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    luis00's Avatar
    luis00 Posts: 12, Reputation: 1
    New Member
     
    #1

    Oct 15, 2010, 01:07 PM
    test coming up!-finding acceleration and tension
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    https://www.askmehelpdesk.com/attachment.php?attachmentid=33409&stc=1&d=12871731 36


    I have a test coming up and I want to be able to understand the following problem:
    Two crates are on a table crate 1 is attached by string A to crate 2 string B

    A large crate (crate 1) has a mass of 4kg and the smaller crate (crate 2) has a mass of 2kg. The coefficient of kinetic friction between each crate and the wooden table is 0.1. A hanging (string C) weight of mass 1 kg causes both crates (crate 1 and crate 2) to accelerate to the right. What's the acceleration of the block and the tension T1 and T2 in strings A and B?

    My attempt:
    I think I would have to set up three different equations to find acceleration?


    F=ma

    Mass 1:

    T-0.1(+9.8)=4a?
    so T-.98=4a


    Mass 2:
    -T1+T2-.98=2a?

    I'm really confused with trying to set up this equation.
    I put -T1 because it's acting to the left and +T2 because it's
    acting to the right.
    -.98 because friction moves in the opposite direction of motion
    so -1.0(9.8)=-.98


    Mass 3:
    1g - T2 = 1a?

    I would appreciate any help, thanks!

    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
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    #2

    Oct 15, 2010, 01:50 PM

    I think I can help! The best thing to do for these is to draw a force diagram for each box to start with. Lets find out the normal force first; n1=m1(g), n2=m2(g). From that we find friction fk1=u(n1) fk2=u(n2).(u is the coefficient of friction). Ok, acceleration; Fnet=m3(g)-fknet, Fnet=ma, ma=m3(g)-fknet, a=(m3(g)-fknet)/m. m=m1+m2+m3. Now for the tensions and this part I'm not 100 percent on but I think it's right(unknown008 will correct me if I'm wrong). T2=Fnet-fknet, T1=T2-fk1. Do you have the correct answers in the back of the book or something? I'm curious myself!
    luis00's Avatar
    luis00 Posts: 12, Reputation: 1
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    #3

    Oct 15, 2010, 04:24 PM
    Comment on kpg0001's post
    OK, I tried solving for a but I don't think I got the correct answer, can you please check, thanks!

    n1=4(9.8)=39.2
    n2=2(9.8)=19.6
    fk1=0.1(39.2)=3.92
    fk2=0.1(19.6)=1.96
    a=1(9.8)-5.88/39.2+19.6+9.8=0.0571

    luis00's Avatar
    luis00 Posts: 12, Reputation: 1
    New Member
     
    #4

    Oct 15, 2010, 04:33 PM
    Is fknet 5.88 I got that number by adding 3.92 + 1.96
    Also is m3(g)=1(9.8) or 1(-9.8) since acceleration is downward?
    When I'm adding m1+m2+m3 is the following correct?
    m1=39.2?
    m2=19.6?
    m3=9.8 or -9.8?

    Sorry for all the questions but I need to know if I'm doing this correctly, sorry there are no answers
    in the back of the book :/
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member
     
    #5

    Oct 16, 2010, 01:01 AM

    Well... all those numbers in a single paragraph!

    I'll do it.

    Consider crate 3 (was that a typing error in your question?).

    It's mass drives the whole system. Hence, we get:







    Consider crate 2 only.







    Consider crate 1 only;







    For friction, remember that
    ------------------------






    Now can you solve them? :)

    I get:

    T_1 = 6.16 N
    T_2 = 9.24 N
    a = 0.56 m/s^2
    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
    Junior Member
     
    #6

    Oct 16, 2010, 11:21 AM

    Darn! I told you he would correct me but I can't be totally wrong. I have a football game to go to, we are playing Arkansas, but I'm coming back to this later.
    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
    Junior Member
     
    #7

    Oct 17, 2010, 01:58 PM

    Yeah I stand corrected. I started completely wrong because I wasn't taking into account the acceleration of the system which changes the tensions. Oh well, I just solved it correctly and it was exactly how you solved it. Thanks unknown008! Oh by the way we whooped Arkansas 65-43 haha! CAM NEWTON FOR HEISMAN!
    luis00's Avatar
    luis00 Posts: 12, Reputation: 1
    New Member
     
    #8

    Oct 21, 2010, 08:08 PM
    Comment on Unknown008's post
    Thank you!!

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