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  • Sep 15, 2009, 11:53 AM
    bflsw09
    1 Attachment(s)
    Finding acceleration and max accerleration
    Consider two masses m1 and m2 connected by a thin string. Assume the following values: m1 = 4.32 kg and m2 = 1.00 kg. Ignore friction and mass of the string.

    What is the acceleration of the two mass?


    Attachment 24590


    What should be the value of mass m1 to get the largest possible value of acceleration of the two masses? What would be the value of that maximum acceleration?
  • Sep 16, 2009, 09:53 AM
    Unknown008

    1. Find the force acting on m2.
    By Newton III, the same force is applied onto m1.
    Find the acceleration of m1 by F=ma.
    m2 will accelerate by the same value.

    2. m2 cannot fall with an acceleration greater than gravity. That means, the acceleration of m1 will be equal to gravity. You can find the force required to make m1 accelerate by the same value as gravity. I think you can finish it now. :)

    EDIT: Sorry, I thought it was easier... :o Take a look at ebaines' answer instead
  • Sep 16, 2009, 10:38 AM
    ebaines

    Quote:

    Originally Posted by Unknown008 View Post
    1. Find the force acting on m2.
    By Newton III, the same force is applied onto m1.
    Find the acceleration of m1 by F=ma.
    m2 will accelerate by the same value.

    Take care that in the equation F=ma the value for "m" is equal to m1 + m2.

    Quote:

    Originally Posted by Unknown008 View Post
    2. m2 cannot fall with an acceleration greater than gravity. That means, the acceleration of m1 will be equal to gravity. You can find the force required to make m1 accelerate by the same value as gravity. I think you can finish it now. :)

    Not quite right. The acceleration of m1 is NOT equal to gravity.

    The acceleration of the combined masses m1 + 2 is calculated from:

    F = ma

    The only force acting on the system is due gravity acting on the mass of m2, but the mass of the system that must accelerate in unison is the sum of m1+m2. So:

    m2*g = (m1 + m2)*a

    From this you can see what value for m1 you'd like to pick to have the greatest value of a.
  • Sep 16, 2009, 10:47 AM
    Unknown008

    Sorry, I thought it was simpler than that. My response to the second part was due to the first part, hence the e.c.f (error carried forward), making the whole thing wrong. That'll teach me...

    The only thing I'm not understanding, is that from your equation, I can put (let's exaggerate) infinity kg and get infinity acceleration... which is not a maximum acceleration... :(
  • Sep 16, 2009, 10:56 AM
    ebaines
    Quote:

    Originally Posted by Unknown008 View Post
    The only thing I'm not understanding, is that from your equation, I can put (let's exaggerate) infinity kg and get infinity acceleration... which is not a maximum acceleration... :(

    I don't follow you. Using the equation

    m2*g = (m1 + m2)*a

    If m2 is, say, 1000 times bigger than m1 you get 1000*g = 1001*a, or a = .999*g. So the acceleration gets close to g if m2 is large compard to m1.

    And at the other extreme - if m1 is 1000 times bigger than m2, you get: 1*g = 1001*a, or a = 0.001*g. So you can see that for really big values of m1 the system hardly moves at all.
  • Sep 16, 2009, 04:36 PM
    bflsw09

    I feel like m1 has to equal 1 because that would give the greatest value for acceleration and not having m1 equal too little for it to be dragged off the table by the weight of m2... am I thinking too hard about this?
  • Sep 16, 2009, 05:31 PM
    bflsw09

    never mind I got it... m1=0 therefore the acceleration is 9.8m/s^2
    thanks again everyone
  • Sep 17, 2009, 07:06 AM
    Unknown008
    Quote:

    Originally Posted by ebaines View Post
    I don't follow you. Using the equation

    m2*g = (m1 + m2)*a

    If m2 is, say, 1000 times bigger than m1 you get 1000*g = 1001*a, or a = .999*g. So the acceleration gets close to g if m2 is large compard to m1.

    And at the other extreme - if m1 is 1000 times bigger than m2, you get: 1*g = 1001*a, or a = 0.001*g. So you can see that for really big values of m1 the system hardly moves at all.

    Oh, how silly I was! Ok if I understood well, then the mass of m1 has to be 0, like the OP said. However, we'll get the same acceleration as that of gravity, that is 9.81m/s^2, which is not what you said earlier.. :confused:

    Rah! I hate when I can't understand something!! :(
  • Sep 17, 2009, 07:16 AM
    ebaines
    Quote:

    Originally Posted by Unknown008 View Post
    Ok if I understood well, then the mass of m1 has to be 0, like the OP said.

    Right. M1 = 0 gives the maximum acceleration, which is g.

    Quote:

    Originally Posted by Unknown008 View Post
    However, we'll get the same acceleration as that of gravity, that is 9.81m/s^2, which is not what you said earlier.. :confused:

    I never said anything about the specific condition of m1 = 0. Since that was the answer to the homework problem, I didn't want to just give the answer away, but rather let the OP work through the final step to the conclusion himself, which he did. So yes indeed, if m1 = 0 you get a = g; that's pretty clear from the equation.
  • Sep 17, 2009, 07:27 AM
    Unknown008

    Ok, let's see if I got it right.

    1. I have to take into account both masses in determining their acceleration.
    2. If one accelerates by 'a', the other mass will also accelerate by 'a' because the forces are equal in magnitude.
    3. The maximum acceleration in such systems cannot be greater than the acceleration due to gravity.
  • Sep 17, 2009, 07:58 AM
    ebaines

    Hello Unknown008 - looking at your 3 statements:

    1. - Right.

    2. - Not right. The reason both masses accelerate at the same rate is because they are tied together by a rope. In general the net forces on the two bodies can't be the same, because if they were the same then they would have different accelerations (unless m1 = m2). But they must have the same acceleration as they are tied together by the rope. For m1 the force acting on it is m1*a in the horizontal direction. For m2 there are two forces: m2*g downward (force due to gravity) and m1*a upward (tension on the rope).

    3. Right.
  • Sep 17, 2009, 08:12 AM
    Unknown008

    *sigh* I keep doing mistakes... never mind, thanks for clearing all this ebaines! :)

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