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Resolving parallel and perpendicular to an inclined plane including friction.
A block mass m kg on a
rough inclined plane at
angle θ to the horizontal.
R is the normal reaction.
P is a pull force acting at an
angle α to the plane.
If the block is moving or
tending to move up the
plane the frictional force Fr
will act down the plane.
P
R
P
α
α
Fr
Psinα
Pcosα
θ
mg
mg
θ
θ mgcosθ
mgsinθ
If the block is in equilibrium (stationary or moving with const velocity):
Parallel to slope : P cosα = Fr + mgsinθ
Perpendicular to slope : R = mgsinθ
If the block is accelerating up the slope with acceleration a then there will be a
resultant force parallel to the slope but perpendicular to the slope the forces will
still balance.
Parallel to slope “F = ma” : The resultant force F is P cosα – mgsinθ - Fr so
P cosα – mgsinθ - Fr = ma
Perpendicular to slope :
R = mgsinθ
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