework Equations




The Attempt at a Solution


I tried as shown and I got mgcos(theta) but I am supposed to be getting mgsin(theta)
 

Answers and Replies

  • #2
Could you please explain the question. Even the picture seems vague.
 
  • #4
SammyS
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Homework Statement


http://postimg.org/image/tlvhadljz/
I am trying to find the component of weight perpendicular to the bar.

Homework Equations




The Attempt at a Solution


I tried as shown and I got mgcos(theta) but I am supposed to be getting mgsin(theta)
It's fairly basic trigonometry. The angle, θ, is obtuse, so you might be better off considering the angle π - θ which is acute. You can then more directly see that you need to use sin(π - θ) by considering opposite/hypotenuse.

Untitled.jpg
 
  • #5

Homework Statement


http://postimg.org/image/tlvhadljz/
I am trying to find the component of weight perpendicular to the bar.

Homework Equations




The Attempt at a Solution


I tried as shown and I got mgcos(theta) but I am supposed to be getting mgsin(theta)
The theta you considered is not the same with the theta that is shown in the figure on the left.
 
  • #6
Delta2
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Oh dear they are right. theta is measured from the vertical to the ground plane (not from the ground plane itself) so the theta we considered is equal to the actual theta minus π/2. So the perpendicular component of weight is mgcos(theta-π/2)=mg(cos(theta)cos(pi/2)+sin(theta)sin(pi/2))=mgsin(theta).
 
  • #7
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thanks everone
 

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