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Is my Free Body Diagram right?

  1. Sep 20, 2003 #1
    Howdy,

    I had Physics two years ago in High School and am now in Engineering Physics. Im trying to have a head start on everything and it is also a project, can anyone check my FDB for correctness?

    Thanks,

    Matt

    [​IMG]
     
  2. jcsd
  3. Sep 21, 2003 #2
    OK first I am assuming that you are settinging your X,Y origin at the left tip of the triangle. Don't do that. Make it easy on yourself by defining the origin as the x axis runs up the incline, and the Y axis runs perpendicular to the axis, you are basically rotating the axes 20 degrees counter clockwise

    the forces here are in some cases redundant, and in other cases wrongly defined try this, using the new coordinate system.

    Starting with the left side of the box and moving clockwise around it to number sides. (left =side 1, bottom = 2, right = side 3, top = side 4)

    Side one - the force exerted down a slope by gravity is always
    mg sin a , where a is the angle the slope is making with the axis, in this case 20 degrees, so the force f = mg * sin(20). That is the only force in this direction, and since you have defined the X direction to be along the slope the force is conpletely in the negative X direction.

    Side two - your new coordinate system has the advantage of placing the Y axis perpendicular to the slope's surface. so to calculate the
    the force on side 2 we need to know the force the block is exerting perpendicular to the slope. The formula for this is mg*cos a, where a is the aforementioned angle. Once again since you have defined the coordinate system on a rotated axis all that force is in one direction. The negative Y direction.


    Side three - this one is a little trickier you have to decompose T into X and Y components. Fortunately our new coordinate system helps us again. The force T will exert along the slope, and thus in our X direction is given by the formula, T(x) = T*cos b were b is the angle T makes with the x axis, in this diagram, 30. To force perpendicular to the slope, and thus in the Y direction is given by T(y) = T*sin b where b is the same angle as before.
    Edit for friction: The frictional force will attempt to oppose the box sliding, and goes on side 3 the frictional force is equal to the static frictional constant,K,(since we are in equilibrium) multiplied by the force of the box pushing Normal(perpendicular) to the slope, so f = K*mg*cos(a)

    Side Four - the force N in the diagram is now entirely in the Y direction, and the only force acting in that direction, so according to newtons third law. ( the equal and opposite one) it has to be equal to the force acting perpendicular to the slope on side 2 minus the T(y) force, but in the opposite direction so it becomes, -1*(mg*cos a - T*sin b). Once again the choice of coordinate systems helps out because all of N is located in the positive Y direction.

    The X and Y force equations work out rather nicely in the system so that.

    F(X) = mgsin(20)-Tcos(30)-K*mg*cos(a) = 0
    F(y) = mgCos(20)-Tsin(30) - N = 0

    That's my two cents hope it helped
     
    Last edited: Sep 21, 2003
  4. Sep 21, 2003 #3
    Thank you for your extensive reply!

    I do, however, have a few questions. I always thought the x axis had to be along the ground, guess not.

    You said, "where a is the angle the slope is making with the axis, in this case 20 degrees, so the force f = mg * sin(20)"

    I thought everything along the x-axis would always be cosine? But then again you say the bottom (y-axis) is mg cos aº, so I am confused why the cosine and sine shouldnt be swapped?

    What about friction? That was the little f, perhaps I shouldve been more clear. I thought big F is for force, little f is for friction

    If sine isnt always for the vertical direction and cosine for the horizontal position, then I'm gonna mess up alot :(
     
  5. Sep 21, 2003 #4
    OK Here we go. As for using a different coordinate system.

    The ONLY control you have when solving physics problems is how you look at them. One of the fundamental ideas in classical physics is that the laws of physics are the same regardless of the vantage point fom which you view them. This means that when evaluating a problem we can look at the problem from whatever position makes it easiest to solve. In the image I am posting, I show your original coordinate system, and the one that I defined when solving the problem.

    as for the sine cosine issue. You should not memorize force components in that manner(sine is always this, and cosine is always this) because by changing the coordinate system(like I did) you can change what trig function you need to use. In the picture I gave the triangle letters on each side A, B, and C. Notice also that there is an angle that is marked I'll call it z. Lets pretend that the hypotenuse C is a force, and lets examine how we can decompose it in your coordinate system.

    C*cosine (z): well cosine = adjacent / hypotenuse so cosine(z) = A/C which gives us,
    C*cosine(z) = C*A/C = A (on the X axis)
    Using the same approach C*sine(z) = B (on the Y axis)

    But If I use the Green triangle with the same approach I end up with
    A on the Y axis, and B on the X axis

    That slope with the axis remark should have been slope's angle with the base sorry. The reason that I said mg*cos(a) is perpendicular to the slope has to do with a geometric proof that I can never remember, but will look up. I do know however that when looking for the gravitational forces on a slope. mg*cos(a) is into the slope, and mg*sin(a) is down the slope. I'll edit the above post to reflect friction, sorry I didn't know what that was. For the record though friction is a force so I would not differentiate if I were you.

    That's my 2 cents
     
  6. Sep 21, 2003 #5
    I'll send you the pic it is too big to post
     
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