Finding the acceleration of a system

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I think I'm missing something in my working of this mechanics question. The working seems to short but I don't know what else to do.




Box $A$ of mass $10kg$ rests on a smooth horizontal table. It is connected to one end of a light inextensible string which passes over a smooth pulley at the same height of the box. Box B of mass $8kg$ is attached to the other end of the string.



Find the acceleration of the system and the tension in the string.




I did:



Tension $ =8g =78.4N$



Then, by $F=ma$,



$78.4 = 10a$, so $7.84 ms^-2 = a$






proof-verification classical-mechanics







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  • 2




    By jumping to the conclusion that the tension is $8g,$ you have also concluded that the hanging mass cannot accelerate, because the forces on it are balanced. Yet the hanging mass should always have the same speed as the sliding mass since the string doesn't change length, and the sliding mass does accelerate, so the hanging mass also must. Contradiction! You resolve the contradiction by admitting your assumption was wrong and the amount of tension is not $8g.$
    – David K
    3 hours ago















up vote
2
down vote

favorite












I think I'm missing something in my working of this mechanics question. The working seems to short but I don't know what else to do.




Box $A$ of mass $10kg$ rests on a smooth horizontal table. It is connected to one end of a light inextensible string which passes over a smooth pulley at the same height of the box. Box B of mass $8kg$ is attached to the other end of the string.



Find the acceleration of the system and the tension in the string.




I did:



Tension $ =8g =78.4N$



Then, by $F=ma$,



$78.4 = 10a$, so $7.84 ms^-2 = a$






proof-verification classical-mechanics







share|cite|improve this question



















  • 2




    By jumping to the conclusion that the tension is $8g,$ you have also concluded that the hanging mass cannot accelerate, because the forces on it are balanced. Yet the hanging mass should always have the same speed as the sliding mass since the string doesn't change length, and the sliding mass does accelerate, so the hanging mass also must. Contradiction! You resolve the contradiction by admitting your assumption was wrong and the amount of tension is not $8g.$
    – David K
    3 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I think I'm missing something in my working of this mechanics question. The working seems to short but I don't know what else to do.




Box $A$ of mass $10kg$ rests on a smooth horizontal table. It is connected to one end of a light inextensible string which passes over a smooth pulley at the same height of the box. Box B of mass $8kg$ is attached to the other end of the string.



Find the acceleration of the system and the tension in the string.




I did:



Tension $ =8g =78.4N$



Then, by $F=ma$,



$78.4 = 10a$, so $7.84 ms^-2 = a$






proof-verification classical-mechanics







share|cite|improve this question















I think I'm missing something in my working of this mechanics question. The working seems to short but I don't know what else to do.




Box $A$ of mass $10kg$ rests on a smooth horizontal table. It is connected to one end of a light inextensible string which passes over a smooth pulley at the same height of the box. Box B of mass $8kg$ is attached to the other end of the string.



Find the acceleration of the system and the tension in the string.




I did:



Tension $ =8g =78.4N$



Then, by $F=ma$,



$78.4 = 10a$, so $7.84 ms^-2 = a$






proof-verification classical-mechanics




proof-verification classical-mechanics






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share|cite|improve this question













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edited 3 hours ago









Alex Pozo

490214




490214










asked 3 hours ago









Cheks Nweze

1397




1397







  • 2




    By jumping to the conclusion that the tension is $8g,$ you have also concluded that the hanging mass cannot accelerate, because the forces on it are balanced. Yet the hanging mass should always have the same speed as the sliding mass since the string doesn't change length, and the sliding mass does accelerate, so the hanging mass also must. Contradiction! You resolve the contradiction by admitting your assumption was wrong and the amount of tension is not $8g.$
    – David K
    3 hours ago













  • 2




    By jumping to the conclusion that the tension is $8g,$ you have also concluded that the hanging mass cannot accelerate, because the forces on it are balanced. Yet the hanging mass should always have the same speed as the sliding mass since the string doesn't change length, and the sliding mass does accelerate, so the hanging mass also must. Contradiction! You resolve the contradiction by admitting your assumption was wrong and the amount of tension is not $8g.$
    – David K
    3 hours ago








2




2




By jumping to the conclusion that the tension is $8g,$ you have also concluded that the hanging mass cannot accelerate, because the forces on it are balanced. Yet the hanging mass should always have the same speed as the sliding mass since the string doesn't change length, and the sliding mass does accelerate, so the hanging mass also must. Contradiction! You resolve the contradiction by admitting your assumption was wrong and the amount of tension is not $8g.$
– David K
3 hours ago





By jumping to the conclusion that the tension is $8g,$ you have also concluded that the hanging mass cannot accelerate, because the forces on it are balanced. Yet the hanging mass should always have the same speed as the sliding mass since the string doesn't change length, and the sliding mass does accelerate, so the hanging mass also must. Contradiction! You resolve the contradiction by admitting your assumption was wrong and the amount of tension is not $8g.$
– David K
3 hours ago











3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










Your calculation is incorrect. In this situation the tension shall be $frac 409 g N$ and the acceleration $frac49 g$.



The equations will be -



Total mass * Net acceleration of system = Total downward force



and,



Net downward force - Tension = Mass of B * Acceleration if B



Mathematically,



$$ 18.a = 8.g$$ and,



$$8g - T = 8.a$$.



Subsisting the value of $a$ from the first equation to the second shall give you $T$.



To gain a better understanding for such problems, note that since the the two bodies are connected to form a single system, they must have the same acceleration, $a$. This acceleration shall be the ratio of the total force acting upon and the total mass if the system (see the first equation), as stated by Newton in his second law.



Now, notice that the same reasoning about Newton's law holds individually for the two bodies. (See second equation with regard to body B). It may be helpful to remember that the since the table is smooth, the only force acting on body A, is the tension in the string.



Note that drawing free body diagrams, is in general, a great help in such problems and shall allow you to quickly go through whatever reasoning (like the above) is required by the physical situation.






share|cite|improve this answer





























    up vote
    2
    down vote













    $F_A,F_B$ forces on $A$ and $B$ resp. $T$ tension, $a$ acceleration.



    1)$ F_A = T = m_A a $,



    2) $ F_B =m_B g -T=m_ B a.$



    Add 2)+1):



    $m_Bg=(m_A+m_B)a$;



    $a= dfracm_Bm_A+m_Bg;$



    $T=dfracm_A m_Bm_A+m_Bg$.






    share|cite|improve this answer





























      up vote
      1
      down vote













      Displacement of the two masses shall be the same, so is speed and so acceleration.



      On one side you have the equilibrium $mg-F=ma$, on the other $F=Ma$: two equations in the unknowns $F$ and $a$.






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        Your calculation is incorrect. In this situation the tension shall be $frac 409 g N$ and the acceleration $frac49 g$.



        The equations will be -



        Total mass * Net acceleration of system = Total downward force



        and,



        Net downward force - Tension = Mass of B * Acceleration if B



        Mathematically,



        $$ 18.a = 8.g$$ and,



        $$8g - T = 8.a$$.



        Subsisting the value of $a$ from the first equation to the second shall give you $T$.



        To gain a better understanding for such problems, note that since the the two bodies are connected to form a single system, they must have the same acceleration, $a$. This acceleration shall be the ratio of the total force acting upon and the total mass if the system (see the first equation), as stated by Newton in his second law.



        Now, notice that the same reasoning about Newton's law holds individually for the two bodies. (See second equation with regard to body B). It may be helpful to remember that the since the table is smooth, the only force acting on body A, is the tension in the string.



        Note that drawing free body diagrams, is in general, a great help in such problems and shall allow you to quickly go through whatever reasoning (like the above) is required by the physical situation.






        share|cite|improve this answer


























          up vote
          2
          down vote



          accepted










          Your calculation is incorrect. In this situation the tension shall be $frac 409 g N$ and the acceleration $frac49 g$.



          The equations will be -



          Total mass * Net acceleration of system = Total downward force



          and,



          Net downward force - Tension = Mass of B * Acceleration if B



          Mathematically,



          $$ 18.a = 8.g$$ and,



          $$8g - T = 8.a$$.



          Subsisting the value of $a$ from the first equation to the second shall give you $T$.



          To gain a better understanding for such problems, note that since the the two bodies are connected to form a single system, they must have the same acceleration, $a$. This acceleration shall be the ratio of the total force acting upon and the total mass if the system (see the first equation), as stated by Newton in his second law.



          Now, notice that the same reasoning about Newton's law holds individually for the two bodies. (See second equation with regard to body B). It may be helpful to remember that the since the table is smooth, the only force acting on body A, is the tension in the string.



          Note that drawing free body diagrams, is in general, a great help in such problems and shall allow you to quickly go through whatever reasoning (like the above) is required by the physical situation.






          share|cite|improve this answer
























            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Your calculation is incorrect. In this situation the tension shall be $frac 409 g N$ and the acceleration $frac49 g$.



            The equations will be -



            Total mass * Net acceleration of system = Total downward force



            and,



            Net downward force - Tension = Mass of B * Acceleration if B



            Mathematically,



            $$ 18.a = 8.g$$ and,



            $$8g - T = 8.a$$.



            Subsisting the value of $a$ from the first equation to the second shall give you $T$.



            To gain a better understanding for such problems, note that since the the two bodies are connected to form a single system, they must have the same acceleration, $a$. This acceleration shall be the ratio of the total force acting upon and the total mass if the system (see the first equation), as stated by Newton in his second law.



            Now, notice that the same reasoning about Newton's law holds individually for the two bodies. (See second equation with regard to body B). It may be helpful to remember that the since the table is smooth, the only force acting on body A, is the tension in the string.



            Note that drawing free body diagrams, is in general, a great help in such problems and shall allow you to quickly go through whatever reasoning (like the above) is required by the physical situation.






            share|cite|improve this answer














            Your calculation is incorrect. In this situation the tension shall be $frac 409 g N$ and the acceleration $frac49 g$.



            The equations will be -



            Total mass * Net acceleration of system = Total downward force



            and,



            Net downward force - Tension = Mass of B * Acceleration if B



            Mathematically,



            $$ 18.a = 8.g$$ and,



            $$8g - T = 8.a$$.



            Subsisting the value of $a$ from the first equation to the second shall give you $T$.



            To gain a better understanding for such problems, note that since the the two bodies are connected to form a single system, they must have the same acceleration, $a$. This acceleration shall be the ratio of the total force acting upon and the total mass if the system (see the first equation), as stated by Newton in his second law.



            Now, notice that the same reasoning about Newton's law holds individually for the two bodies. (See second equation with regard to body B). It may be helpful to remember that the since the table is smooth, the only force acting on body A, is the tension in the string.



            Note that drawing free body diagrams, is in general, a great help in such problems and shall allow you to quickly go through whatever reasoning (like the above) is required by the physical situation.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 3 hours ago









            Devashish Kaushik

            36714




            36714




















                up vote
                2
                down vote













                $F_A,F_B$ forces on $A$ and $B$ resp. $T$ tension, $a$ acceleration.



                1)$ F_A = T = m_A a $,



                2) $ F_B =m_B g -T=m_ B a.$



                Add 2)+1):



                $m_Bg=(m_A+m_B)a$;



                $a= dfracm_Bm_A+m_Bg;$



                $T=dfracm_A m_Bm_A+m_Bg$.






                share|cite|improve this answer


























                  up vote
                  2
                  down vote













                  $F_A,F_B$ forces on $A$ and $B$ resp. $T$ tension, $a$ acceleration.



                  1)$ F_A = T = m_A a $,



                  2) $ F_B =m_B g -T=m_ B a.$



                  Add 2)+1):



                  $m_Bg=(m_A+m_B)a$;



                  $a= dfracm_Bm_A+m_Bg;$



                  $T=dfracm_A m_Bm_A+m_Bg$.






                  share|cite|improve this answer
























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    $F_A,F_B$ forces on $A$ and $B$ resp. $T$ tension, $a$ acceleration.



                    1)$ F_A = T = m_A a $,



                    2) $ F_B =m_B g -T=m_ B a.$



                    Add 2)+1):



                    $m_Bg=(m_A+m_B)a$;



                    $a= dfracm_Bm_A+m_Bg;$



                    $T=dfracm_A m_Bm_A+m_Bg$.






                    share|cite|improve this answer














                    $F_A,F_B$ forces on $A$ and $B$ resp. $T$ tension, $a$ acceleration.



                    1)$ F_A = T = m_A a $,



                    2) $ F_B =m_B g -T=m_ B a.$



                    Add 2)+1):



                    $m_Bg=(m_A+m_B)a$;



                    $a= dfracm_Bm_A+m_Bg;$



                    $T=dfracm_A m_Bm_A+m_Bg$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 3 hours ago

























                    answered 3 hours ago









                    Peter Szilas

                    8,7482619




                    8,7482619




















                        up vote
                        1
                        down vote













                        Displacement of the two masses shall be the same, so is speed and so acceleration.



                        On one side you have the equilibrium $mg-F=ma$, on the other $F=Ma$: two equations in the unknowns $F$ and $a$.






                        share|cite|improve this answer
























                          up vote
                          1
                          down vote













                          Displacement of the two masses shall be the same, so is speed and so acceleration.



                          On one side you have the equilibrium $mg-F=ma$, on the other $F=Ma$: two equations in the unknowns $F$ and $a$.






                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Displacement of the two masses shall be the same, so is speed and so acceleration.



                            On one side you have the equilibrium $mg-F=ma$, on the other $F=Ma$: two equations in the unknowns $F$ and $a$.






                            share|cite|improve this answer












                            Displacement of the two masses shall be the same, so is speed and so acceleration.



                            On one side you have the equilibrium $mg-F=ma$, on the other $F=Ma$: two equations in the unknowns $F$ and $a$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









                            G Cab

                            15.3k31136




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