Force and Motion-1

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CHAPTER-5

• Force and Motion-1

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Ch 5-1 Physics

• Physics Study of cause of acceleration of an
  object
• Cause Force
• Force is said to act on the object to change its
  velocity

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Ch 5-3 Newtons First Law

• If no force acts on a body,
• then the bodys velocity cannot change i.e.
  the body cannot accelerate.
• If body is at rest, it stays at rest, if it is in
  motion, it continues to move with the same
  velocity

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Ch 5-4 Force

• A force F is measured by the acceleration a it
  produces.
• Principal of superposition of forces
• When two or more forces are acting on a body
  resultant force or net force Fnet can be obtained
  by adding the forces individually Fnet ? F
• If no net force acts on a body (Fnet 0), the
  body velocity cannot change.

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Checkpoint 5-1

• Which of the figures six arrangements correctly
  show the vector addition of forces F1 and F2 to
  yield the third vector, which is meant to
  represent net force Fnet?

• Ans c, d and e

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Ch 5-5 Mass

• Mass is an intrinsic characteristic of a body
• Mass of a body relates a force acting on a body
  to the resulting acceleration
• The ratio of the masses of two bodies m0 and mx
  is equal to inverse of their acceleration a0 and
  ax when same force is applied to both bodies
• mx/m0 a0/ax

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Ch 5-6 Newtons Second Law

• The net force Fnet acting on a body is equal to
  the product of the body mass m and its
  acceleration a
• Fnet ma a Fnet / m
• Acceleration component along a given axis is
  caused only by sum of forces component along that
  axis
• ax Fnet,x /m ay Fnet,y /m az Fnet,z /m
  
• SI unit of force Newton (N)
• 1N 1kg.m/s2

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Checkpoint 5-2

• The figure here shows two horizontal forces
  acting on a block on a frictionless floor. If a
  third horizontal force F3 also acts on the block,
  what are the magnitude and direction of F3 when
  the block is
• (a) stationary
• (b) moving to the left with a constant speed 5
  m/s?

• ?F F3 5-30
• Then F3 -53 -2 N
• in both cases (a) and (b)

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Checkpoint 5-3
The figure shows overhead views of four
situations in which two forces accelerate the
same block across a frictionless floor. Rank the
situation according to the magnitudes of (a) the
net force on the block and (b) The acceleration
of the block, greatest first

• Calculate Fnet F1 F2
• (a) (1) Fnet538
• (2) Fnet5-32
• (3) Fnet ?(5232)?346
• (4) Fnet?(53cos?)2(3sin?)2
• gt 6
• Ans (1),(4), (3), (2)
• Acceleration of block a
• a Fnet/ constant block mass
• The order of acceleration is same as that of Fnet
• Ans (1),(4), (3), (2)

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Ch 5-7 Some Particular Forces

• Gravitational Force Fg
• The force with which a body is pulled towards
  earth. Downward force Fg m(-g)
• Weight
• Equals the magnitude of a net force required to
  prevent the body from falling freely, as measured
  on the ground.
• weight W Fg

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Ch 5-7 Some Particular Forces (FN and T)

• Normal Force FN
• When a body presses against a surface, the
  surface pushes the body with a normal force FN,
  that is ? to the surface
• For a block resting on a table
• FN mg
• Tension Force T
• It is the force applied to a body through a
  cord, pulled taut opposite to object. For a mass
  less cord , the pulls at both ends of the cord
  has same magnitude T

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Checkpoint 5-3

• (a) For constant speed
• Fnet0 FN mg0
• FN mg
• (b) For increasing speed
• Fnetma FN mg
• FN ma mg
• FN gt mg

• In figure shown below, is the magnitude of the
  normal force FN greater than, less than, or equal
  to mg if the block and table are in an elevator
  moving upward
• (a) at constant speed
• (b) at increasing speed

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Checkpoint 5-4
(a) For constant speed Fnet0 T mg0 T
mg 75 N (b) For increasing speed Fnetma
T mg T ma mg T gt
mg (c) For decreasing speed Fnet-ma T mg
T mg - ma T lt mg

• The suspended body in the figure (c) weighs 75 N.
  Is T equal to greater than, less than 75 N when
  body is moving upward at
• (a) constant speed
• (b) at increasing speed
• (c) at decreasing speed

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Ch 5-8 Newtons Third Law

• Newtons Third Law
• When two bodies interact, the forces on the
  bodies from each other are always equal in
  magnitude and opposite in direction.
• For the book and crate
• FBC-FCB
• FBC and FCB are called third-law force pair
• FBC and FCB are also called action and
  reaction forces pair

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Ch 5-9 Applying Newtons Laws

• Problem-solving strategy
• Draw a simple and neat diagram of the system
• Isolate the object whose motion is being
  analyzed. Draw free-body diagram for this object
  for the external forces acting on the object
• Establish convenient coordinate axes for each
  object. Find component of forces along these
  axes. Apply Newtons Second Law ?Fma in
  component form.
• Solve the component equations for the unknowns

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Ch 5-9 Applying Newtons Laws

• For (a)
• ?FyFN-FgsMay0 FN Fgs
• ?FxMaxTMa TMa
• For (b)
• ?Fx0
• ?FyT-FgHT-mg-ma
• Tm(g-a) and TMa
• Solve for a
• a mg/(mM)

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Ch 5-9 Applying Newtons Laws Sample Problem 5-6

• Find T1, T2 and T3 ? For (a)
• ?FyT3-Fg0 T3 Fgmg and ?Fx0
• For (b)
• ?FxT2cos 47?-T1cos 28?0
• T2cos 47?T1cos 28? (1)
• ?Fy T2 sin 47?T1 sin 28?-T30
• T2 sin 47?T1 sin 28?T3 .(2)
• Solve eq. (1) and (2) to get T1 and T2

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Sample Problem 5-7

• Find T and FN if ? 27º
• Resolving force Fgmg along and the plane and
  perpendicular to the plane
• T-mg sin? 0 T mg sin ?
• Also FN-mg cos ? 0
• Then FN mg cos ?

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Checkpoint 5-7

• In the figure , horizontal force F is appled to a
  block on a ramp.
• (a) Is the component of F that is perpendicular
  to the ramp Fcos? or Fsin?
• (b) Does the presence of F increases or decreases
  the magnitude of the normal force on the block
  from the ramp?

• Ans
• (a) Fsin? ? to ramp
• (b) ?F ? to ramp
• N-mgcos?- Fsin? 0
• Then N mgcos? Fsin?
• Fsin? increases the magnitude of the normal force
  on the block from the ramp

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Checkpoint 5-8
Answer In free fall ?F N-Fg-Fg Apparent
weight N N -FgFg 0
In the figure , what does the scale read if the
elevator cable breaks sp that the cab falls
freely that is what is the apparent weight of
the passenger in free fall?
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Sample Problem 5-9

• Find a?
• Introduce action and reaction par forces FAB and
  FBA
• Fapp(mAmB)a
• a Fapp/(mAmB)
• Then FBA mBa