Weight Component: Find Force Components for Box on Inclined Surface

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Homework Help Overview

The discussion revolves around understanding the components of weight acting on a box sliding down an inclined surface at a 30-degree angle. Participants are exploring the relationship between the weight force and its components along and perpendicular to the incline.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the calculation of weight components using trigonometric functions, specifically questioning the use of sine and cosine in relation to the incline angle. Some are drawing diagrams to visualize the problem and are discussing the geometric interpretation of the forces involved.

Discussion Status

The discussion is ongoing, with participants providing insights and asking for further clarification on the concepts of force components. There is an exploration of different interpretations regarding the setup of the problem and the application of trigonometric functions.

Contextual Notes

Some participants mention potential inaccuracies in their diagrams and coordinate setups, indicating that assumptions about the angle and force directions are being questioned. There is a focus on ensuring a correct understanding of the physical situation before proceeding with calculations.

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Homework Statement


Well I am trying to get help on weight compent. It is a box sliding on an inclined surface with angle 30 degrees.


Homework Equations



w= mg

The Attempt at a Solution



I am trying to find the components for weight force acting on the box. Book says w=mgcos(30) is when its perpendicular to surface, and w= mgsin(30) when its parallel to surface. Can someone explain how to do this?
Thanks
 
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I am trying to find the components for weight force acting on the box. Book says w=mgcos(30) is when its perpendicular to surface, and w= mgsin(30) when its parallel to surface. Can someone explain how to do this?
Thanks
spdowind,welcome to PF! Are you familiar with the method for calculating vector components by drawing a right triangle? Draw a sketch. The weight force acts straight down vertically...that's the hypotenuse of the right triangle. The components are the magnitudes of the legs of that right triangle, one leg parallel to the plane surface, and the other perperpendicular to it. You'll have to do a little geometry and some basic trig to calculate these components. Please show some attempt at this for further assistance.
 
1.jpg

See i am trying to find the horizontal forces for the block sliding down.
I did:
Fx= m* ax
mg*sin(angle)- Tension force - Friction Force= mass * Acceleration.

I just don't understand why I use mg* sin ( angle) instead of mg*cos(angle) when I do the weight x-component minus tension force and friction force.
 
well from my understand again,my graph is a lil off the angle is wrong.
What I do understand is:

The weight is split into 2 components like my graph, but one is horizontal to the sliding surface , one is perpendicular to it, the one horizontal is the X compoent, which means parallel to surface means mgsin(angle)? Am I on the right track? Because I set up my x-y coordinates horizontal to sliding surface and perpendicular to surface.

Thanks
 

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