Component of weight perpendicular to bar

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Homework Help Overview

The discussion revolves around determining the component of weight that acts perpendicular to a bar, with a focus on the trigonometric relationships involved. The problem is situated within the context of physics, specifically dealing with forces and angles in mechanics.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the angle θ and the components of weight, with some attempting to clarify the correct trigonometric functions to use. Questions arise regarding the interpretation of the angle in relation to the bar and the ground plane.

Discussion Status

There is an ongoing exploration of the correct approach to identifying the perpendicular component of weight. Some participants provide insights into the trigonometric relationships, while others question the assumptions made about the angle θ. The discussion reflects a mix of agreement and differing interpretations without reaching a definitive conclusion.

Contextual Notes

Participants note that the angle θ is measured from the vertical to the ground plane, which may differ from the angle considered in previous calculations. This distinction is crucial for understanding the components of weight being discussed.

theone
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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)
 
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Could you please explain the question. Even the picture seems vague.
 
theone said:

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
 
theone said:

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.
 
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).
 
thanks everone
 

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