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Homework help

  1. Feb 28, 2006 #1
    homework help!!!!

    Hi! everyone I am new tp physics and I need help. We had just finished a part on Newtons 1st and 2nd laws and were given a question as follows.

    An object is sliding down a hill, at an angle 30 degrees to the horizontal. The object has mass 50Kg. If the acceleration of the object is 0.6 ms^-2, what is the friction force acting on the object.

    So, I first began by constructing the free body force dig. I marked the weight Mg downwards, the normal reaction R upwards, frictional force backwards and a force P forward. So firstly did I mark it correctly?

    Now I know that F=ma does apply here but I am not sure about the other component. All I could come up with is
    Psin30+mg = R and F= Pcos30.
    I know something is wrong here but I don't know what!. Can anyone help me please?
     
  2. jcsd
  3. Feb 28, 2006 #2
    [tex]\large F_{\text{friction}=\mu mg\cos\theta=(\tan\theta-\frac{a}{g\cos\theta})mg\cos\theta=...[/tex]
     
    Last edited: Feb 28, 2006
  4. Feb 28, 2006 #3
    Thnks for the reply but I can't understnd the equation you have given. Could you explain the equation or is there any other way to find the friction by resolving the individual forces?
    Thanks again.
     
  5. Feb 28, 2006 #4

    Hootenanny

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    Welcome to PF! Your diagram is nearly right. The gravity should be acting straight down and frictional force acting backwards parallel to the inclined plane as you say. Make sure the normal reaction force is acting upwards perpendicular to the inclined plane. However, your question does not mention an additional accelerating force so you have to assume gravity is the only force acting and therefore you donot have a force P.

    You need to startby summing all the forces acting to pull the object down the hill, in this case it is only gravity, but you need to resolve the force parallel to the inclined plain. However, there is a frictional force acting to oppose the force of gravity so using Newton's law [itex] F = ma [/itex] we can say:
    [tex]F_{net} = ma \Rightarrow F_{gravity} - F_{frictional} = ma[/tex]
    Don't forget to resolve the force of gravity (using trig). The frictional force doesn't need resolving because it always acts parallel to the plain of motion.

    Hope this helps.
     
  6. Feb 28, 2006 #5
    O.k, I am still a bit confused because the question states that the object is sliding down the hill at an angle of 30 degrees, so shouldn't there be a gravitational force acting in that direction?. I was kinda thinking that while the weight (Mg) acted directly downwards, gravity (g) would act at an incline of 30 degerees to the horizontal. Or am I still going about it the wrong way??

    That also leaves me with the resolving of the forces. Thanks for pointing put that Friction does not need to be resolved. So left with gravity I end up with R=mg+gsin30. This is according to my current diagram anyway. So is this correct?
     
  7. Feb 28, 2006 #6

    dfx

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    You can resolve the WEIGHT (force) which is a vector and hence calculate the component of the weight (force) acting down the incline. GRAVITY, however, always acts toward the center of the earth, and hence vertically "downward". Draw the incline on a huge circle representing the earth if you still find it hard to grapple with. By any chance are you doing M1 (UK math syllabus)?
     
  8. Feb 28, 2006 #7
    No I am not doing M1..so I am not too sure about my resolving. So given the question above is my resolving that I have done correct..please help!!
     
  9. Feb 28, 2006 #8

    Hootenanny

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    You should end up with [itex]R = mg\cos 30[/itex] but why you want to find [itex]R[/itex] I don't quite know as you are only really concerned with the components parallel to the inclined plane. You want to calculate the force of gravity acting parallel to the plane.
     
  10. Feb 28, 2006 #9
    right, and isn't the frictional force the only component that is parallel to the inclined plane? therefore do we need to resolve it. Plus does mg sin 30 equal to the component of the weight (force) acting down the incline?
     
  11. Feb 28, 2006 #10
    consider the forces acting along the incline.when you consider mg acting downward,u cannot additionally consider g.So,do the following-

    Resolve mg along the incline which would be mgsin30.So,net force acting along the incline=mgsin30-F where F is frictional force opposing the movement.

    Finally,net force=ma
    i.e.,mgsin30-F=ma.
    Deduce F.
     
  12. Feb 28, 2006 #11

    Hootenanny

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    Yes, you are right! Now all you have to do is substitute in the formula we have just defined and the values into:
    [tex]F_{gravity} - F_{frictional} = ma[/tex]
    Show the equation first, then put the numbers in a work through until you have an answer for the frictional force. Post your working and asnwer here.
     
  13. Mar 1, 2006 #12
    Got it.I would end up getting mg sin 30-F=ma. Therby rougly i would end up getting about 330 N of friction...thanks guys
     
  14. Mar 1, 2006 #13

    Hootenanny

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    Yep, you got it.
     
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