Motion on a Smooth Inclined Plane

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Motion on a Smooth Inclined Plane
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Consider the forces acting on an object moving on
an inclined plane of inclination q to the
horizontal.
The simplest case is one in which the object is
sliding on the inclined plane under the action of
the weight and normal reaction only.
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Mass of the object m Acceleration due to
gravity g Weight of the object W mg
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The weight mg is resolved into two
components. One is along the inclined plane. One
is perpendicular to the inclined plane.
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The weight mg is resolved into two
components. The component mg sin q is along the
inclined plane. The component mg cos q is
perpendicular to the inclined plane.
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Motion takes place along the inclined plane.
The resultant force on the object is mg sin q
acting down-slope.
By Newtons Second Law of Motion, the
acceleration a of the object also points
down-slope.
No matter the object is moving up-slope or
down-slope, the resultant force and hence the
acceleration points down-slope.
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Motion takes place along the inclined plane.
F ma mg sin q ma a g sin q
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For object sliding on a smooth inclined plane

• The acceleration depends on the inclination of
  the plane only. It does not depend on the mass.
  Objects of different masses slide on the inclined
  plane with the same acceleration.
• The acceleration always points down-slope,
  independent of the direction of motion (velocity)
  of the object.

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There is NO motion in the direction perpendicular
to the inclined plane.
By Newtons First Law of Motion, the forces in
this direction should balance each other. Hence
the normal reaction R mg cos q The normal
reaction depends on the weight of the object and
also it decreases with the inclination q. The
steeper the slope, the smaller is the normal
reaction.
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Example 1
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A trolley of mass 0.5 kg is released from a
point O on a smooth runway inclined at 30o to
the horizontal. Assume g 10 m s-2
Find the time and the velocity of the trolley
when it reaches a point A 2 m from O.
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A trolley of mass 0.5 kg is released from a
point O on a smooth runway inclined at 30o to
the horizontal. Assume g 10 m s-2
The acceleration a is uniform and independent
of the mass of the trolley.
a g sin 30o 10 0.5 5 m s-2
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A trolley of mass 0.5 kg is released from a
point O on a smooth runway inclined at 30o to
the horizontal. Assume g 10 m s-2
The trolley reaches A with velocity v at time
t.
S ut0.5at2 2 0t 0.5 5 t2 t 0.894 s
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A trolley of mass 0.5 kg is released from a
point O on a smooth runway inclined at 30o to
the horizontal. Assume g 10 m s-2
The trolley reaches A with velocity v at time
t.
v2 u2 2aS v2 02 2 5 2 v 4.47m s-1
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Example 2
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A trolley of mass 0.5 kg is projected with
speed 4 m s-1 from O up a smooth runway inclined
at 30o to the horizontal. Assume g 10 m s-2
Take up-slope direction as positive. Draw the
acceleration-time graph, velocity-time graph and
the displacement-time graph of the trolley.
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A trolley of mass 0.5 kg is projected with
speed 4 m s-1 from O up a smooth runway inclined
at 30o to the horizontal. Assume g 10 m s-2
Acceleration a is negative because it points
down-slope which is in the negative direction.
a - g sin 30o - 5 m s-2
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(No Transcript)
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v4(-5)t
1.6
0.8
- 4
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v4(-5)t
1.6
0.8
- 4
S 4t0.5(-5)t2
1.6
1.6
0.8