dynamics of material point

Jul 2008
179
1
The two blocks shown in the figure are initially at rest. Despise the masses of the pulleys and the friction between different parts of the system. Get (a) the acceleration of each block and (b) the tension in cable.

Answer:
(a) aA = 2,56 m/s^2; aB = 1,28 m/s^2
(b) 677 N
 

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May 2009
102
44
Mumbai,India
consider the fixed pulley on the wedge
Let the block on the plane be at a distance xAfrom the pulley and let the movable pulley be at a distance xB

Now the idea here is to determine the accleration correlation between the two blocks.

For the block
A

T-(mA)g/2=(mA)aA.........(1)

T is the tension in the string.
for the moving pulley,
Forces are 2T upward and a downward tension T1
If the pulley is assumed to be light then
T1=2T

An upward tension of magnitude 2T thus acts on the hanging block, by newton's third law.

for the block B
(mB)g-2T=(mB)aB.........(2)

Now for the relation between aA and aB,
We have to use the principle of thread length conservation,
i.e. the total length of the thread doesnot change.

The total length of the string
is xA+2(xB)+d1(this is the length wound over the pulley and doesnot
change)=constant

so, differentiating this equation twice yields,(with respect to time)

aA=-2aB
therefore magnitudewise,
aA=2aB.......(3)

so use (1),(2),(3) to get the answers.
 
Jun 2009
79
9
Hi
consider the fixed pulley on the wedge
Let the block on the plane be at a distance xAfrom the pulley and let the movable pulley be at a distance xB

Now the idea here is to determine the accleration correlation between the two blocks.

For the block
A

T-(mA)g/2=(mA)aA.........(1)

T is the tension in the string.
for the moving pulley,
Forces are 2T upward and a downward tension T1
If the pulley is assumed to be light then
T1=2T

An upward tension of magnitude 2T thus acts on the hanging block, by newton's third law.

for the block B
(mB)g-2T=(mB)aB.........(2)

Now for the relation between aA and aB,
We have to use the principle of thread length conservation,
i.e. the total length of the thread doesnot change
All of the above I understand,Akshay.
What I don't get is ,how you get,
from this,
so, differentiating this equation twice yields,(with respect to time)
what is diffentiation exactly?is that like a math term?:eek:
(i checked it on wiki,but i ended up on this article with all these alien-like symbols)
Derivative - Wikipedia, the free encyclopedia

I hope someone can help.
Thank you
 

physicsquest

PHF Helper
Feb 2009
1,426
474
Sometimes wikipedia goes over the top.

Yes it is a math term. What it shows is the rate of change of one variable
(x in our case) with respect to (w.r.t.) the rate of change of another variable (time in our case).
Thus when we differentiate displacement x w.r.t. time, we get velocity
v = change in x / change in t. This is the first order derivative.

Similarly when we differentiate v w.r.t. t, we get the accleration

a = change in v / change in t. This is the second order derivative

If a term is a constant (like the length of the string here), then it cannot change with respect to time. Thus the derivatives of both orders are zero and you get aA=-2aB.
 
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Jun 2009
79
9
Yes it is a math term. What it shows is the rate of change of one variable
(x in our case) with respect to (w.r.t.) the rate of change of another variable (time in our case).
Thus when we differentiate displacement x w.r.t. time, we get velocity
v = change in x / change in t. This is the first order derivative.

Similarly when we differentiate v w.r.t. t, we get the accleration

a = change in v / change in t. This is the second order derivative
Oh,so that's what differentiation is.I get it now,
but now I'm having another problem,
i get the answer when I solve for aA and aB by just substituting their magnitudes,but why is it wrong to substitute it as , aA = negative 2 aB?
doesn't the direction of acceleration count here?

Thank you
 
May 2009
102
44
Mumbai,India
Well actually the sign convention is taken care of in the equations for Newton's second law. Hence I didn't consider that in the relation aA=-2aB I just took the magnitude. What this shows is that the accelerations are in opposite directions( by this I mean if one body goes towards the pulley,the other goes away, which is obvious.)

What you see here is actually the rate at which the two objects approach the pulley.These happen to be equal in magnitude of the accelerations of the bodies. (because the fixed pulley is equivalent to an inertial frame such as the earth.) So as one object is moving away and the other moving towards, the signs are opposite. Or in other words, the length of thread between one block and the pulley increases while that between the pulley and the other block decreases.

You can arbitrarily assume directions for accelerations and use the relation aA=2aB(relation betwen the magnitudes of the accelerations.)

if either aA or aB comes out to be negative, its direction is opposite to your assumed direction. If it comes positive,it is in the same direction.

Because clearly their directions are of motion are not what you'd generally call opposite. One object moves slant while the other moves vertically.
 
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