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## Physics 04, Problems based on Constrained Relations

Motion on inclined plane. Motion on inclined plane. What is inclined plane? See, the
triangular prism structure, this is called inclined plane. And if we see from this side, what we
can see is, this one, the cross section view of inclined plane sides, okay. This is the base, this
is the height and this particular angle, which is most important. It is known as angle of
inclination. This is the angle we are talking about. So, this kind of structure will be
represented like that. We will discuss about motion of block. Actually, the block is kept here
and the block will slip along the inclined plane. This is the structure which we will get. This is
the structure we will get always in books, text books, tests. So, this is the story of inclined
plane. Sometimes, we fix this on horizontal surface on this inclined plane, we make this type
of structure. This represents the surface. So, this inclined plane is placed on the surface,
okay. Any situation based on any topic of Newton’s laws of motions, its basic analysis starts
from, in the first step we will draw a free body diagram. What is our planning? WTN, so, let’s
say mass of the block is m. So, on this block, mg, downward force will be applied, yes or no.
And which other force will be applied. T, T is there and, where will the normal force be
applied. I told we have to apply normal on the block. Keep your hand on the other surface
and push it. Where will it get pushed? Perpendicular, away. So there will be normal on the
block. So, there are two forces presently acting on the block, one is the weight and the other
one is the normal force. This is the normal on the block due to wedge, got my point. Wedge
is the triangular plane or inclined plane. Call it wedge or triangular plane or triangular
wedge, is one and the same thing, done. And we can see that, we can see that this block can
move along this direction. Otherwise the inclined plane will come down or it may go
upwards, depending upon the other forces available in the systems, okay. First of all we will
see what happens. First I will tell you one thing which will almost come into use of every
problem of inclined plane. Listen carefully, what I did here is, I drew a line here like this. This
angle is 90 degree, sorry, this angle is theta, this angle is 90 degree, so what this angle will
be? It will be 90 minus theta, so, this angle is theta, yes or no. if this angle is theta, can I
make two components of this weight. One is mg cos theta and other one mg sin theta. What
we will take from the vector, only components, mg cos theta and mg sin theta, done. Now,
see for yourself, if this can move in this direction then this force should be balanced or not.
In that case, one thing is very much clear that Normal is equal to mg cos theta, okay and
which force is unbalanced, mg sin theta, so the resultant force on this side is mg sin theta.
So, that will provide some a acceleration to the block, yes or no. We will make FBD, we will
draw acceleration and we will write resultant force is equal to mass into acceleration. So,
what is the final acceleration? It is g sin theta. And we have to make a component, this is a
bright idea. It will be more useful to you in future that is why, keep this as a standard in your
mind. If a block, is going along the inclined plane is going up or down, we will comprehend it
with one force that will be mg. Remember mgcos theta and mg sin theta. Use your head and
write straight away mg cos theta and mg sin theta. If this theta is given then then this will be
mg cos theta and this will be mg sin theta. Suppose the examiner uses his head and gives
you this theta then what will happen. Cos will become sin and sin will become cos. So just
remember this FBD always….okay, mg cos theta mg sin theta. N is equal to mg cos theta is
always valid. But acceleration g sin theta (6:53) but there are only two forces, WTN and
Normal. (07:03) constant acceleration g sin theta. But if there is some force along with this
so definitely g sin theta (07:18). But the motive behind giving you this example are these two
things which you should stick it inside your mind mg cos theta and mg sin theta (7:29-7:45).
Find the acceleration of the block? Find the acceleration of the block? Done, there is one
stationary triangular wedge, on that we placed a pulley and connected two blocks with it
with a string. We are releasing the system from rest, you have to calculate the acceleration
of the block. What will be our first step? Free body diagram, let’s see if everybody is
connected correctly in first step, 60, T after that T now mg sin theta. How much is it? 60, mg
cos

2016-06-13T14:17:20+00:00 Categories: Archive - 2015-16||0 Comments
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