Can someone please help me with this physic problem?

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NubianPrincess

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This is the question:
A box slides down a 30 degree ramp with an acceleration of 1.20m/s^2. Determine the co-efficient of kinetic friction between the box and the ramp.
The solution to this problem is <a href="http://physics.baruch.cuny.edu/summer2k2/sol0403.gif" target="_blank">here</a>.
I understand the theories involved, but my problem is I am unclear how to determine the components of the forces when the object is on an incline. I dont understand how to know that the component in the x direction is = Fg*sin30.
Please help me if you can, please? 🙂I understand where im getting stuck - the method seems to be to break the force of gravity down into x and y components, but I thought that the x componet would involve cos, not sin?????
 
Look at the 'triangle' in the diagram.

The component parallel to the inclined plane is the force you need to calculate friction coefficient.
 
You can reason this out from simple geometry (similar triangles).

Sorry I can't be of more help but it's really hard to teach someone this stuff w/o drawing the diagrams.
 
•••quote:•••Originally posted by NubianPrincess:
but I thought that the x componet would involve cos, not sin?????[/QB]••••If you rotate 'x-component' about 90 degrees, it becomes 'y-component'.

Look at the triangle carefully. That triangle seems somewhat inverted, right?
 
In my opinion, if you reason these things out, you'll do much better on any test if you know the "why's"... with that said:

Look at the angle in the diagram (30 degrees). cos = adjacent/hypotoneuse. If you want the adjacent side to that 30deg, you have the "y" component of Fg (cos(30) goes vertically down in this case, NOT sin(30)). So, Fg*cos(30) is the Normal force (force perpendicular to the plane).

sin = opposite/hyp. So following the logic given before, Fg*sin(30) is the "radial" (or the "x") component down-and-along the plane.

You know that Fg*sin(30) - mu*Fg*cos(30) = ma.

Let me know if this helps or not. I hope I helped. 🙂
 
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