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.the problem about tension forces

  1. Sep 15, 2003 #1
    .....the problem about tension forces....

    ok so the problem states:
    three blocks are pulled along a frictionless surface by a horizontal force ( as shown in the figure ill describe)A)what is the acceleration of the system? B) what are the tension forces in the light strings?(hint: can T(1) equal T(2)? investigate by drawing free body diagrams of each block seperately)
    so the drawing has 3 blocks each increasing in size relative to the last one m(1) the first block is 1.0 kg.there is a string attatching blocks m(1) and m(2) labeled T1 and arrows pointing them toward each other. block 2 m(2) is 2.0 kg and block 3 m(3) is 3.0 kg. the string attatching them is labeled T(2) and the arrows again are pushing them toward each other. off of m(3) there is an arrow pointing -> labeled F= 18.0 N. there is an arrow -> that shows a goes in that direction i suppose.
    so first i drew freebody diagrams for each. then i labeled F(g) which always points down. then i labeled F(a)-> on each. so i guess you apply Newton's second law to it somehow. im unsure how to solve for acceleration and how to answer the second part. would you plug in force and mass for each to get their acceleration?then add them? any help anyone could give me would be GREATLY appreciated! thanks!
  2. jcsd
  3. Sep 15, 2003 #2


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    (Arrgg... your formatting is rather foobarised...)

    Solving for acceleration is a peice of cake - consider the whole system, and use F=ma using F as your force, and m as your total mass and a as your acceleration. (Since you can presume the acceleration is the same for the entire system) You don't add the acceleration, no. That's not how it works.

    For the tension... that's a little more fun...:wink: Ignore gravity for now and consider all the forces acting on each block in the horizontal sense. For example, for m(3) you have... 3.0 * (acceleration from previous part) = 18 - T(2)

    Do all of this, and you have 3 equations with 2 unknowns, which can be solved very simply. Note that the best method is not to insert the acceleration as calculated previously, but to equate the accelerations in each part of the system... This would give you a more reliable result.
  4. Sep 15, 2003 #3

    sorry about the html! i have to change my settings..... thanks, you helped me alot!
  5. Oct 7, 2003 #4

    I have a very similar problem. Finding the acceleration is easy. However, I am having trouble with the tension could anyone elaborate a little more. I Have F =9.0N M1=1.0 kg M2=3.0 kg M3=2.0kg
    acceleration works out to 1.5 m/s2

    Now how do i determine the tension

    I have an eror going form m1 to m2 and vice versa labeled t1
    and an arrow going from m2 to m3 and vice versa labeled t2

    i just dont get the tension part

    any help
  6. Oct 7, 2003 #5


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    Hello Physicsit, welcome
    I have the impression you are new to PF
    I got your message but chose to reply here.
    Let me know what you have in mind and if it is
    homework we can stay here or if it is more
    general we can go to a more appropriate forum.
  7. Oct 7, 2003 #6
    Thank you for the reply

    Thank you for replying.

    It is homework. I need to determine the acceleration of system and the tension of t1 and t2
    t1 t1 t2 t2 F
    m1 ----> <---- m2 ---> <--- m3 ------>

    m1= 1.0kg
    m2= 3.0 kg
    m3= 2.0 kg
    F=9.0 N

    I know that

    and so 9.0 N = 6.0kga
    so a =1.5 m/s2

    however I am really lost on the tension thing. I am not surehow to put a formula form
  8. Oct 7, 2003 #7


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    Re: Thank you for the reply

    Phys...it lets assume the wires are not elastic so theres no stretch to complicate things and lets neglect the wires mass so the whole system mass is 6 kilo and the force to the rite is 9 N so you as you say the acceleration of the whole system to the rite must be 1.5 m per s per s, And this is also (because no stretch) the acceleration of each separate weight!

    So you can solve for what the net force on m1 must be----what it takes to accelerate it to the rite by 1.5 m per s per s (that you have figuired out) so that is t1----the force on m1

    when you have found t1 then you will easily see what t2 must be since t2 - t1 is the net force on m2 and it is obvious what that must be to give the acceleration (that the whole system has and also each separate mass)

    Please show that you can find what the tension t1 is.

    this will give me a chance to catch my breath----i am reading/writing in several threads and have sort of decided I am never going to do HW help---there are so many here who are very good at it

    I will check out what you did when you have found the force t1
    (in this case where the wires dont stretch at all)
    but probably someone else will step in and help

    cordial good wishes,


    Last edited: Oct 7, 2003
  9. Oct 7, 2003 #8
    Here is what I know

    I know the force for m1 is equal to 1.5 N to the right
    I know the total force for m2 is equal to 4.5 N not sure on direction
    I know the total force for m2 is equal to 3 N

    Where i get lost is there are two forces working on m2

    one to the left balalcing out the force on m1
    I believe is 1.5 N to the left
    <--- m2

    and one to the right balacing out the force on m3
    I believe is 3 N to the right
    m2 ---->

    the two forceson m2 puzzle me

    I have no way to relate

    force of m1 force of m2 and force of m3

    to identify t1 and t2

    I am lost

    does any of this info help solve the problem
  10. Oct 7, 2003 #9
    forgot this

    THey are weightless frictionless light strings
  11. Oct 8, 2003 #10

    Maybe this will help clear things up here is the exact question:
    I think everyone understands the problem but just in case thier is any confusion

    Three blocks connected by light (i.e., weightless) strings rest on a frictionless surface (see figure below). If a horizontal force of 9.0 N is applied to the right-end block, (a) what is the accelera-tion of the system? (b) What are the tensions in the strings?

    I have attached the associated picture i had to split it in two becauseit was two large. The thing I have a problem with is the two forces of of m2

    Attached Files:

  12. Oct 8, 2003 #11


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    Re: Thank you for the reply

    Hello Physit, I had to be away for a few hours but I am still around and I see your inference that t1, the force on m1, is
    equal to 1.5 N (to the rite)

    now the "t1" string pulls m2 to the left with the same force 1.5 N that it pulls m1 to the rite

    so m2 is getting pulled t2 to R and 1.5 N to the L and the net force on it is the differnce (t2 - 1.5 N)

    We say F = ma about m2 and realize that the net force on it must be its m2 mass (3 kilo) multiplied by its accel 1.5 m/s/s
    and that is 4.5 N.

    So (t2 - 1.5N) = 4.5 N
    all that means is we have two alternative grips on the force, we know on one hand it is (t2 - 1.5 N) and on the other hand it has to be 4.5 N because of F=ma law.
    So t2 must be 6 N by algebra.

    The "t2" string is pulling 6 N to the rite on m2
    but it is also pulling 6 N to the left on m3 !!!!
    that is how string does

    so now we are down to short strokes and we are cutting to the chase because we look at m3 and we know its accel is 1.5 m/s/s
    and its mass 2 kilo
    so we know the NET force on it must be 3 N because of F=ma law

    (you see we are up to our old tricks again and doing same with m3 as already with m2)

    now that actually solves the problem because it says t2 is 3 N

    but we can follow thru as a kind of check that no arithmetic mistakes have crept in and verify this 3 N from another angle:
    we also know the net force on m3 is the force on rite, 9 N, minus the force on the left from the "t2" string which we know is 6 N
    and that is 9N - 6N = 3 N
    that confirms the 3 N answer gotten just before

    this very same problem was probably being done when the British Navy was under sail and Admiral Nelson caught the French with their pants down in Aboukir and destroyed their Mediterranean fleet in a couple of hours

    so you see by solving it you and I are part of a venerable time-honored tradition (physics textbook writers have a bunch of problems which are over 100 years old that they inevitably use because they are creatures of custom and personally I approve of this.)

  13. Oct 8, 2003 #12
    Still lost

    I am still lost

    this is so easy and i just dont get it

    is this correct


    or t1=1.5N


    or 3.0N

    if not could you please provide a formula for t1 and t2
  14. Oct 8, 2003 #13


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    Two hints: force (tension) in a string works in both directions, so dont worry about direction in your calculations. Its meaningless. Looking at the system from left to right (acceleration force to the right), you have 3 forces with each adding the acceleration force of the object to its left to the tension in the string to that object's left. Calculate the forces on all 3 strings this way and the third should equal the fore given to you.
  15. Oct 8, 2003 #14
    SO is what i said wrong

    T1 does not equal 1.5?

    T2 does not equal 4.5?

    I have been at this for 12 hours non stop trying to research it.

    I just dont understand it

    I even found a full answer explanation at


    It just makes no sense to me

    I know that tnesion is merely force
    However, i dont understand the fact that there are two forces working on each string
    I dont know how to clauclate this
    I know there has to be some kind of simple formula

    My eyes are popping out of my head
    Im overly tired [zz)] and i am begining to think ill never get it:frown:
    Last edited by a moderator: Apr 20, 2017
  16. Oct 8, 2003 #15
    maybe this is right

  17. Oct 8, 2003 #16


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    Re: SO is what i said wrong

    A string can and does only have one tension force in it. Breaking it apart into two forces is simply to make the math easier.

    The tension on the string between m1 and m2 is equal to the acceleration force on m1:

    f = ma = 1.5*1 = 1.5N, like you said.

    The tension on the string between m2 and m3 is equal to the acceleration force on m2 PLUS the force on the string between m1 and m2:

    f = ma+1.5 = 1.5*3+1.5=6N.

    Now like I said in my last post - use this result to calculate the force in the 3rd string. Yes, I know you were given that force, but it'll help you understand how they are related.

    Another thought experiment to help clarify. Suppose you take your weights on strings apparatus and pick it up, holding it vertically by the same string you were pulling it from. You would calculate the forces in the strings exactly the same way - you just wouldn't have to calculate the acceleration first: its the acceleration due to gravity or 9.8m/s.
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