ATTWOOD'S MACHINE (rope + pully)

In summary: I think we can use Newton's second law, F=mat=0.9sd=0.36mv1=0m/sIn summary, the teacher said that the heavier mass falls towards one end of the rope due to the force of gravity. They used the equation d=v1t+1/2at2 to calculate the acceleration and found that it came out to be .889m/s2. Mass A has an acceleration of .889m/s2 and Mass B has an acceleration of .889m/s2. The sum of the forces acting on the masses is .3g.
  • #1
twenty5
93
0

Homework Statement


Determine the force of gravity acting on the masses (we want to end up with 9.8m/s2---> My situation : A rope with 2 different masses hanging on each side, is held by a pulley. The heavier mass falls towards one end. (teacher said something about the mass having to cancel out afterwards) or so.

Homework Equations


d=v1t +1/2at2
I think we can use Newton's second law, F=ma

t=0.9s
d=0.36m
v1=0m/s

The Attempt at a Solution


I used d=v1t +1/2at2 and got acceleration which came out to be .889m/s2

I was wondering what steps I can take now
 
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  • #2
How does the measured acceleration relate to the acceleration due to gravity (g)?

You'll need to apply Newton's laws to analyze the system. Apply F = ma to each mass.
 
  • #3
Doc Al said:
How does the measured acceleration relate to the acceleration due to gravity (g)?

You'll need to apply Newton's laws to analyze the system. Apply F = ma to each mass.

awesome ^^ i just tried that juuust right before you replied haha thanks Now I know I am sort of on the right track^^
 
  • #4
ok I am stuck again xD
 
  • #5
Show exactly what you did and where you got stuck.
 
  • #6
Doc Al said:
Show exactly what you did and where you got stuck.

ok So,
step 1:
d = v1t +1/2at2
got acceleration = .889m/s2

Mass A:

FA=ma
FA=0.889a

Mass B:
FB=ma
FB=0.889a

I got stuck right there, like what can I do with it? can I find net force because If I subtract FA and Fb, it'd be 0N O_O, but since I am 100% sure that the Net force isn't 0, what do I do? because Newtons second law states that, if the object is at rest, or at a constant velocity, the net force is 0. but the mass is acceleration towards the floor =P so it cannot be 0
 
  • #7
ehhh... anything? I am still stuck there lol...been working on it and still working on it ^^
 
Last edited:
  • #8
You must first identify the forces acting on each mass, find the net force on each (symbolically), and then apply Newton's 2nd law (∑F = ma) to each. You'll get two equations (one for each mass) which you will combine.
 
  • #9
I don't think you can finish it without knowing something about the masses.
The wording is inconsistent, too - it asks for a force, then says it should have units of acceleration. Could you check the question again and make sure it has been copied word for word?
 
  • #10
Also:
twenty5 said:
ok So,
step 1:
d = v1t +1/2at2
got acceleration = .889m/s2
This makes sense, assuming you solved for the acceleration correctly.

Mass A:

FA=ma
FA=0.889a

Mass B:
FB=ma
FB=0.889a
But I don't know what you're doing here. Each mass is different, so call them m1 and m2. Your equations will look something like:

∑F(on m1) = m1*a
∑F(on m2) = m2*a

Note that if m1 has an acceleration of a downward, then m2 must have an acceleration of a upward.
 
  • #11
MMMM, i just found out that we can use a value for mass...

m1 = .1kg
m2 = .2kg
and now I need to figure out how to get g = 9.8
 
  • #12
Oh, now you can do it knowing the masses!

Sum of forces = ma, where m is the total mass.
Put in an mg for each force.
 
  • #13
ok I'll try that ^^
 
  • #14
ok I got... for the sum of forces is the following...

FA= mg
FA= .1g N

FB= mg
FB= .2g N

FNET = FB - FA
FNET = .1g N

Fsum of forces = .3g
.1g = .3g?

and now I don't know how this will get me to 9.8 ^^

mmmmmmmmmm

I also tried doing...

Ftotal force = ma, where I subbed in the a that I obtained from before...

0.1g = 0.3*0.889
g = 2.6m/s2...

But If I heard the teacher correctly, I believe I heard I'm supposed to get 9.8 or something lol
 
Last edited:
  • #15
sum of forces = ma
.2g - .1g = .3*0.889
you can easily find the ACCELERATION of gravity.
No, it isn't working out to 9.8 as you would expect . . . on Earth.
Your question says find the FORCE of gravity - not sure what that means in this case since there are 2 different masses so two different forces of gravity. Check the wording again?
 
  • #16
Delphi51 said:
sum of forces = ma
.2g - .1g = .3*0.889
you can easily find the ACCELERATION of gravity.
No, it isn't working out to 9.8 as you would expect . . . on Earth.
Your question says find the FORCE of gravity - not sure what that means in this case since there are 2 different masses so two different forces of gravity. Check the wording again?

sorry, the question just says

Attwood's Machine - Measure Acceleration, calculate g.

I also did the experiment physically too =P

I made a pulley with a rope and a mass on either end. and I just let it drop, timed it, and measured the distance =P
 
  • #17
Are you sure the values of time and distance are accurate? Given those two masses g should be equal to 3a.
 

Related to ATTWOOD'S MACHINE (rope + pully)

1. What is Attwood's Machine?

Attwood's Machine is a simple mechanical device that consists of a rope wrapped around a pulley and attached to a weight. It is used to demonstrate the principles of mechanical advantage and the conservation of energy.

2. How does Attwood's Machine work?

The weight is suspended from the rope, which is wrapped around the pulley. When the weight is pulled down, the rope on one side of the pulley is shortened while the other side is lengthened. This creates a mechanical advantage, allowing the weight to be lifted with less force. The pulley also serves to redirect the force, making it easier to lift the weight.

3. What is the principle of mechanical advantage?

Mechanical advantage is the amplification of force achieved by using a machine, such as Attwood's Machine. It allows a smaller input force to lift a larger weight by using a combination of leverage and redirection of forces.

4. How does Attwood's Machine demonstrate the conservation of energy?

Attwood's Machine demonstrates the conservation of energy by showing that the work done on one side of the machine is equal to the work done on the other side. The weight may be lifted with less force, but the distance it is lifted is greater. This trade-off between force and distance shows that the total energy remains the same.

5. What are some practical applications of Attwood's Machine?

Attwood's Machine can be used in various applications, such as lifting heavy objects with less force, creating a block and tackle system for lifting objects, and in elevators to lift and lower the car. It is also used in physics education to demonstrate the principles of mechanics.

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