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Inclined Plane

  1. Aug 17, 2006 #1

    danago

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    Gold Member

    Hey. In physics, we are studying motion, and the three laws of motion.

    At the moment im having trouble with inclined planes. Take the following example for instance:

    "A frictionless plane is inclined at 50 degrees to the horizontal. A mass of 65kg is released on the slope. What is its acceleration down the slope?"

    From what i understand, the weight/gravity force should be resolved into its horizontal and vertical components relative to the slope (slope being horizontal). In this case, the weight force becomes:

    [tex]
    W = \left( {\begin{array}{*{20}c}
    {65g\cos 40} \\
    { - 65g\sin 40} \\
    \end{array}} \right)
    [/tex]

    To balance the vertical component, so the object doesnt move up and down, the normal force therefore has the same magnitude but opposite direction as the vertical component of the weight force. Therefore:

    [tex]
    N = \left( {\begin{array}{*{20}c}
    0 \\
    {65g\sin 40} \\
    \end{array}} \right)
    [/tex]

    The net force then ends up being the horizontal component of the weight/gravity force...[tex]
    \sum F={65g\cos 40}[/tex]

    Since i want to find acceleration, i then divide the force being applied to the object by its mass, 65kg. So the acceleration ends up being [tex]a={g\cos 40}=7.5 m/s/s[/tex]

    However, apparently the answer is like 12.8 m/s/s. My teacher says that i need to create a right angled triangle from the information, where 9.8 m/s/s (gravitational acceleration) is the side opposite to the 50 degree angle, and then solve for the hypotenuse, which should give me the actual acceleration i need.

    Could someone please tell me where ive gone wrong in my calculations. Thanks,
    Dan.
     
    Last edited: Aug 17, 2006
  2. jcsd
  3. Aug 17, 2006 #2
    This may be a very unproductive answer. See above and always draw an image. That will help you more than anything.
     
  4. Aug 17, 2006 #3
    You've mixed up the components of the normal force. Redraw the diagram and check again.

    Also the angle in the question does not match the one in the answer. Moreover, the answer must be less than 9.8ms-2, because the maximum occurs when the object is falling straight down.
     
  5. Aug 17, 2006 #4

    Doc Al

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    Staff: Mentor

    Your work looks good to me. (Note to others: cos(40) = sin(50) :wink: )

    Sounds good.

    How can that be? The acceleration down the incline can only be a fraction of the acceleration due to gravity; it can't be greater!

    I don't see that you did anything wrong.
     
  6. Aug 17, 2006 #5
    :redface: Note to self : Draw the diagram and check again.
    Sorry about that, danago, if I misled you.
     
  7. Aug 17, 2006 #6

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    This is wrong. Your answer makes much more sense logically, and probably mathematically as well. Consider this... if the object accelerates faster than g, and has to travel a longer distance (the length of the slope vs. straight down), doesn't it have more energy than a falling object released from the same height when they both hit the ground?

    That's certainly in violation of a principle or two of physics
     
  8. Aug 17, 2006 #7

    danago

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    Gold Member

    Hmmm well im glad my working seems to be ok. Now ive just gotta figure out what my teacher was trying to do.

    Thanks very much for the replies everyone.
     
  9. Aug 18, 2006 #8

    andrevdh

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    Homework Helper

    What your teacher thought (which is not right!)

    Is that


    sin(50o) = g/a

    so

    a = g/sin(50o) = 12.8
     

    Attached Files:

    Last edited: Aug 18, 2006
  10. Aug 18, 2006 #9

    danago

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    Gold Member

    yea. He drew diagrams and showed us that, but it didnt seem right to me. Im still lost as to why he did it like that though :s
     
  11. Aug 18, 2006 #10

    andrevdh

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    Homework Helper

    Well as one goes from straight down to some incline the mass will experience a component of g. The way he drew it is that g is a component of the acceleration along the incline, which is not correct.
     

    Attached Files:

    Last edited: Aug 18, 2006
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