How do you evaluate the tension force if the coefficient of friction between the objects K and L is $0.6$?











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How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?




So the system is accelerating, whence we have to consider that



$$sum F_x = m_1a$$



$$F_k - T = 2a $$



$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$



For the object L,



$$sum F_x = m_2a$$



$$F - F_k = 6a $$



$$10 - 12 = 6a implies a = -dfrac{1}{3}$$



Plugging $a$ into the first equation



$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$



However, there won't be an integer solution from what I got above. Could you assist me?










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  • 1




    Who says the solution must be integer?
    – Sean Roberson
    Nov 17 at 15:19










  • @SeanRoberson The correct answer seems to be $10$ according to my answer key.
    – Enzo
    Nov 17 at 15:23















up vote
1
down vote

favorite












enter image description here




How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?




So the system is accelerating, whence we have to consider that



$$sum F_x = m_1a$$



$$F_k - T = 2a $$



$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$



For the object L,



$$sum F_x = m_2a$$



$$F - F_k = 6a $$



$$10 - 12 = 6a implies a = -dfrac{1}{3}$$



Plugging $a$ into the first equation



$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$



However, there won't be an integer solution from what I got above. Could you assist me?










share|cite|improve this question


















  • 1




    Who says the solution must be integer?
    – Sean Roberson
    Nov 17 at 15:19










  • @SeanRoberson The correct answer seems to be $10$ according to my answer key.
    – Enzo
    Nov 17 at 15:23













up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here




How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?




So the system is accelerating, whence we have to consider that



$$sum F_x = m_1a$$



$$F_k - T = 2a $$



$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$



For the object L,



$$sum F_x = m_2a$$



$$F - F_k = 6a $$



$$10 - 12 = 6a implies a = -dfrac{1}{3}$$



Plugging $a$ into the first equation



$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$



However, there won't be an integer solution from what I got above. Could you assist me?










share|cite|improve this question













enter image description here




How do you evaluate the tension force if the coefficient of friction
between the objects K and L is $0.6$?




So the system is accelerating, whence we have to consider that



$$sum F_x = m_1a$$



$$F_k - T = 2a $$



$$mu mg - T = 2a implies 0.6 times 2 times 10 - T = 2a implies 12-T = 2a$$



For the object L,



$$sum F_x = m_2a$$



$$F - F_k = 6a $$



$$10 - 12 = 6a implies a = -dfrac{1}{3}$$



Plugging $a$ into the first equation



$$12-T = 2 times -dfrac{1}{3} implies 12-T = -dfrac{2}{3} $$



However, there won't be an integer solution from what I got above. Could you assist me?







physics






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asked Nov 17 at 15:14









Enzo

956




956








  • 1




    Who says the solution must be integer?
    – Sean Roberson
    Nov 17 at 15:19










  • @SeanRoberson The correct answer seems to be $10$ according to my answer key.
    – Enzo
    Nov 17 at 15:23














  • 1




    Who says the solution must be integer?
    – Sean Roberson
    Nov 17 at 15:19










  • @SeanRoberson The correct answer seems to be $10$ according to my answer key.
    – Enzo
    Nov 17 at 15:23








1




1




Who says the solution must be integer?
– Sean Roberson
Nov 17 at 15:19




Who says the solution must be integer?
– Sean Roberson
Nov 17 at 15:19












@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 at 15:23




@SeanRoberson The correct answer seems to be $10$ according to my answer key.
– Enzo
Nov 17 at 15:23










2 Answers
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To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.



Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.



Since the tension balances the frictional force, the tension is $10text{N}$.






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    It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.



    1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.



    2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.



      Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.



      Since the tension balances the frictional force, the tension is $10text{N}$.






      share|cite|improve this answer

























        up vote
        0
        down vote













        To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.



        Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.



        Since the tension balances the frictional force, the tension is $10text{N}$.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.



          Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.



          Since the tension balances the frictional force, the tension is $10text{N}$.






          share|cite|improve this answer












          To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.



          Frictional force is a reactive force whose maximum value is $2text{kg}times 9.8text{N/kg}approx19.6times 0.6approx 11.8gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10text{N}$.



          Since the tension balances the frictional force, the tension is $10text{N}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 23:35









          John Douma

          5,13611319




          5,13611319






















              up vote
              0
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              It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.



              1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.



              2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.






              share|cite|improve this answer



























                up vote
                0
                down vote













                It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.



                1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.



                2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.



                  1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.



                  2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.






                  share|cite|improve this answer














                  It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.



                  1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.



                  2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 20 at 20:06

























                  answered Nov 17 at 21:39









                  Rafa Budría

                  5,4001825




                  5,4001825






























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