The
problem.
An inclined plain with the length of 1 m and the
height of 50 cm has a pulley at its top. A block of 280 grams, which is on the
plain, is connected to a freely hanging 120 grams block by means of a cord
passing over the pulley.
Compute
the distance the each block will pass in 2 s starting from rest.
Consider
two situations: the coefficient of kinetic friction is
a)
μ = 0.05, b) μ = 0.1
The tips
Types
of motion
|
“Check this out”
→
↓ |
trajectory is a strait line |
trajectory is no a strait line |
|
speed is constant |
a linier motion with a
constant velocity (LMCV), acceleration is 0 (you can use all the other
formulas , which are right for this type of motion) |
usually it is a circular
motion with a constant speed (CMCS), there is a centripetal acceleration (you can use all the other
formulas , which are right for this type of motion) |
|
speed is no constant |
usually a linier motion
with a constant acceleration (LMCA) (you can use all the other
formulas , which are right for this type of motion) |
usually a projectile motion
(PM), the acceleration is g = 9.8 m/s2 (you can use all the other
formulas , which are right for this type of motion) |
The
second Newton’s law: Fnet
= m∙a
The
properties of a right triangle:
a2
+ b2 = c2
a/c = sin
(α)
b/c = cos
(α)
c a/b = tan (α)
a
┐ α (
b
Weight = mass∙g (g = 9.8 m/s2)
The reasoning
As usual, first, let’s analyze the text of the problem.
What objects
do we have? What bodies do we have in this situation? An inclined plain, a
block, a pulley, a cord, one more block. And do not forget about such an
important object as Earth. We are talking about a real situation.
What kind of characteristics, parameters, properties can we assign to each of
these objects?
The blocks have a mass. Let’s take a
capital letter M for the mass of the
first block, and m notates the mass
of the second one. Let’s l is the
length of the plain, h is the
height.
What can we say about the cord and the pulley?
As usual, we assume that the cord dose not has any mass, the length of the cord
is fixed and there is no any friction between the cord and the pulley.
After considering the objects, we go over
to physics processes.
What kind of processes can we describe in
this situation?
We know already, there are three the most
common physics processes: a motion, an
interaction and a transformation.
Let’s talk about motion first.
There should be a motion in this
situation. It is possible the system is in equilibrium. But, we have some key words like “the distance” or “pass
in 2 s”. These words describe motion. They mean that the position of the blocks
is changing during a time. The changing
of the position is motion.
What type of motion should we consider for
this problem? What type of motion do we have?
Remember, we have 4 basic types of motion (take a look on the page I gave you). What
type of the motion do we have to use for our system? We can definitely
eliminate the circular motion and the projectile motion. Now, the bodies start
the motion from rest. It means, the speed of the blocks is changing from 0 to
some non-0 number, the speed is no constant. So, we should eliminate LMCV.
The only motion we can use is LMCA, a linier motion with a constant acceleration.
Let’s fix it on the board. OK. Now.
What can we say about interactions? We have a lot of interactions over here.
Usually, every interaction produces at
least one force.
First, it is obviously; Earth is acting on the blocks. The
result of this acting is the weight. So, the first block has the weight, let’s
use the capital W for it, and the second block has the weight w.
Are there some other interactions? Yes
there are. There is an interaction between
the block and the cord. It means, there is a force, which is exerted by the
cord and acting on the block. Let’s use capital T for this force. There
is one more interaction between this block and the cord. We have the force; let
mark it with the letter P, which is acting from the cord on
the block.
Do not forget; any force is a vector, it
has a direction.
Talking about the cord, we have to recall,
this is a special cord, with no
mass, with fixed length, without any friction. That cord has the same magnitude
of a tension at any point. It means, the forces T and P have
the same magnitude. T = P.
In the end, we have two more bodies, which
are interacting. The block and the plain. This interaction leads to a pair of forces. The first one is the
normal force N, which is keeping the block on the surface. A normal force
works against a weight, it is directing perpendicularly to the surface.
The last force is a friction force. This force is produced by the surface, but its
direction is parallel to the surface. A friction force is a most mystic force
among all the named. First, we should recall that there are two kinds of
friction forces; the static friction force and the kinetic friction force.
Second, it is not obviously some time how
to find the direction of a friction
force. For example, the direction of the kinetic friction force has to be
opposite to the velocity of the body. But, the problem is; we do not know in
what direction the bodies will be moving, when we let them go. More ever, if
the static friction were big enough, there would be no any motion.
So. There is a friction force, F. But, the direction of this force
is still an enigma. We cannot begin
solving our problem without solving this enigma. Hence, the first step of the
solution is finding the direction of the friction force.
Any ideas?
Let’s say, we do not have any friction at
all in the system. Can we find the velocity of the blocks for the frictionless system? Yes, we can.
When we find the velocity of this block with no friction, the direction of the
friction force will be opposite to the velocity. So, the idea is, we have to
solve this problem first without a friction (we could say that μ = 0). I’m
erasing the friction force. And, the goal is to find the direction of the
velocity of this block.
By the condition of the problem, the block
starts the motion from rest. And, as
we said already, we have LMCA. By properties of this type of motion, if the
motion is starting from rest, a velocity and acceleration have the same
direction. Hence, we need to find the acceleration now.
So what do we have? We have bodies, we
have forces and we look for acceleration. Bodies,
forces, acceleration. It is sort of hypnosis.
What do these key words mean? They mean we
have to use Newton’s laws of Motion.
Let me recall the most important second
Newton’s law of Motion.
The net force is equal to the mass
multiplied by the acceleration.
We must apply this law to each of the
bodies, or at least to one of them.
It is very convenient using a free-body-diagram. I want to indicate
the plain. And, we have three vectors for the first block: N, W, T, and two vectors
for the second one: P, w.
It is obviously, the acceleration of the
first block should be parallel to the plain. Let’s assume, that the
acceleration is pointing to the top of the plain. What if we made a mistake,
what if the real acceleration goes to the bottom? We’ll find the answer later.
Now we can write down the second Newton’s
law.
The net force, which is the sum of all
these forces, N + T + W is equal to the mass M times A.
To calculate some quantities we have to
convert the vectors into some numbers. We
need to choose X and Y-axis. The most convenient way is to direct the
X-axis along the acceleration, and the Y-axis along the normal force. Now we
can write the second Newton’s law in components, which are numbers.
We have three forces, the acceleration and
two components for each of the vectors. Let’s make the table.
|
vectors\components |
X |
Y |
|
N |
Nx = 0 |
Ny = N |
|
T |
Tx = T |
Ty = 0 |
|
W |
Wx = -Wsin(α) |
Wy = -Wcos(α) |
|
A |
Ax |
Ay = 0 |
The X-component of the normal force is 0.
The y-component is equal to the length of the vector, that is, to the magnitude
of the force.
The X-component of the “pulling” force T equals
to its magnitude. But, the Y-component is 0.
The more zeros, the better.
The acceleration looks like the force T.
But, we are not sure about the direction of the A, it can be opposite.
So, we can write just Ax and 0.
I want to recall, we are looking for the
acceleration. Hence, we want to find this number. It is possible to get the
negative number for Ax. If so, what would be that mean? The negative number for Ax means, the real acceleration is directed in opposite
direction, that is, to the bottom of the plain.
OK, we have one more force to break it
down into components. The weight.
Unfortunately, we have to keep two components of the weight. This
is the X-component, and that is the Y-component. W, Wx and Wy form the right triangle, which I want to
draw separately. We have two legs and the hypotenuse. The lengths of the legs
are equal to the magnitude of the components. By properties of a right
triangle, we can write the following formulae. We can solve these equations for
the components. Because, the components are directed against the axis, they
have to be negative. It means Wx = -Wsin(α) and Wy = -Wcos(α).
Now we can write down the second Newton’s
law in components.
First, we have to rid off the small arrows over the vectors, and write down instead
of them small letters x and y, which indicate the components.
Instead of one vectorial equation we have now two scalar equations.
Nx + Tx + Wx = M*Ax and
Ny + Ty + Wy = M*Ay .
The second step is; we have to substitute the values of the components, taking them from the table and plugging into the equations.
It leads to following equations. 0 + T - Wsin(α) = M*Ax
and
N + 0 - Wcos(α) = M*0
After taking in account all zeros, we got
T - Wsin(α) = M*Ax and
N - Wcos(α) = 0
If you have any questions, you have to stop me at any time and ask your question.
OK, I
have a question, who remember yet what are we looking for?
We want to find the acceleration (which is
necessary to find the velocity, which is necessary to find the direction of the
friction force).
May be we can find Ax already? How can we check it out? We can just plug all the
numbers into the equation. This is the first time during the solution, when we
have thoughts about numbers. It is because Physics is more than just Algebra or
Geometry. Physics is a special way of thinking about Nature.
OK. Let’s go to the numbers.
What must we have done before using any
numbers? We have to convert all the numbers into the International System of units. So, we have the mass of the first
block is 280 grams, which is exactly .28 kg.
…
What numbers can we use right now to
compute the acceleration? We do not have T, no W, no alfa. We can plug into the
equation just 0.28.
Ax is unknown. It’s obviously; we need more numbers, that means, we
need more information. Let’s start with alfa. Alfa is a geometrical quantity. We have two more geometrical
quantities; l and h. Probably there is a connection between all this
geometrical stuff. We need a good sketch. The plain, the length is 1 m. The
height is .5 m. this is a right angle, So, we have a right triangle. The
hypotenuse is 1, the leg is .5. We can find any angles for this triangle. The
question is; what angle do we need to find? What of these two angles is equal to alfa?
Let’s see. This is a right angle, which is
90 degrees. Hence, this angle is 90 – alfa. I can make one more right triangle.
This angle is 90 – (90 – alfa) which = alfa.
By the property of a right triangle, while
the leg is opposite to the angle, the leg divided by the hypotenuse equals
sin(alfa).
Plug the numbers. We’ve got sin(alfa) =
.5, that means alfa = 30 degrees.
OK. We have found one more number for this
equation. Next letter is W. W indicates
the weight of this block. Weight is a special word; there is a definition
for a weight. A weight equals mass multiplied by the acceleration due to the
gravity. We know the mass, and we know g. Hence, we can calculate the weight.
W = .28*9.8 = 2.744 N.
After all the numerical considerations, we
have now the equation:
T – 2.744*sin(30) = .28*Ax
We cannot solve this equation to find Ax,
because we do not know T.
The second equation is not helping,
because it dose not content T.
We still have a deficit of information. We
need some additional source of information. What could it be?
We have the second body. We can apply the
second Newton’s law to this body. It should be helpful. At least, we do not
have any more resources.
There are just two forces, which are
acting on the small block. The second Newton’s law for the block has to be
written as following: P + w = m a
The direction of the acceleration a
cannot be chosen at our will. There is a connection
between the motions of the blocks. When that block goes to the top of the
plain, this block goes downward. Why? Because of the gravity and because of the
cord.
We have to direct the acceleration
downward. Let’s choose the X-axis downward as well. These to vectors have the
positive components, but that have the negative one. We can write –P + w = m*ax
We know the mass m = .12 kg, and we can
find easily the weight w = .12*9.8 = 1.176 N. What can we say about P? We
already have talked it of. P = T. The magnitudes of these forces are equal. Therefore,
we can write T instead of P.
The
last letter, which we have to talk of
is ax. ax is the acceleration of the second block. This
block is connected by the cord to that one. And, the length of the cord is
fixed. It means, that when the first block pass any distance, the second one
must pass exactly the same distance in exactly the same time. Therefore, the
magnitudes of the velocities and the accelerations of the blocks are the same.
Ax = ax.
(By the way, can I write A =
a?)
Let’s make our last substitution, let’s
write Ax instead of ax.
After all these manipulations, we have two
the following equations
T – 2.744*0.5 = .28 * Ax
-T + 1.176 = .12*Ax
We do not need to know T. Let’s eliminate
it simply by addition of the equations.
T – T = 0. -2.744*.5 + 1.176 = -.196 .28 + .12 = .4 In the end of all
computations we have A x = -.196/.4 = -.49 m/s2
It dose not matter what number did we get.
The point is this number is negative.
It means, the acceleration of the first block is directing to the bottom of the plain. So do the velocity. And, at last, we can say with sure that the direction of the friction force is opposite to the velocity, that is, the friction force is directing to the pulley. Finally, we’ve got it. I want to draw the friction force on the picture, and on the free-body-diagram. Now, we can start solving the original problem.
But it will be your home work.