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A block of mass $m$ is placed on an inclined plane (a ramp). If a constant force $f$ is applied to the ramp so that it is accelerating horizontally at a proper rate, the block will remain at the same height. But what then is the force that cancels the component of the block weight parallel to the plane, (i.e. $mgsinθ$), and prevents the block from sliding along the inclined plane?

Note: all surfaces are frictionless.


Notice that in-order for the block(of mass $m$) and the inclined plane(wedge of mass $M$) to move together, they must have a common horizontal acceleration given by: $$a=\frac{F}{M+m}$$ And thus for the block of mass $m$ it's horizontal acceleration must be equal to this, so there is a resultant force on the small block, acting horizontally(which I'll call $F_m$ which is given by $F_m=ma$ where a is the common horizontal acceleration of the block and wedge).

Indeed there is no force opposing the component $mgsin(\theta)$ and you can see below that it is not required to be cancelled as it itself becomes a component of the resultant force $F_m$ which has components $N-mgcos(\theta)$ and as expected $mgsin(\theta)$:

Note: Diagram showing the forces on only the block of mass $m$.

Another diagram requested to view the force diagram in another way which will give the same end result:

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    $\begingroup$ The fact is the block would indeed slide down the plane if $Fsin(\theta)$ was the only resultant force on the block, but notice here that $F_m$ does not "only have components down the incline". Here there is a component that perpendicular to the plane that is $N-mgcos(\theta)$. Maybe you have wrongly assumed that $N$ and $mgcos(\theta)$ are equal but they are not as in an accelerated system there will be higher $N$ needed to keep the object stationary relative to the plane. $\endgroup$ – Tausif Hossain Sep 3 '18 at 12:10
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    $\begingroup$ Tausif, I figured it out using 3 strategies: My approach - 1) (Use Inertial Forces) Set the inertial force opposite and equal to the accelerating force. 2) (No Inertial Forces). Accelerate the block, so that the down incline component of the external acceleration equals the down-incline component of gravity. Your approach, 3) (No Inertial Forces) Solve for the required acceleration using the Normal vectors as the sides of the triangle. a) $mg = Ncos\theta$, and $F_m=Nsin\theta$. b) use the ratio of triangles, $tan\theta = \frac{ma}{mg}$. c) Solve for acceleration: $a=gtan\theta$. $\endgroup$ – Thomas Lee Abshier ND Sep 3 '18 at 23:14
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    $\begingroup$ Finally your reasoning is now done right and indeed you can now be correct in saying $a=gtan(\theta)$. $\endgroup$ – Tausif Hossain Sep 4 '18 at 8:29
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    $\begingroup$ Thanks Tausif- great discussion! I learned several important principles about acceleration problems. There are (at least) 3 strategies to solve the problem in an inertial frame: a) Substitute Normal forces in a trigonometric relationship to compute the unknown, even though they do not produce motion in a frictionless and restrained system. b) Equate (rather than oppose) external and gravitational components downhill. c) Use the inertial force of mass, which arises with the acceleration of mass, to oppose the gravitational force. That was a learning experience. Thanks so much! $\endgroup$ – Thomas Lee Abshier ND Sep 4 '18 at 13:48
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    $\begingroup$ Very glad I could help, it is especially great now how you understand to equate forces instead of just trying to cancel them out. $\endgroup$ – Tausif Hossain Sep 4 '18 at 13:54

This question is correctly answered here, but it deserves an additional conceptual examination because of how clearly this problem illuminates the need to explicitly consider and include the Inertial Force in this force vector diagram.

Newton's third law: All forces in the universe occur in equal but oppositely directed pairs. A Force applied to any mass will produce the same force in reaction. Note that the Inertial Force has a different character than the Field forces (e.g., gravity, electric, magnetic). The Inertial Force does not exert the sustained force of a potential field. Rather, the Inertial Force exerts a reactive opposing force only during the instant while accelerating and transferring kinetic energy.

The problem:

  • What is the $F_{external}$ on the block required to accelerate the mass and wedge at the rate to suspend the block at a point on the wedge?

The Solution:

  • Gravitational force $F_{gravity-downhill} = mg sin\theta$ accelerates the block down the frictionless surface of the wedge.
  • Maintaining a static position on the block, requires balancing the downward acceleration of gravity with the upward acceleration of an external force.
  • Horizontal acceleration of the wedge and block provides a vector component of upward Force: $F_{inertial-uphill}$.
  • Solution requires determining the magnitude of $F_{external}$ which contributes the force vector component of equal magnitude and opposite direction to the downhill gravitational force.

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The block does accelerate downward the inclined plane. This acceleration is caused by $mgsin( \theta )$. The block also accelerates along the normal. This acceleration is caused by the normal and $mgcos( \theta )$.

There are no other relevant forces.

This seemingly peculiar view is a result of choosing a coordinate system along the plane and normal to the plane while we already know that the overall acceleration is along the horizontal axis.

If you do select the horizontal and vertical axis as your coordinate system, than only along the horizontal axis you obtain acceleration while there is zero acceleration along the vertical axis.

  • $\begingroup$ I'm trying understand your reasoning. As I understand it, you have chosen the Normal as the y axis, and the Inclined Plane as your x axis. A horizontal Force is applied to the wedge producing a horizontal acceleration vector. The horizontal acceleration magnitude is chosen to produce an x axis component equal and opposite to the downward gravitational acceleration of $gsin\theta$. I do not see how an acceleration to the right, will equalize the gravitational force, which is also to the right. It seems like a force opposing $mgsin\theta$, such as the Inertial Force, is needed. $\endgroup$ – Thomas Lee Abshier ND Aug 27 '18 at 20:08
  • $\begingroup$ The component along x-axis is $gsin(\theta)$. Why do you say it is opposite to it? In general, you never need an inertial force to solve a problem. In some cases, it is convenient to solve a problem while "sitting on" the accelerating coordinate system, in which case you use the virtual force. $\endgroup$ – npojo Aug 27 '18 at 20:47
  • $\begingroup$ I agree, that the acceleration along the positive x-axis due to gravity is $gsin\theta$ in this chosen coordinate system. I was saying that to attain static positional equilibrium on the x-axis, we must apply a force balancing the positive x-axis force with an oppositely directed force. Applying a positive x-axis force on the wedge produces an acceleration in the positive x-axis direction, which will not balance/oppose/zero/negate the gravitational force. Thus the need for considering the Inertial Force - a negative x-axis directed reactive force, which leaves the block with $0$ net force. $\endgroup$ – Thomas Lee Abshier ND Aug 27 '18 at 22:07
  • $\begingroup$ You say: "to attain static positional equilibrium on the x-axis". Notice that you are not static on the x-axis from the point of view of the inertial frame (ground). Think of an x-axis fixed to the ground at one point and you will see that the block "shadow" slides along this axis. You are static on the x-axis if the x-axis accelerates with the block. In this frame you indeed need to use the inertial force. Adopting the accelerating frame seems to be more natural to you, but you should be aware that it is not the natural inertial frame. $\endgroup$ – npojo Aug 28 '18 at 4:07
  • $\begingroup$ I unintentionally described the accelerating/non-inertial frame while attempting to shift to the (x-incline/y-normal) frame you were describing. Still, gravity $mgsin\theta$ pulls down, and the Inertial Force, $mgtan\theta$ pushes up from the wedge regardless of the frame chosen. Forces arising from acceleration are called "fictitious" since they arise from inertia, not a field. Still, this force is real and relevant as it balances $mgsin\theta$. How would you explain the equilibrium of forces otherwise? It appears you were attempting to do so by changing frames. How is that possible? $\endgroup$ – Thomas Lee Abshier ND Aug 28 '18 at 6:45

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