Three masses pulley question


by artyboy
Tags: acceleration, block, friction, mass, tension
artyboy
artyboy is offline
#1
Oct18-11, 10:30 AM
P: 3
1. The problem statement, all variables and given/known data

The three blocks are attached via a massless, frictionless pulley system, as
shown. The frictionless plane is inclined at an angle  = 60 degrees. When released
from rest, the 20kg block will start to slide down the plane. Find the tension in the
string.
The image is attached.
2. Relevant equations
Fnet = ma

3. The attempt at a solution

For the 20 kg block I got, ma = mgsinθ - T
2kg block- ma = 2T - mg
3kg block: ma = 2T - mg
The only problem is that the acceleration of all of the masses is different. How do I find the acceleration of each? Is it a ratio of the tensions, but then the 2kg and 3kg would have the same acceleration which wouldn't make sense?
Attached Thumbnails
pulley.png  
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collinsmark
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#2
Oct18-11, 09:18 PM
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Quote Quote by artyboy View Post
1. The problem statement, all variables and given/known data

The three blocks are attached via a massless, frictionless pulley system, as
shown. The frictionless plane is inclined at an angle = 60 degrees. When released
from rest, the 20kg block will start to slide down the plane. Find the tension in the
string.
The image is attached.
2. Relevant equations
Fnet = ma

3. The attempt at a solution

For the 20 kg block I got, ma = mgsinθ - T
2kg block- ma = 2T - mg
3kg block: ma = 2T - mg
I think that looks about right so far.
I'd label my a's and m's though (something like a1, a2, and a3, m1, m2, m3 or whatnot), to avoid confusing them with each other.
The only problem is that the acceleration of all of the masses is different. How do I find the acceleration of each? Is it a ratio of the tensions, but then the 2kg and 3kg would have the same acceleration which wouldn't make sense?
It's not a ratio of tensions, no. It's a matter of geometry. The configuration of the system.

Suppose for a moment that the 2 and 3 kg masses are held in place (not allowed to accelerate) when the system is released from rest. In this situation, the 20 kg mass won't accelerate either. [Edit: the point being that the acceleration of the 20 kg mass is dependent on the acceleration of the other two masses -- and taking this a step further the acceleration of any of the three masses is dependent on the other two.]

Now let the 3 kg mass (and 20 kg mass) move freely when the system is released from rest, only holding onto the 2 kg mass. Now the 20 kg mass and 3 kg mass can both accelerate, but they won't both accelerate at the same rate. Look at the geometry of the system and determine a relationship between the 3 kg mass' acceleration and the 20 kg mass's acceleration. Once you figure that out, hold on, because you're not quite finished with this yet.

Do the same thing except hold the 3 kG mass in place instead of the 2 kg mass.

Now get crazy and hold onto the 2 and 3 kg masses together (or one in each hand, it's up to you) and lift both masses up such that they both accelerate at the same rate, say 1 m/s2. What's the acceleration of the 20 kg mass this time? You should be able to figure out a relationship (i.e. an equation) between the three accelerations. This equation is doesn't have anything to do with the tension by the way (so the equation is not going to have a T in it), it's just based on the configuration of the system.

That gives you your fourth simultaneous equation. Which is nice because you have four unknowns, a1, a2, and a3 and T. Four equations, four unknowns. The rest is algebra.


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