Net force on ramp












0












$begingroup$


I'm having difficulty understanding the free body diagram for the forces acting on a ramp when a block is sliding down it.



Say a block of mass $m$ is accelerating down a ramp with mass M with inclination $theta$, where the coefficient of kinetic friction is between the block and the ramp is $mu_k$. Assume the surface the ramp rests on is frictionless.



I can easily determine the free body diagram of the forces acting on the block:



$$F_{gx}=mgsin(theta)$$
$$F_{gy}=mgcos(theta)$$
$$F_n=F_{gy}=mgcos(theta)$$
$$F_k=mu_kF_n=mu_kmgcos(theta)$$



But I'm having difficulty establishing the free body diagram and the magnitude of the forces acting on the ramp.



How would both FBD's change if the ramp is now being pushed with a constant force $F$ pointed parallel to the flat surface?
Any help would be appreciated.






physics classical-mechanics







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I'm having difficulty understanding the free body diagram for the forces acting on a ramp when a block is sliding down it.



    Say a block of mass $m$ is accelerating down a ramp with mass M with inclination $theta$, where the coefficient of kinetic friction is between the block and the ramp is $mu_k$. Assume the surface the ramp rests on is frictionless.



    I can easily determine the free body diagram of the forces acting on the block:



    $$F_{gx}=mgsin(theta)$$
    $$F_{gy}=mgcos(theta)$$
    $$F_n=F_{gy}=mgcos(theta)$$
    $$F_k=mu_kF_n=mu_kmgcos(theta)$$



    But I'm having difficulty establishing the free body diagram and the magnitude of the forces acting on the ramp.



    How would both FBD's change if the ramp is now being pushed with a constant force $F$ pointed parallel to the flat surface?
    Any help would be appreciated.






    physics classical-mechanics







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm having difficulty understanding the free body diagram for the forces acting on a ramp when a block is sliding down it.



      Say a block of mass $m$ is accelerating down a ramp with mass M with inclination $theta$, where the coefficient of kinetic friction is between the block and the ramp is $mu_k$. Assume the surface the ramp rests on is frictionless.



      I can easily determine the free body diagram of the forces acting on the block:



      $$F_{gx}=mgsin(theta)$$
      $$F_{gy}=mgcos(theta)$$
      $$F_n=F_{gy}=mgcos(theta)$$
      $$F_k=mu_kF_n=mu_kmgcos(theta)$$



      But I'm having difficulty establishing the free body diagram and the magnitude of the forces acting on the ramp.



      How would both FBD's change if the ramp is now being pushed with a constant force $F$ pointed parallel to the flat surface?
      Any help would be appreciated.






      physics classical-mechanics







      share|cite|improve this question









      $endgroup$




      I'm having difficulty understanding the free body diagram for the forces acting on a ramp when a block is sliding down it.



      Say a block of mass $m$ is accelerating down a ramp with mass M with inclination $theta$, where the coefficient of kinetic friction is between the block and the ramp is $mu_k$. Assume the surface the ramp rests on is frictionless.



      I can easily determine the free body diagram of the forces acting on the block:



      $$F_{gx}=mgsin(theta)$$
      $$F_{gy}=mgcos(theta)$$
      $$F_n=F_{gy}=mgcos(theta)$$
      $$F_k=mu_kF_n=mu_kmgcos(theta)$$



      But I'm having difficulty establishing the free body diagram and the magnitude of the forces acting on the ramp.



      How would both FBD's change if the ramp is now being pushed with a constant force $F$ pointed parallel to the flat surface?
      Any help would be appreciated.






      physics classical-mechanics




      physics classical-mechanics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 21:00









      Anson PangAnson Pang

      6215




      6215






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          enter image description here



          I will let you figure out the actual projections into the coordinate system of the forces shown above. Just a few things to consider




          • Newton's third law: if the incline exerts a force $color{red}{bf N}$ on the block, then the block exerts a force $color{red}{-{bf N}}$ on the incline. Same goes for the friction force $color{orange}{{bf F}_f}$


          • There's a normal force $color{brown}{bf N''}$ which appears as a consequence of the fact that the incline rests on a surface


          • There's no friction force between the incline and the surface it rests on, but you can add in the diagram if needed.


          • I also included a external force $color{green}{{bf F}_{rm ext}}$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If the ramp exerts a force of friction $F_k$ on the block whose direction opposes its acceleration, does that mean that the block exerts an equal force in the opposite direction on the ramp?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:33












          • $begingroup$
            @AnsonPang yes, forgot to add that one
            $endgroup$
            – caverac
            Dec 7 '18 at 22:43










          • $begingroup$
            Is the force of gravity acting on the incline $Mg$, or $(M+m)g$?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:47










          • $begingroup$
            @AnsonPang It is $M$, because the mass of the incline is just $M$
            $endgroup$
            – caverac
            Dec 7 '18 at 22:52










          • $begingroup$
            So the forces acting on the incline are gravity, the opposing force to $F_k$ that I mentioned, and $-N$?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:54



















          0












          $begingroup$

          If the block is sliding down the ramp then the horizontal force on the ramp is.......



          $F_{rx} = cos(theta)(mgsin(theta) - u_kmgcos(theta))$






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            enter image description here



            I will let you figure out the actual projections into the coordinate system of the forces shown above. Just a few things to consider




            • Newton's third law: if the incline exerts a force $color{red}{bf N}$ on the block, then the block exerts a force $color{red}{-{bf N}}$ on the incline. Same goes for the friction force $color{orange}{{bf F}_f}$


            • There's a normal force $color{brown}{bf N''}$ which appears as a consequence of the fact that the incline rests on a surface


            • There's no friction force between the incline and the surface it rests on, but you can add in the diagram if needed.


            • I also included a external force $color{green}{{bf F}_{rm ext}}$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              If the ramp exerts a force of friction $F_k$ on the block whose direction opposes its acceleration, does that mean that the block exerts an equal force in the opposite direction on the ramp?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:33












            • $begingroup$
              @AnsonPang yes, forgot to add that one
              $endgroup$
              – caverac
              Dec 7 '18 at 22:43










            • $begingroup$
              Is the force of gravity acting on the incline $Mg$, or $(M+m)g$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:47










            • $begingroup$
              @AnsonPang It is $M$, because the mass of the incline is just $M$
              $endgroup$
              – caverac
              Dec 7 '18 at 22:52










            • $begingroup$
              So the forces acting on the incline are gravity, the opposing force to $F_k$ that I mentioned, and $-N$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:54
















            1












            $begingroup$

            enter image description here



            I will let you figure out the actual projections into the coordinate system of the forces shown above. Just a few things to consider




            • Newton's third law: if the incline exerts a force $color{red}{bf N}$ on the block, then the block exerts a force $color{red}{-{bf N}}$ on the incline. Same goes for the friction force $color{orange}{{bf F}_f}$


            • There's a normal force $color{brown}{bf N''}$ which appears as a consequence of the fact that the incline rests on a surface


            • There's no friction force between the incline and the surface it rests on, but you can add in the diagram if needed.


            • I also included a external force $color{green}{{bf F}_{rm ext}}$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              If the ramp exerts a force of friction $F_k$ on the block whose direction opposes its acceleration, does that mean that the block exerts an equal force in the opposite direction on the ramp?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:33












            • $begingroup$
              @AnsonPang yes, forgot to add that one
              $endgroup$
              – caverac
              Dec 7 '18 at 22:43










            • $begingroup$
              Is the force of gravity acting on the incline $Mg$, or $(M+m)g$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:47










            • $begingroup$
              @AnsonPang It is $M$, because the mass of the incline is just $M$
              $endgroup$
              – caverac
              Dec 7 '18 at 22:52










            • $begingroup$
              So the forces acting on the incline are gravity, the opposing force to $F_k$ that I mentioned, and $-N$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:54














            1












            1








            1





            $begingroup$

            enter image description here



            I will let you figure out the actual projections into the coordinate system of the forces shown above. Just a few things to consider




            • Newton's third law: if the incline exerts a force $color{red}{bf N}$ on the block, then the block exerts a force $color{red}{-{bf N}}$ on the incline. Same goes for the friction force $color{orange}{{bf F}_f}$


            • There's a normal force $color{brown}{bf N''}$ which appears as a consequence of the fact that the incline rests on a surface


            • There's no friction force between the incline and the surface it rests on, but you can add in the diagram if needed.


            • I also included a external force $color{green}{{bf F}_{rm ext}}$







            share|cite|improve this answer











            $endgroup$



            enter image description here



            I will let you figure out the actual projections into the coordinate system of the forces shown above. Just a few things to consider




            • Newton's third law: if the incline exerts a force $color{red}{bf N}$ on the block, then the block exerts a force $color{red}{-{bf N}}$ on the incline. Same goes for the friction force $color{orange}{{bf F}_f}$


            • There's a normal force $color{brown}{bf N''}$ which appears as a consequence of the fact that the incline rests on a surface


            • There's no friction force between the incline and the surface it rests on, but you can add in the diagram if needed.


            • I also included a external force $color{green}{{bf F}_{rm ext}}$








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '18 at 23:42

























            answered Dec 7 '18 at 21:55









            caveraccaverac

            14.5k31130




            14.5k31130












            • $begingroup$
              If the ramp exerts a force of friction $F_k$ on the block whose direction opposes its acceleration, does that mean that the block exerts an equal force in the opposite direction on the ramp?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:33












            • $begingroup$
              @AnsonPang yes, forgot to add that one
              $endgroup$
              – caverac
              Dec 7 '18 at 22:43










            • $begingroup$
              Is the force of gravity acting on the incline $Mg$, or $(M+m)g$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:47










            • $begingroup$
              @AnsonPang It is $M$, because the mass of the incline is just $M$
              $endgroup$
              – caverac
              Dec 7 '18 at 22:52










            • $begingroup$
              So the forces acting on the incline are gravity, the opposing force to $F_k$ that I mentioned, and $-N$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:54


















            • $begingroup$
              If the ramp exerts a force of friction $F_k$ on the block whose direction opposes its acceleration, does that mean that the block exerts an equal force in the opposite direction on the ramp?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:33












            • $begingroup$
              @AnsonPang yes, forgot to add that one
              $endgroup$
              – caverac
              Dec 7 '18 at 22:43










            • $begingroup$
              Is the force of gravity acting on the incline $Mg$, or $(M+m)g$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:47










            • $begingroup$
              @AnsonPang It is $M$, because the mass of the incline is just $M$
              $endgroup$
              – caverac
              Dec 7 '18 at 22:52










            • $begingroup$
              So the forces acting on the incline are gravity, the opposing force to $F_k$ that I mentioned, and $-N$?
              $endgroup$
              – Anson Pang
              Dec 7 '18 at 22:54
















            $begingroup$
            If the ramp exerts a force of friction $F_k$ on the block whose direction opposes its acceleration, does that mean that the block exerts an equal force in the opposite direction on the ramp?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:33






            $begingroup$
            If the ramp exerts a force of friction $F_k$ on the block whose direction opposes its acceleration, does that mean that the block exerts an equal force in the opposite direction on the ramp?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:33














            $begingroup$
            @AnsonPang yes, forgot to add that one
            $endgroup$
            – caverac
            Dec 7 '18 at 22:43




            $begingroup$
            @AnsonPang yes, forgot to add that one
            $endgroup$
            – caverac
            Dec 7 '18 at 22:43












            $begingroup$
            Is the force of gravity acting on the incline $Mg$, or $(M+m)g$?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:47




            $begingroup$
            Is the force of gravity acting on the incline $Mg$, or $(M+m)g$?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:47












            $begingroup$
            @AnsonPang It is $M$, because the mass of the incline is just $M$
            $endgroup$
            – caverac
            Dec 7 '18 at 22:52




            $begingroup$
            @AnsonPang It is $M$, because the mass of the incline is just $M$
            $endgroup$
            – caverac
            Dec 7 '18 at 22:52












            $begingroup$
            So the forces acting on the incline are gravity, the opposing force to $F_k$ that I mentioned, and $-N$?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:54




            $begingroup$
            So the forces acting on the incline are gravity, the opposing force to $F_k$ that I mentioned, and $-N$?
            $endgroup$
            – Anson Pang
            Dec 7 '18 at 22:54











            0












            $begingroup$

            If the block is sliding down the ramp then the horizontal force on the ramp is.......



            $F_{rx} = cos(theta)(mgsin(theta) - u_kmgcos(theta))$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              If the block is sliding down the ramp then the horizontal force on the ramp is.......



              $F_{rx} = cos(theta)(mgsin(theta) - u_kmgcos(theta))$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                If the block is sliding down the ramp then the horizontal force on the ramp is.......



                $F_{rx} = cos(theta)(mgsin(theta) - u_kmgcos(theta))$






                share|cite|improve this answer









                $endgroup$



                If the block is sliding down the ramp then the horizontal force on the ramp is.......



                $F_{rx} = cos(theta)(mgsin(theta) - u_kmgcos(theta))$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 21:56









                Phil HPhil H

                4,1632312




                4,1632312






























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