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    Hey guys, i was reading an example.

    "A box of mass 2 kg is pushed up a rough plane by horizontal force of magnitude 25N. The plane is inclined to the horizontal at angle of  10^\cdot . Given that the coefficient of friction between the box and the plane is 0.3, find the acceleration of the box."

    so they draw a diagram for me and i understand everything that is going on, but then they seem to be finding the normal reaction, though they phrase it as "resolving the perpendicular to the acceleration", they label the force as R, and i can tell by the way that they work it out that its the normal reaction, my only promblem is the fact that they do this  R-2gcos10^\cdot -25cos80^\cdot.

    So why are they including the horizontal force and calculating it as if it was a force on the inclined plane? and even it was in the inclined plane why does it affect the normal? It would be perpendicular to it and perpendicular forces can not affect one another.
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    25N is a "horizontal force" so needs to be resolved perpendicular and parallel to the plane. It's not acting up the slope of the plane.
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    (Original post by vc94)
    25N is a "horizontal force" so needs to be resolved perpendicular and parallel to the plane. It's not acting up the slope of the plane.
    Exactly yet the book includes it into the equation as if it was acting up the slope which is confusing me.
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    Actually i think i understand now the since the force is horizontal to the plane it is acting at an angle on the box, if we were resolving in the direction of acceleration up the slope then it would be 25cos10N degrees but since we are resolving perp to the plain then we use 25sin10N or 25cos80N.
 
 
 
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