Hello all. Trying to brush up on my physics but I'm stuck on this problem. http://phy-page-g5.princeton.edu/~p...""between the masses and horizontal surfaces" Problem #2. Diagram included. I think I've gotten a and b but I can't seem to get c, which I know what the answer is in advance. T = (u + 1)g/[2/m(3) + 1/(2m(1)) + 1/(2m(2))] if -x1 is the acceleration of m1, x2 is the acceleration of m2, and -x3 is the acceleration of m3, i got x(1) + 2x(3) + x(2) = 0, which is pretty simple. from newtons 2nd law, i arrived at T - um(1)g = -m(1)x(1) um(2)g - T = m(2)x(2) 2T - m(3)g = -m(3)x(3), which i combine to get: ug(m(1) + m(2)) - (m(1)x(1) + m(2)x(2)) - m(3)g = -m(3)x(3) = m(3)(x(1) + x(2))/2 , by combining the first two to get an expression for 2T and substituting into the 3rd equation. but getting from here to an expression for T is seemingly impossible. Please Register or Log in to view the hidden image!
Try rearranging each of these equations in terms of x<sub>1</sub>, x<sub>2</sub>, and x<sub>3</sub> respectively: then applying x<sub>1</sub> + x<sub>2</sub> + 2x<sub>3</sub> = 0.
to steal a line from the video game You Don't Know Jack- "remember this clue, it could save your life some day" the most important thing to remember is that everything sums to zero. just start putting things together, and the answer should fall out in your lap.
Thanks guys I was able to figure it out last night. This one is stumping me too: Image here: http://hep.physics.wayne.edu/~harr/courses/5200/w99/problem6-4.gif Consider the "pedagogical machine" drawn below. All surfaces are frictionless; the pulley is frictionless; and the pulley and rope are massless. Find the force F which must be applied such that the relative positions of all the masses remain constant. Hint: If all the masses equal M, then F = 3Mg. The drawing is a little unclear but there's supposed to be a string coming from the pulley to M3. Qualitatively, here is what I figure. The tension in the rope must equal the weight of m3 in order for m3 not to move up or down. therefore m2 is being pulled to the right by a force equivalent to the weight of m3. any force acting on m1 must thus provide m1 with an acceleration of (m3/m2)g in order for m1 not to move relative to m2. now say using newtons 2nd law, i write F = m1a (I am assuming there is no contact between m3 and m1, although i am not sure how, qualitatively, m3 is able to accelerate horizontally without any horizontal force. i have tried including a horizonal force but i am still getting a wrong answer.) if i instead assume there is contact (by a force N) between m1 and m3, i can write F - N = m1a I am no longer sure if this is even correct. Any help on how to solve this?
This looks good. I'd spell it out like so: F = force on m1 T = tension in the rope a = acceleration of the system T = m3.g T = m2.a m3.g = m2.a a = (m3/m2)g The easiest way to attack this is to realise that F is effectively acting on the whole system (m1 + m2 + m3). But, you can break it down and consider all the horizontal forces. Don't forget the horizonal force between m1 and m2 through the pulley! N = m3.a m1.a = F - N - T
Thanks again guys. For some reason it seems as if as soon as I post my questions on this thread, I figure out the answer.. Maybe verbalizing my reasoning helps. But. I have some more questions. (These are all from Kleppner's 'Introduction to Mechanics' by the way, a book I hear is very fun for physics majors Please Register or Log in to view the hidden image! ) This time, we are asked to find the acceleration of M1 in the case where F is zero. But this one is beyond me. I can't fathom how M1 can even accelerate if there are no horizontal forces on it (or are there?) Also if you can help me get started on problem 3b here: http://www.core.org.cn/NR/rdonlyres...0C5F20-6600-4AFE-93E5-B7D029A5060A/0/ps05.pdf To begin, I think the force diagram of some small bit of the rope would have the force of the pulley on the bit, directed perpendicular and outwards, and then two tensions directed to either side of the bit. firstly, how can we determine the magnitudes of the tension on either side?
Don't forget the pulley. Or equivalently, you can address it as a centre of mass problem - as m2 moves right, m1 and m3 move left.
right now the chapter hasn't covered centers of mass so i'm assuming that isn't necessary. what are you saying about the pulley? isn't it massless? and on that note, what exactly would the force diagram for m1 look like for the earlier variation of this problem where a force was pushing it? wouldn't the force F be the only horizontal force acting on m1?
The rope is applying a force to the pulley, which in turn applies a force to m1. The pulley applies a force of T down on m1, and a force T to the left on m1. I don't know if I'm explaining it well... I'm a bit short of time right now. Back in a few hours.
My point was that doesn't the pulley transmit 0 force since it is masless? I'm pretty sure it's assumed to be masless. My answer for the previous variation was correct even though I didn't account for any pulley forces, but that was when M1 actually had a horizontal force. if you're right though I can see where the applied force of sqrt(2)*T comes from. I'm just confused because my solution was correct previously even without accounting for the pulley. Let me see if I can come up with an explanation and you tell me if I'm right. I remember that ropes in problems are often considered massless but still transmit force. I'm not very sure how this is possible realistically but.. I think it might have something to do with simplifying things and just that. If we are to take that as it is, then it is easy to see why the pulley should also be able to transmit force. Speaking of ropes, I have some questions. Why is it that in problems where two masses of different weight are hanging from a pulley by rope, the tension in the rope is the same for each mass? That is, the tension is uniform throughout the rope? Does this have something to do with the fact that the rope is massless? Because I remember the book saying this: "although a string may be under considerable tension, if the tension is uniform, the net string force on each small section is zero and the section remains at rest unless external forces act on it. if there are external forces on the section, or if the string is accelerating, the tension generally varies along the string." isn't the pull of the masses of different weight considered an external force since it makes the rope accelerate? and if the rope on such a pulley is accelerating, why is the tension still uniform in the string?
What do you mean by transmitting force? Yes. It's a reasonable assumption provided the mass of the rope is much smaller than the masses it is attached to. The tension will be the same everywhere in the rope, provided the pulley is frictionless. This does have to do with the fact that the rope is massless. Consider a rope hanging from the ceiling, with no mass attached. Consider the force the top half of the rope exerts on the bottom half. If the rope is hanging without moving, then the upwards force exerted by the top half must exactly balance the weight of the lower half of the rope. Similarly, the upwards force exerted by the top 1/4 of the rope must balance the weight of the lower 3/4 of the rope. So, the tension in the rope must increase with height, with zero tension at the lower end and tension equal to the rope's mass at the upper end. If we make the rope massless, obviously there is no weight to support, so the tension is zero everywhere. If the mass per unit length of the rope is very small, then the tension will only increase a little from bottom to top, and that increase might be small enough to ignore. This is especially true when largish masses are attached to the rope - i.e. when the mass hung from the rope is much larger than the mass of the rope itself. In that case, the tension can be considered equally transmitted throughout the rope, since the top of the rope only has to support the weight hung from the rope, and we neglect the weight of the rope itself. Does that make sense?
I was thinking if we wrote out Newton's 2nd law for the pulley, the tension force on it would have to equal 0 since its mass is 0. Your reply makes sense. Thanks.
