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Help with angular acceleration

  1. Jun 2, 2006 #1
    there are two blocks on a trapizoid one on the top and one on the side with the slope a wire that goes throu a pully connects the two. determin the the acceleration of the two blocks.
    block one =m1 block two = m2 pully =m with radius r Ufriction =.36
    what i have so far is
    sum of forces(along x axis) on block one is t (for tension) - ufm1g=m1a(acceleration)
    sum of forces (along yaxis) on block two is tsin(thata) +ufm2gsin-mg=m2a

    is this correct
     
  2. jcsd
  3. Jun 2, 2006 #2
    This is not angular acceleration by the way. You might want to check that sum of the y forces again, because you must remember that (I'm assuming block 2 is on an incline) that there is a component of gravity which is pulling the block downwards and is certainly not equal to m*g. Break the force exerted due to gravity up into components vertical and horizontal to the incline.
     
  4. Jun 2, 2006 #3
    for the sum of forces along y sin 30 mg for the force of gravity i just cant get the tension not sure if it is just t or tcos30
     
  5. Jun 2, 2006 #4
    so would it be t -sin30*m*g + ufcos30*m*g?
    also is the tension a and b both positve because the are going in the same direction or would tension b be negitive cant make sense of it
     

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    Last edited: Jun 2, 2006
  6. Jun 3, 2006 #5
    Remember what an ideal pulley does... it merely changes the direction of a force, therefore you would not break it up into components based on its angle relative to the rope's orientation on the other side of the pulley. You're getting closer to the answer though; your picture has not been authenticated yet so I can't give you great answers.
     
  7. Jun 3, 2006 #6
    thanks for the help finally got it to give me the correct answer since i had it from the book just couldnt get it to come out . this is what i have
    sum forces x = t1 - uf*m1*g =m1 *a
    sum forces y =t2 +uf*m2*g*cos()-m2*g*sin() =m2*-a
    that i dont understand is why for the y the weight is mgsin and the friction is mgcos (that is assuming i am doning it right when i plug in values i get the right answer for the sum)
     
  8. Jun 3, 2006 #7
    The easiest way to prove it to yourself is to take it to extremes. It's not that the *weight* is mg*sin(theta) it's that the *component of gravity pulling the block downward* is mg*sin(theta). This is the force that acts opposed to the tension in the rope and the friction force at the surface. Consider the situation where theta = 0. Does it make sense that there would be no part of gravity that is able to pull a block down a flat, horizontal surface?

    Now, think about what determines friction force at a contact point on a surface. It's the coefficient of friction multiplied by the normal force at that point, right? What is the component of gravity that is perpendicular to the incline? Take the angle to extremes again. What if theta was 90 degrees (and therefore mg*cos(theta) = 0, that is, the surface of the plane is vertical. Does it make sense then that gravity would produce no potential for a friction force because that force runs parallel to the surface at that inclination?
     
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