What is the component of weight along a ramp?

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Homework Help Overview

The discussion revolves around understanding the component of weight acting along a ramp, specifically how to resolve the gravitational force into components parallel and perpendicular to the ramp's surface. The subject area includes vector resolution and trigonometry in the context of physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the concept of resolving vectors, particularly the gravitational force, into components along the ramp. Questions arise about the correct setup of the triangle used for vector resolution and the appropriate trigonometric functions to apply. Some participants express confusion about the relationship between the ramp angle and the components of weight.

Discussion Status

The discussion is ongoing, with participants attempting to clarify their understanding of the problem. Some guidance has been provided regarding the setup of the triangle and the use of trigonometric functions, but there is still uncertainty and differing interpretations among participants.

Contextual Notes

Participants mention specific values for weight and angles, but there is confusion regarding the calculations and the geometric representation of the problem. The original poster expresses difficulty in visualizing the triangle necessary for resolving the weight components.

heartOFphysic
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Ramp.jpg


(see image for question)

I just don't know what this question means! What is the weight "parallel" to the ramp? Please any help on this would be appreciated. I can resolve vectors, but in this particular question, I just don't quite get what to resolve- it's confusing me a little bit, can someone please clarify this question and assist me?

Thanks!
 
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The weight is mg so you'll need to resolve this vector (which obviously points straight down) into a component which is at 30 degrees to the horizontal (a little hint is that it always involves multiplying by sine or cosine of the angle of the ramp, you can draw out triangles and such but I find it more useful to figure out what you want to happen at 0 and 90 degrees to pick which trig function to use).
 
Triangle.jpg


(see image)
Yes so I get that! and the Hypotenuse is parallel to the ramp. so using Sine= Opp/Hyp

we get sin30= 7200N/Hyp

Hyp= 7200/sin30

so Hypotenuse = 14400N? so I get double the weight not half the weight! :cry:
 
Your triangle is set up wrong, you need one where the right angle is the formed between the ramp and its normal vector and the hypotenuse is the force of gravity. More simply just figure out whether sine or cosine is the appropriate choice and multiply the gravitational force by the trig function you chose of 30 degrees.
 
I'm not quite sure what you mean, I can't picture this triangle at all- apologize for my ignorance!

I know the only way I can arrive at half the weight is by the calculation
sin30 X 7200 = 3200N

or Cos60 X 7200 = 3200N
 
anyone?
 
attachment.php?attachmentid=42308&stc=1&d=1325259090.jpg


check img

and you know component of any vector (of magnitude a) along any other line making angle b with it is a cos(b)
 

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