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Inclined plane and work help

  1. Jul 20, 2008 #1
    Edit: I'll fix the symbols once I figure out how to do them properly. Sorry about that.

    Is there a tutorial ob how to use the math symbols? I can't find one anywhere.

    1. The problem statement, all variables and given/known data
    A man pushes a box up an inclined plane making an angle of 30 degrees with the horizontal. The mass of the box is 60kg, and the coefficient of kinetic friction between the box and the ramp is 0.45. How much work must the man do to push the box to a height of 2.5m at constant speed. Assume that the man pushes the box in a direction parallel to the surface of the ramp.

    Angle = 30 degrees

    Mass = 60kg

    [tex][tex]\mu[/tex]_{}k[/tex]

    Height = 2.5m

    2. Relevant equations
    1/2m[tex]^_{1}v{2}[/tex] + mg[tex]y{1}[/tex] = 1/2m[tex]^{2}v{2}[/tex] + mg[tex]{2}[/tex]

    [tex]\Delta[/tex]K = W

    K = 1/2 mv^2


    3. The attempt at a solution

    I worked on this sort of aimlessly for about 45 minutes. I'm honestly embarrassed to post anything because I am teaching myself this material and have no idea what I'm doing; however, I'll post something if necessary
     
    Last edited: Jul 20, 2008
  2. jcsd
  3. Jul 20, 2008 #2
    Re: Work

    ah hah! I solved it. I was able to find the force in the x direction by taking the sum of the forces in the x direction and setting them equal to 0. Then I multiplied that by the distance traveled along the hypotenuse of the ramp and voila.

    2616 N
     
  4. Jul 20, 2008 #3
    Re: Work

    zoner7, do not be ashamed to post anything online. this site is made to get help/advises from others. Looking at your errors can really help you understand the subject more. So, next time, to be ashamed, just post them.
    I also want to give you my up most respect for teaching yourself about physics. It takes real dedication to do so. Congrats!!!!!!! :)

    Ok now, lets get to the problem. I am going to break it down in order to make it more understandable.
    1) let us say that there was no more friction on the ramp. That means that the total work that the person would need to do to bring the box 2.5m up the ramp would be W=60kg*9.81m/(s^2) * 2.5m=1471.5J. That is using the equation W=Force * distance. The work is being done on only gravity. If you ever did centripetal force, you would know that the centripetal force of anu object orbiting earth is directed towards the center of earth. It is that force that you are tryin to over come with your box. Because the force due to gravity is acting straight down on the box, you want to overcome it to push it 2.5m up. That is what work against gravity is.
    Sure you can calculate the amount of work using the lenght of the path that the box will be pushed up on (A.K.A the hypothenus) but that is just a waist of time because you have to use a whole bunch of trigonometry for that ****** if you want to see how i would do it using that method, just tell me so in a PRIVATE MESSAGE*******

    2)ok, now we are going to put friction back into the problem. The friction is basicaly defined as a resistance to an action. the action is the bo moving up the ramp. The friction is the force that tries to restrain the box from moving up the ramp. A very important equation for friction in a mechanical system is : u=Ff/Fn. u id the coefitional friction, Ff is the frictional force, and Fn is the normal force. To find the amout of friction tha will be applied to the box, you need to reconfigure the equation and make it look like: Ff=u*Fn. You already have u and you can solve for Fn. Because the box is not being pushed straight up, the normal force of the box is not to it's fullest. If the box was being pushed staright up than the normal force would equal the force due to gravity on the box(A.K.A weight). BUt because it is being pushed up at 30 degrees the normal force exposed by the box will be Fn=cos(30degrees)*60kg*9.81m/(s^2)=509.74. **** Fn=cos(angle of ramp)*weight******

    Now that you have the normal force, you can find the frictional force that will be acting of the box. Ff=509.74N*0.45=229.28N. The total work agaisnt friction that you will have to do is: W=229.28N*(2.5m/sin(30degrees))=1146.4J *********W=F*d,,,,,,,hypothenus=opposite/sin(angle)*****


    Now you have the amount of work you will need to bring the box up without friction, the amount of work that will b edone by friction. But WAIT!!!!!!!!!! you still need one more piece of data. If you put a box up a ramp with no friction, the box is going to slide down the ramp. So we can say that friction helps maintain the box at it's position when not moving. But the friction does not always have to keep a box at a spot. the "sliding" force can over come friction and make th ebox go down the ramp. So now, we are goin to find what the sliding force is.

    3)*****Fs=sin(angle of ramp)*weight***** Fs= sin(30 degrees)*(60kg*9.81m/(s^2))=294.3N. This sliding force, if the force due to gravity at an angle of 30 degrees for that particular object. Now look, that force is greater than the frictional force. That only means that if the object was placed on the ramp and no one pushing, the box would slide down. So now, we need to find by how much the force of friction will decrease the "sliding" force. We do :294.3N-229.28N = 65.02N. that resulting force is going to push the box down the ramp. So now we need to know how much work it wil take to take care of that that "sliding" force up to when it reaches the top of the ramp. So we do: W=65.02N*5m= 325.1J

    So now, we have all the work will will need to bring the box to the top. All we have to do now is to add them up. So....: total work=1471.5J + 1146.4J + 325.1J=2943J

    2943Joules is the answer considering that there is no acceleration.

    sry that it is that long :)
    hope you had fun reading it haha :)
    private message me if you have any questions
     
  5. Jul 20, 2008 #4
    Re: Work

    zoner7, you were kinda close with that answer. Your units are off. Work is expressed in Joules and not Newtons
     
  6. Jul 20, 2008 #5
    Re: Work

    heh... yea. I just noticed that looking back. I suppose that happens when you try to work on physics at 1 in the morning
     
  7. Jul 20, 2008 #6
    Re: Work

    did you se the mistake that you made or calculation that you did not do in order to solve the problem?
     
  8. Jul 20, 2008 #7
    Re: Work

    Thank you for help! I wanted to mention that I think you may have made a little miscalculation somewhere. The answer in the back of the book is suggesting 2.6 x 10^3 J. I know these aren't always right, but I did get the same answer as it. Your answer would be strikingly similar if you removed the work of that last force: 2943J - 325.1J = 2618J. Since the book rounds, this would make sense.

    I'd try to compare our answers more; however, it is tough to follow without a picture (I'm a very visual person).

    Thank you so much again for the help, though. Perhaps we can figure out the cause of the discrepancy between our answers and the book's.
     
  9. Jul 20, 2008 #8

    tiny-tim

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    Hi zoner7! :smile:

    Yes, I get 2616 (joules :wink:) also (different method). :smile:
     
  10. Jul 20, 2008 #9
    Re: Work

    ah, yes. I see my mistake.
     
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