in this video we're looking at a block that's being dragged along the ground
we have friction acting we're told that it starts from rest we're given of course the mass of the
block the tension in the rope that we're using to drag it and the angle for that rope and then we're
asked a whole series of questions that requires a force analysis on this block so before i even
read the questions i'm going to look at the components of the tension i know i'm going to
need those eventually and that tension is a force vector with a length of 150 newtons i'm pointing
38 degrees above the horizontal and i want to decompose that into its y component that's the
part that points up and it's x component that's the part that points to the right
and that x component is going to be a 150 times cosine of 38. this gives me about 118.2 newtons
the y component is given by 150 sine of 38. this gives me about 92.3 newtons all right so that's
out of the way and then i'll read part a of the question i want to get the normal force on the
block and the most common mistake that you can make here is to say the normal force is equal
to the weight or mg the reason that's not going to work is that we have this additional force
tampering with the vertical directions so that y component of the tension is messing with
the sum of forces in the y direction so let's draw everything else that's going on in the y direction
certainly mg force of gravity on this thing is pulling straight down and we'll go ahead and
just put the numbers in the diagram that's 18.5 times about 9.8 this gives me a force of 181.3
newtons and then we have our normal force that's the force the floor is exerting and
that's pushing up and i'm just going to label that for now as a little n
now certainly the block is not going to jump off the surface and that's because the vertical
component of tangent is less than the weight and it's not going to sink into the surface in
other words the y acceleration is zero and that means the forces are balanced in the y direction
so it's fair to start this way i can say f up equals f down and i have a normal force
pointing up and i have a 92.3 newton component of the tension pointing up and i have gravity
pointing down and when i enforce this idea that the y-direction forces are balanced i can find
my normal force and this comes out to 89.0 newtons next i'm asked to find the kinetic friction force
on the block well we're going to drag the block along to the right and the friction force acts
opposite the direction of the slipping so i'll put that in the diagram i'm going to call it fk and i
remember that's given by mu k the coefficient of kinetic friction times the normal force now that
i have the normal force i can very quickly get the magnitude of my friction force that's mu k times
the normal force i was given mu k in the problem that's 0.44 and my normal force is 89.0 this gives
me a force of about 39.2 newtons now in part c i'm asked to compute the acceleration of the block
well this is what's going on in the x direction so we apply newton's second law in the x direction
i can subscript the net force with an x and the acceleration with an x just to emphasize that and
the net force in the x direction if we just adopt the most common coordinate system where rightward
is positive the net force is going to involve a positive 118.2 so that's the x component of the
tension pulling to the right and then a negative for the kinetic friction force a negative 39.2
so my net force is 118.2 minus 39.2 and that's equal to the mass of this object 18.5 kilograms
times the unknown acceleration and we crunch the numbers we do the subtraction and then divide by
18.5 and this gives us about 4.27 meters per second squared finally in part d we're going
to combine this acceleration result with our knowledge of kinematics and we're asked to get
the distance the block slides in the first two seconds of this motion again it started from rest
so we just plug into our first kinematics equation this is the kinematics equation describing the
position of an object as a function of time we'll go ahead and put the origin right there at the
initial position so my initial x value is zero it started from rest so my initial velocity is zero
and all i've got is this one half a t squared term so one half times 4.27 times t
squared we were told it runs for two seconds so t is two and this gives me a sliding distance
of about 8.54 meters and we're done if you enjoyed this video or at least found it useful
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