Dismiss Notice

Interview Feedback: Visit Interview Feedback to view and submit interview information.

Interviewing Masterclass: Free masterclass on interviewing from SDN and Medical College of Georgia

Can someone please help me with this physic problem?

Discussion in 'Pre-Medical - MD' started by NubianPrincess, Jun 16, 2002.

  1. NubianPrincess

    NubianPrincess Perpetually Bored
    7+ Year Member

    Joined:
    Aug 10, 2001
    Messages:
    12,640
    Likes Received:
    2
    Status:
    Medical Student
    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?????
     
  2. Legend

    Legend Super Senior Member
    10+ Year Member

    Joined:
    Mar 29, 2001
    Messages:
    860
    Likes Received:
    1
    Status:
    Attending Physician
    Look at the 'triangle' in the diagram.

    The component parallel to the inclined plane is the force you need to calculate friction coefficient.
     
  3. moo

    moo 1K Member
    10+ Year Member

    Joined:
    Jul 4, 2000
    Messages:
    1,423
    Likes Received:
    6
    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.
     
  4. Legend

    Legend Super Senior Member
    10+ Year Member

    Joined:
    Mar 29, 2001
    Messages:
    860
    Likes Received:
    1
    Status:
    Attending Physician
    </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by NubianPrincess:
    but I thought that the x componet would involve cos, not sin?????[/QB]</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">If you rotate 'x-component' about 90 degrees, it becomes 'y-component'.

    Look at the triangle carefully. That triangle seems somewhat inverted, right?
     
  5. Flack Pinku

    Flack Pinku U lookin at my glasses??
    7+ Year Member

    Joined:
    Jun 5, 2002
    Messages:
    2,033
    Likes Received:
    0
    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. :)
     

Share This Page