Direction of acceleration of an incline

[marksyncm]

Object with mass $m$ is sliding down a sloped (incline = $\alpha$) object of mass $M$. The coefficient of kinetic friction acting between $m$ and $M$ is $\mu$. There is no friction between object $M$ and the ground.

In the drawing above, the red vectors above are the forces acting on the big object $M$. I realize I should have drawn the vectors extending out from object $M$, but I did it this way to make the angles clearer. So what I have is that there are two forces acting on the sloped object $M$ in the horizontal direction:

1) $\mu F_g cos^2\alpha$ to the left - the horizontal component of the opposite friction force exerted by $m$ onto $M$
2 $F_gcos\alpha sin\alpha$ to the right - due to the weight of mass $m$.

Two questions:

2) Do the above two forces correctly (and comprehensively) describe the forces acting on object $M$ in the horizontal direction?

2) Does this mean that it is possible, given a small enough angle $\alpha$, for the inclined object $M$ to accelerate to the left (ie. assuming $\mu F_g cos^2\alpha > F_gcos\alpha sin\alpha$)?

Thank you

 

[A.T.]

There is no friction between object $M$ and the ground.

If M is not fixed, can you really assume that the normal force between m and M is mg*cosα ?

 

[marksyncm]

If M is not fixed, can you really assume that the normal force between m and M is mg*cosα ?

$M$ only moves horizontally. Does its horizontal movement impact the normal force between $M$ and $m$? Perhaps I misunderstood your question.

 

[A.T.]

$M$ only moves horizontally. Does its horizontal movement impact the normal force between $M$ and $m$?

How does it affect the acceleration direction of m, if M can move out of the way? How does your normal force formula follow from of Newtons laws?

Regarding your second question: If m is initially moving, while M is initially static, then it is possible that both will end up moving to the left. But if both are initially static, then they must move in opposite direction due to momentum conservation.

 

[marksyncm]

How does it affect the acceleration direction of m, if M can move out of the way? How does your normal force formula follow from of Newtons laws?

As far as I understand, the normal force should be equal to the net force exerted by $m$ onto $M$, perpendicular to its surface. Do we mean to say that when $M$ accelerates due to $m$'s friction, $M$ will exert a horizontal force onto $m$, and that this force will have a component that will impact the normal force?

 

[jbriggs444]

As far as I understand, the normal force should be equal to the net force exerted by $m$ onto $M$, perpendicular to its surface.

Yes.

Do we mean to say that when $M$ accelerates due to $m$'s friction, $M$ will exert a horizontal force onto $m$, and that this force will have a component that will impact the normal force?

The suggestion is that if M moves out of the way, the acceleration of m that keeps it in contact with the surface of M will be different than if M does not move out of the way.

Different acceleration, different force needed to produce that acceleration.

 

[marksyncm]

The suggestion is that if M moves out of the way, the acceleration of m that keeps it in contact with the surface of M will be different than if M does not move out of the way.

Different acceleration, different force needed to produce that acceleration.

Thanks. Still not sure I follow. When you say "the acceleration of m that keeps it in contact," what exactly do you mean? I am not sure I understand what it means for acceleration to keep two bodies in contact. As I'm visualizing it, $m$ is sliding down $M$, and $M$ is moving left or right - after a while, $m$ will just slide off $M$.

 

[jbriggs444]

Thanks. Still not sure I follow. When you say "the acceleration of m that keeps it in contact," what exactly do you mean? I am not sure I understand what it means for acceleration to keep two bodies in contact. As I'm visualizing it, $m$ is sliding down $M$, and $M$ is moving left or right - after a while, $m$ will just slide off $M$.

While they are sliding against one another there is a "constraint" in effect. They are in contact. The only permissible accelerations for either object are those that conform to the constraint.

 

[marksyncm]

Thanks, now I understand what you meant by "keeping it in contact." I still can't deduce how this affects the normal force, however. For starters, what direction of acceleration should I assume for $M$?

 

[A.T.]

As far as I understand, the normal force should be equal to the net force exerted by $m$ onto $M$, perpendicular to its surface.

That is the definition of "normal force". But how did you arrive at the formula for its magnitude? I assume you have simply used the one from similar problems with a static incline. You should ask yourself how this formula was derived, and whether the assumptions used in that derrivation are still true in your case with the sliding incline.

To get some intuition, ignore all friction and compare the extreme cases:
1) M >> m: Static incline
2) m >> M: How does m accelerate here? Is your formula still consistent with that acceleration?

 

[marksyncm]

I think I see the issue. I assumed that $N- mgcos(\alpha)=0$. Now that the incline $M$ is moving, it is exerting a force $ma$ onto block $m$, which has a component of $masin(\alpha)$ perpendicular to the incline surface. This makes $N = mgcos(\alpha) + masin(\alpha)$.

Is this correct?

 

[A.T.]

Is this correct?

Consider case 2) from post #10. Does your formula make sense there?

Have you tried to express the kinematic constraint mentioned by jbriggs444 in post #8 as an equation?

 

[jbriggs444]

Thanks, now I understand what you meant by "keeping it in contact." I still can't deduce how this affects the normal force, however. For starters, what direction of acceleration should I assume for $M$?

If block M is also sitting on a table (frictionless or otherwise) then there is a constraint between the table and block M that prevents any acceleration into the table. Block M must accelerate horizontally along the table surface.

Given the absence of any information about forces toward or away from our line of sight (i.e. into or out from the graph paper) we can reason from symmetry that there is no motion in that direction either. The remaining direction is horizontally right or left. Pick a direction to be positive and start writing down equations. It does not matter whether you choose a direction such that M's acceleration turns out to be negative. Let the math give you the answer you seek using the sign convention you choose.

Edit: Personally, I'm a first quadrant guy. Right positive x, up positive y.

 

[marksyncm]

Thanks to you both, I must rest now but will give this another serious shot tomorrow based on your suggestions.

 

[neilparker62]

http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node80.html if it's any help ?

 

[marksyncm]

I decided to approach the problem from a non-inertial frame of reference as it seems easier (to me) for this problem:

For starters, I would like to see if I finally managed to get the normal force right.

Assuming the incline $M$ is accelerating to the right, the sliding object $m$ is experiencing a "pseudo-force" $ma$ to the left. Therefore, the sum of the forces acting perpendicular to the incline is $N + masin(\alpha) - mgcos(\alpha) = 0 \rightarrow N = mgcos(\alpha) - masin(\alpha)$.

Do I have this part right?

Thank you for your patience.

 

[marksyncm]

To add, for now I'm assuming above that block $m$ is stationary with respect to block $M$.

 

[A.T.]

Assuming the incline $M$ is accelerating to the right, the sliding object $m$ is experiencing a "pseudo-force" $ma$ to the left. Therefore, the sum of the forces acting perpendicular to the incline is $N + masin(\alpha) - mgcos(\alpha) = 0 \rightarrow N = mgcos(\alpha) - masin(\alpha)$.

Looks OK to me: In the rest frame of M, m can only accelerate parallel to the incline, so the forces on m normal to the incline must cancel.

 

[Tom Hammer]

The net force on M is to the left, not to the right.

 

[marksyncm]

The net force on M is to the left, not to the right.

$M$ is the underlying incline, while $m$ is the sliding block. Is the net force on the underlying incline to the left?

 

[Tom Hammer]

Yes. There are two forces on M. One, the force of gravity on M is perpendicular to the horizontal, so it has no component to the left or the right. m is sliding to the left. The force of friction ffMm (the force of friction of M on m is up the incline, in the opposite direction of the velocity of m. ffmM, the force of m on M is down the incline, since it is the reaction force (Newton´s third law.) This force has a horizontal component, which makes M slide to the left.

 

[jbriggs444]

Yes. There are two forces on M. One, the force of gravity on M is perpendicular to the horizontal, so it has no component to the left or the right. m is sliding to the left. The force of friction ffMm (the force of friction of M on m is up the incline, in the opposite direction of the velocity of m. ffmM, the force of m on M is down the incline, since it is the reaction force (Newton´s third law.) This force has a horizontal component, which makes M slide to the left.

The normal force of m on M has a horizontal component that you have ignored. That component is responsible for the leftward motion of m. Newton's third has something to say about the resulting motion of M.

 

[neilparker62]

Just a hunch: could this be seen as a 2 variable feedback system and analysed as such? I just don't have the Maths (or perhaps have forgotten the Maths!) for doing this. Any control theory guys out there ...?

 

[jbriggs444]

Just a hunch: could this be seen as a 2 variable feedback system and analysed as such? I just don't have the Maths (or perhaps have forgotten the Maths!) for doing this. Any control theory guys out there ...?

The usual implicit assumption is that transient deviations from an equilibrium state die down. One solves for the equilibrium state and does not look at the feedback loop that leads there.

If the block were "chattering" down the slope (something like squeaky chalk on a chalkboard) without settling into a stable slide, the solution would be vastly more difficult and could involve control theory.

 

[neilparker62]

Something like this perhaps. I am basing the matrix algebra on this solution obtained for a similar problem in which there is no friction.

$$ \begin{pmatrix}
M+m & mcos(θ) \\
mcos(θ) & m
\end{pmatrix} \begin{pmatrix} \ddot x \\ \ddot x' \end{pmatrix}=\begin{pmatrix} 0 \\ mgsin(θ) \end{pmatrix}$$
Determining the inverse matrix yields the actual solution you can see at the above mentioned reference. Perhaps we can adapt to solve this particular problem where there is friction ?