Today we'll take one more numerical that
is a question number three let us see the
question in graphically then we'll saw I have
an inclined plane  like so and the object
is placed in this is inclined plane like
this
the object is mass of four kilogram is
released the object of mass for kilogram
is released on this point and this
surface hence on friction this is some
fiction say neither point right I'll
consider that this angle is a say I will
take 30 D okay so this is an angle 30
degree there's a block of mass 4 kg is
released and the surface is some
friction so now the question is there
can be various types of questions that
can be put one is whether the block will
slide or not slide so forth that is a
question that you will find if block
slides
so we'll see the blocks lines are not
first that will take second so because I
have taken this from my own mind so I
don't know that slider box will take
that after that I will make other
questions okay so let us see so this
know if the block has to move down
because it will never move up that is
for zero if you keep playing like this
indefinitely either it will stay there
on the fall down like this right though
but if you put it here to not never move
up so that part is very clear so you see
first method is moving down a lot what
force is responsible for e2 for this
downward motion that will see here let
us draw a simple diagram here this is
the weight mg that is acting on this
block I resolved this block force into
two components one L of this and another
this this this angle is 30 or theta I
will call this our two theta that is 30
so you take your general angle theta so
this moment is M G cos theta and this
component is m-g sine-theta so what
we've done is we have resolved this mg
the weight into two components one along
the puzzle along the incoming plane and
another perpendicular to it now because
the block is in contact with the engine
plane that will be perpetration so and
so normal direction with another
recovery is suppose this is a direction
this is a normal reaction force that
will act a normal to this surface so
obviously it should not be in this
direction it should not be in the
direction of mg as in the case of
earlier cases because you know that when
I do
is placed on the Hollanders office so
wait is that word and normal reaction is
awkward but this is not the case in case
of enjoying plane the normal reaction is
not in the direction of weight in my pee
direction is always in the direction of
the inclined plane the direction of the
2002 the direction of the surface in
which it is it so in case of a
horizontal surface
Nagaraju and weight will be in the same
direction but in case of impact plane
and you can see here the backward
reaction is not in the direction
opposite direction of mg because mg side
is the combination is not like this part
in the case here okay the surface here
it is normal to the surface here also
this is normal to the surface now then
we have to find what is the operational
force maximum value of frictional force
that will find the maximum value of
frictional force that is f10 yielding
friction how much it is that mu has into
n what is me waist 0.5 is the question
what is n now n is equal to mg cos theta
okay now n is equal to nd cos theta you
can see this tie line is that line this
and this mg cos theta will be equal
because when the block slides or doesn't
slide there is no movement in this
direction I can call this as X direction
and this as y direction
so there is no movement of the block to
the Y direction so therefore these two
forces in this direction must be same
therefore n is equal to mg sine theta
okay now putting the value we get 0.5
and then how much M is for G stay so
fall into G for Newton for T so right
for 10:40
cos theta theta is 60 if it is 1330
prostatic so how much is now now cos 30
is root 3 by 2 okay for 30 is root 3 by
2 this is
five is for state that you are
multiplying seem of maturity now this is
for pencils
this is Ruth rebooting is 1.73 into ten
so probably it will be 70 by three
students always telling you that the
maximum
six mangoes this is 17.3 Newton so no
the so this is the maximum up to which
it can prevent now if I put a block on
the inclined plane it has a tendency to
slide it has the tendency to slide whoo
what is the force which is making it to
slide it is mg sine theta and you can
see here this is our force this is a
component of gravitational force which
is acting along the inclined plane and
it is responsible for the downward
motion of this block now what is its
value that also will to find out because
this plays a role in the down motion now
let us find out that downward force that
is acting along the block is m-g
sine-theta let us find out that M is
for
this comes out to be
- so is ready
20 different forces acting downward and
fiction has the maximum of 17.3 that
means the body will slide down because
this reputational force that is pulling
it in this direction is more than the
maximum well your section force
therefore the Freebody diagram for these
will be you can see here section in this
direction will be 17.3
and decided to go
because of design theta21 act in the
direction
so therefore the difference is downward
so it will move downward the first thing
is if the blocks lines are not yes the
answer is yes it will it's like okay
the block will slide now the next
question we can make is if it slides
find the acceleration of the block okay
so let us find the acceleration of the
block find the acceleration of the
blocks down the plane now this was the
situation 17.3 Newton is acting off
because of frictional force 20 Newton
forces acting downward because of the
gravitational force its mass is 4 kg so
therefore along this line and on this
line the acceleration is net force
Tomas
that is 20-17 point 3/4 which is this
two pointer this is point seven two
points in okay this is two point seven
divided by 4 that is there whatever is
the case that is two point seven back
for meter per second square this is the
answer to this question so therefore
this is how you find out acceleration
and now if I say how much extra force is
needed to prevent the block from sliding
down
then obviously you say it is two point
seven because if I applied two point
seven in this direction there will be no
motion of this block if I say if I apply
a force of 20 Newton in more than 20
Newton in the upward direction suppose I
apply a force of 25 Newton then what
will happen to the friction with the
question
what is efficient if 25 urban force
is a plan of the plane
to find it up right off the plane 20
Newton is always there because of
gravity now what will happen now if the
interesting point is that then I applied
25 Newton force and this is 20 the net
force is try it with an upward so
because the net force is wracked with
Nepal that there is a tendency of this
body moving in our direction therefore
the friction will act in the downward
direction so in this case the friction
will be in the downward direction so
therefore friction will act in this
direction and can the body move the
answer is obviously known because the
friction we have calculated that its
next value last time it over 17.3% is
point 3 so therefore it can provide now
what is the friction in this case will
be the friction that is acting - is this
for the friction that will add in this
case is only 5 Newton because this will
prevent because 20 is already there
trial will be there 25 25 so there is no
acceleration in this case the friction
that will act
in this case is 5 Newton they're on the
plane so you must remember them big no
acceleration then what should be the
maximum force you apply for there is no
acceleration then obviously you can say
20 Newton is there it will be always
there and FN is 70.3 so if you add then
there is 37.3 so if the force applied in
this direction in this direction is more
than 37.3 then only the block will start
moving upward okay
thank you
