why is the Fn acting on the block when the block is on a horizontal surface larger than when it is on the incline?
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Fn is devided in two components on a incline:-
=>1) mgsinθ working downward and parallel to the inclined
=>2) mgcosθ working Normal to the inclined
As cosθ<1 between 0* to 90*, thus for any angle Normal on a inclined will be less than mg i.e. normal at horizontal surface.
=>1) mgsinθ working downward and parallel to the inclined
=>2) mgcosθ working Normal to the inclined
As cosθ<1 between 0* to 90*, thus for any angle Normal on a inclined will be less than mg i.e. normal at horizontal surface.
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The simplest way to think of it is that when on a plane horizontal surface the force on the block is simply the weight of the block in kg multiplied by gravity. On an incline theres two forces acting on the block one force that is horizontal and another that is vertical. The two can be drawn as a right angle with respect to the block and incline, in the middle of the right angle (45 degrees) is the force that is allowing the block to slide down the incline depending on the friction, otherwise the block stay put and this third force (the 45 degrees) isn't present. I hope this made sense. Basically while on a horizontal surface that would be the maximum force possible, on an incline the two forces are ratios
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Fn = m*g*cosΘ
When Θ = 0 (horizontal surface), cosΘ = 1.0, its maximum value.
When Θ = 0 (horizontal surface), cosΘ = 1.0, its maximum value.