
#1
Jan1707, 06:13 PM

P: 150

1. The problem statement, all variables and given/known data
A mass, [tex]m_1[/tex]=5 kg, resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, [tex]m_2[/tex]=10 kg. Find the acceleration of each mass and the tension in the cable. 2. Relevant equations 3. The attempt at a solution For mass 1: Sum of x = F=[tex]m_1[/tex]a where F= Forward Force Sum of y= n  [tex]m_1[/tex]g=0 where g=9.8 m/s² How can I determine the acceleration using the data for mass 1? For mass 2: Sum of Y = [tex]T_2[/tex][tex]m_2[/tex]g where T= tension Hence, [tex]T_2[/tex]= 10 kg(9.8+a) Then, what do I do? 



#2
Jan1707, 06:28 PM

HW Helper
P: 858

ok let's hope I have got the picture correct. I have got things wrong twice already this week when answering questions......
firstly, no friction... greatly simplify the problem. so only one force on mass 1 that would actually cause movement/acceleration assuming the cable is massless/nonstretchable etc... also a perfect pulley of course... then you must have Tension on m1 equal Tension caused by m2.. so only one T. T is of course equal to something times g (what is that something eh?) ok.. forgot to read the question.. ok you want to answer acceleration for each mass for m2 ... easy for m1 ... use relation from the only one T you have for T ... once you know acceleration on m2 or m1... easy 



#3
Jan1707, 06:40 PM

P: 150





#4
Jan1707, 07:18 PM

HW Helper
P: 858

Determine acceleration/tension of masses
If I have got your picture correct... then your expression
T_2 = 10(9.8+a) is not quite right. 



#5
Jan1707, 07:20 PM

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P: 858

my picture (for crosschecking): gravity pulls m2, m2 pulls T, T pulls m1 and m1 accelerate....that's it




#6
Jan1707, 07:21 PM

P: 312

Acceleration is always the same for both blocks, and in most cases tension will also be the same (the only time tensions do not equal eachother is when the pulley is massive and its rotation is taken into account)
Because there is no friction, the forces in the y for block 1 are of no concern. For your second equation, however, the best choice for positive is the direction of acceleration. Once the signs are straightened out, plug tension into the first equation and solve for a 



#7
Jan1707, 07:37 PM

P: 150

I'm sorry I cannot put up the diagram, but I think you have the wrong picture. It's not the atwood machine, so the tension along the cord is not the same, since we are not dealing with one system where gravity is acting on it. Mass one moves due to the forward force, and mass two due to the gravitational force, and the normal force (which is the tension along the cord).




#8
Jan1707, 08:26 PM

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#9
Jan1707, 08:27 PM

P: 312

The forward force IS the tension in the rope. Especially with no friction, the rope is the only thing accelerating m1
One way to think of it is that the 2 tensions are internal forces. Therefore, they must cancel out because nothing inside the system is moving (remember, the pulley isnt rotating). In this respect, the problem is like an atwood machine because the entire thing can be called a system which gravity acts on. If anything, its a simplified atwood machine because only one force accelerates m1 



#10
Jan1707, 08:32 PM

HW Helper
P: 858

The total forward force for m1 can only come from the tension T, since there is no friction. For m2, there are two forces acting, T and gravity. My claim is that the two T's are the same because of all those assumptions I mentioned, nonstretchable, massless string, frictionles + massless pulley.etc. (think about it, you will notice that the two's T in your diagram actually points in towards/away from each other, and if they are not the same, there will be a net force, and something will happen to the string!) Now total force on m1 is F1=m1 a... and you want to find this a. 


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