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Calculate the forward driving force which accelerates the car down the slope.

  1. Jul 23, 2005 #1
    There is a 760N heavy car on a plane which is inclined to the horizontal at an angle of 6.8 degrees. Calculate the forward driving force which accelerates the car down the slope.
    I know the answer is 760sin6.8. I know that the forward driving force has a direction which is parallel to the slope.The thing I want to know is why the 760N vector is resolved in to its perpendicular component? We can have an infinite no. of pairs of components in which one of the component has a direction which is parallel to slope.
    why cant we resolve it like that:


    |*...... unknown vector parallel to the slope
    |......*
    |..........*
    |_________*. <-----------This angle is 6.8deg.
    unknown vector pointing towards left

    I hope that you guys understand what I am saying. Thanks in advance for any help.

    :smile:
     
  2. jcsd
  3. Jul 23, 2005 #2
    I have alwasy been unable to add diagrams so I am going to try and explain it.

    Imagine a plane (6.8 degrees to the horizontal) and the horizontal plane itself. Now imagine the weight from the car (represented as an arrow from the car - perpendicular to the horizontal plane). The horizontal plane and the weight arrow will form a right angle. You want to know the component of the weight that acts down the slope. Imagine that the weight arrow is the hypotenuse of another triangle. Another of the sides is perpendicular to the inclined plane (and comes from the same point as the weight arrow). Then join the two so that you have a right-angled triangle. The angles will means that you will work out the side that joins the two you know (e.g. the line parallel to the inclined plane). Work this side out for 760 sin 6.8.

    The Bob (2004 ©)
     
  4. Jul 23, 2005 #3

    VietDao29

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    Homework Helper

    Whenever you put something on the slope (light enough for the slope to support, otherwise the slope will be broken), you will see it accelerates down. Because the (frictionless) slope can only exert a force that is perpendicular to its surface. So it will cancel out the perpendicular force to its surface. And the force that's parallel to its surface remains intact (this force will accelerate the object down the slope). That's why you must break the [tex]\vec{P}[/tex] into 2 forces.
    Hope you get it,
    Viet Dao,
     
    Last edited: Jul 23, 2005
  5. Jul 23, 2005 #4

    mukundpa

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    Homework Helper

    If I got your problem correectly you want to know that why we resolve a vector in two mutually perpendicular direction, not in the directions making acute or obyuse angle...right?

    [q is for theeta]

    We know that the projection (effectiveness) of a vector A in a direction making angle q is A*cos q which is zero if q is 90. thus we take the components in the direction q and 90 - q i.e. A*cos q and A*cos (90 - q) [=A*sin q] so that both components have no effect in each other's direction, otherwise we have to take component of other component in that direction.
     
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