Pushing an incline so mass on ramp accelerates upward

In summary, the problem is that the FBD's attached appear to be incorrect. The force required to keep the block stationary is not given by the equation, but rather by finding the maximum force that would cause the block to have no vertical acceleration.
  • #1
matineesuxxx
77
6
This seems to be a slight variation of a pretty standard problem, however I didn't have any luck finding any seemingly helpful information. I am mainly wondering if I am not getting the FBDs correct?

Homework Statement



A block of mass [itex]m[/itex] is sitting on a movable ramp of mass [itex]M[/itex] with an incline of [itex]\theta[/itex] degrees to the horizontal. The system is initially held at rest. Once released, a force, [itex]F[/itex], pushes the vertical side of the ramp such that the block accelerates upward along the incline. There is no friction anywhere; Write an expression for the force.

Homework Equations



[itex]\sum \text{F} = ma[/itex]

The Attempt at a Solution



The FBD's are in the attached image.

I seem to be arriving at a system with one more variable than equations, so I am definitely not understanding something, and I believe it has to do with the FBD? Here is what I come up with:

As the inclines acceleration ([itex]a_{\text{I}}[/itex]) is purely on the horizontal plane, then the components of the blocks acceleration ([itex]a_B[/itex]) are:

[itex]a_x = a_{\text{B}}\cos \theta - a_{\text{I}}[/itex] and [itex]a_y = a_{\text{B}}\sin \theta[/itex].

for the little mass, m:

[itex]\sum \text{F}_x = N_1 \sin \theta = m(a_{\text{B}} \cos \theta - a_{\text{I}}) [/itex]

[itex]\sum \text{F}_y = N_1 \cos \theta - mg = ma_{\text{B}}\sin \theta[/itex]

and for the incline:

[itex]\sum \text{F}_x = \text{F} - N_1\cos \theta = a_{\text{I}}M[/itex]
I'm pretty stuck and have no idea what it is that I'm missing, so I would really appreciate it if someone could give me a hint.
 
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  • #2
Forgot to include the FBD.

[edit] The rightmost FBD is for the incline.
 

Attachments

  • FBD.jpg
    FBD.jpg
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  • #3
Is the ramp moving vertically or is it rotating via the external force?

Take a ping pong ball and put it on a board. Angle it and, keeping the angle the same, move the whole system upwards. What happens?

Now, same situation, but hold the board by one end and swing it up over your head (you might want to be outside) what happens?
 
  • #4
BiGyElLoWhAt said:
Is the ramp moving vertically or is it rotating via the external force?

Take a ping pong ball and put it on a board. Angle it and, keeping the angle the same, move the whole system upwards. What happens?

Now, same situation, but hold the board by one end and swing it up over your head (you might want to be outside) what happens?

I think that I may have stated the problem in an unclear way. So the incline is resting on the ground and the force is pushing it along the horizontal and there is no rotation in the ramp. I want to find an expression for the magnitude of the force so that the block slides up the ramp leftward as the ramp slides to the right.
 

Attachments

  • Prob2.jpg
    Prob2.jpg
    5.7 KB · Views: 458
  • #5
Oh, ok. I thought the ramp was moving upward, not rightward. So initially, you can just look at the acceleration of the ramp, and see what conditions need to be met in order for the block to move upward.

Can you tell me what these are? Basically you have force of Earth on block and force of ramp on block.

Hint* think components or rotate your axes.
 
  • #6
Maybe a better question would be which way should the net force be directed?
 
  • #7
Are you trying to find the minimum value of F to accelerate the mass up the incline?
 
  • #8
BiGyElLoWhAt said:
Maybe a better question would be which way should the net force be directed?

Well, if I draw the acceleration vectors for the block and the incline - the block with respect to the incline is up and to the left and the incline with respect to the ground is only to the right - then the resulting acceleration vector for the block with respect to the ground is up and to the right. That would then be the direction of the net force, no?

Doc Al said:
Are you trying to find the minimum value of F to accelerate the mass up the incline?

yes, I am.
 
  • #9
matineesuxxx said:
yes, I am.
Good. And that was my hint. :wink:
 
  • #10
Doc Al said:
Good. And that was my hint. :wink:

Ah, ok. Well, my first attempt at this problem was to find the maximum force with which to push the incline such that the block stays still, since any force greater than that would cause the block to accelerate upward, but I am actually thinking that its not the way I should go
 
  • #11
matineesuxxx said:
Ah, ok. Well, my first attempt at this problem was to find the maximum force with which to push the incline such that the block stays still, since any force greater than that would cause the block to accelerate upward, but I am actually thinking that its not the way I should go
I would say that finding the force that would cause the block to have no vertical acceleration is the way to go.

Did you find that force?
 
  • #12
Doc Al said:
I would say that finding the force that would cause the block to have no vertical acceleration is the way to go.

Did you find that force?

I did, and here is what I got:

[itex] \text{F} = g(m+M)\frac{\sin \theta - \mu_{s} \cos \theta}{\cos \theta + \mu_{s} \sin \theta}[/itex].

My logical check was that if [itex]\mu_{s} = 0[/itex] then [itex]\text{F} = g(m+M)\tan \theta [/itex] which is what I got when I solved the no friction situation.

for [itex]\mu_{s} \neq 0 [/itex] and if we let [itex] \theta = 0 [/itex] then we can view the system as two rectangular blocks stacked together and i get $$\text{F} = g(m+M)\mu_{s}$$ which is also what i got when i solved that situation seperatly.

Finally, if [itex]\mu_{s} = 0 [/itex] and [itex] \theta=0[/itex] then we have the two block situation again, but no friction, so there is no way you can push the bottom block so that the top block has no acceleration with respect to the bottom block, which is what i get from that equation as well, [itex] \text{F} = 0 [/itex].Is that correct?
 
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  • #13
matineesuxxx said:
I did, and here is what I got:

[itex] \text{F} = g(m+M)\frac{\sin \theta - \mu_{s} \cos \theta}{\cos \theta + \mu_{s} \sin \theta}[/itex].
Is that correct?

That's not quite what I'm getting, but we might be starting from 2 different places and getting to 2 different solutions. How did you start?

My N2L is ##\Sigma \vec{F} = m \vec{a} = \vec{N} + \vec{w} = (mg + \frac{m}{M}F_{on ramp})<cos( \theta), sin( \theta)> + mg<cos( \theta), -sin(\theta)>##

do you see where I got everything? The ##\frac{m}{M}F_{onramp}## is the force on the ramp, divided by the mass of the ramp (the acceleration on the ramp) and since the block isn't moving wrt the ramp (solving for the force needed to keep the change in height 0) the system moves as one, so the acceleration of the ramp will be the acceleration of the block. The force is then ##m\vec{a}_{oframp}##
 
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  • #14
What your equation looks like is you're treating the F on ramp as a force on the system, not just the ramp. (M+m)g
 
  • #15
matineesuxxx said:
I did, and here is what I got:

[itex] \text{F} = g(m+M)\frac{\sin \theta - \mu_{s} \cos \theta}{\cos \theta + \mu_{s} \sin \theta}[/itex].
Yes, that's good. (If you get rid of the friction! Why complicate things with friction?)
 
  • #16
BiGyElLoWhAt said:
What your equation looks like is you're treating the F on ramp as a force on the system, not just the ramp.
F is both: It acts on the ramp and on the "ramp + block" system.
 
  • #17
BiGyElLoWhAt said:
That's not quite what I'm getting, but we might be starting from 2 different places and getting to 2 different solutions. How did you start?

If we are finding the maximum force with which to push the incline so the block has no vertical acceleration, then the block/ramp system is one mass with only one acceleration in the positive x direction, so i get [itex]\text{F}=m_Ta = (m+M) a \implies a = \text{F}/(m+M) [/itex]. So now we only need consider the forces on the block,
$$\sum \text{F}_x = N(\sin \theta - \mu_s \cos \theta) = ma\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
$$\sum \text{F}_y = N(\cos \theta + \mu_s \sin \theta) - mg = 0 \;\;\;(2)$$

so dividing (1) by (2) and substituting in a, I get my answer.

Doc Al said:
Yes, that's good. (If you get rid of the friction! Why complicate things with friction?)

Who likes friction, eh? ;)
 
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  • #18
Doc Al said:
Yes, that's good. (If you get rid of the friction! Why complicate things with friction?)

Also, the reason I thought I shouldn't tackle it in this way was because the answer the book provides is $$\text{F} = \mu_{\text{k}}(m+M)g\sin \theta \cos \theta $$, but now that I think about it, their answer says if there is no friction in the system then a force of 0 N will cause the block to accelerate up the ramp. what??
 
  • #19
What book are you using?
 
  • #20
Doc Al said:
What book are you using?

Physics for Scientists and Engineers An Interactive Approach by Nelson.

One of the profs at the university I go to co-authored the book, so we used it in the advanced physics class. Dropped it after two days, so I'm trying to get ahead of the game before I take it again in the fall.
 
  • #21
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  • #22
SammyS said:
My suggestion for solving this problem is to find the force, F, necessary so the block doesn't move up or down the ramp, i.e. the vertical component of the block's acceleration is zero. (Yes, this was you initial strategy.)

There's no need to consider friction at all.
Exactly. (Which is what I have suggested above.) Makes it much easier if you don't have to bother with friction.
 

1. How does pushing an incline cause a mass on a ramp to accelerate upward?

Pushing an incline causes the mass on the ramp to accelerate upward due to the force applied by the push. This force is directed along the incline and is resolved into two components, one parallel to the ramp and one perpendicular to the ramp. The component parallel to the ramp is responsible for the acceleration of the mass.

2. Why does the mass on the ramp accelerate faster when pushed on an incline compared to a horizontal surface?

The mass on the incline accelerates faster because the incline provides an upward force component in addition to the force applied by the push. This upward force component reduces the net force required to accelerate the mass and allows it to accelerate at a faster rate.

3. Can the angle of the incline affect the acceleration of the mass on the ramp?

Yes, the angle of the incline can affect the acceleration of the mass on the ramp. As the angle of the incline increases, the component of the applied force directed along the incline decreases, resulting in a decrease in the acceleration of the mass. Conversely, as the angle of the incline decreases, the component of the applied force along the incline increases, resulting in an increase in the acceleration of the mass.

4. Is the acceleration of the mass on the ramp affected by the mass of the incline?

No, the acceleration of the mass on the ramp is not affected by the mass of the incline. The acceleration of an object is dependent on the net force acting on it, and the mass of the incline does not contribute to this force. Therefore, the mass of the incline does not affect the acceleration of the mass on the ramp.

5. How does friction play a role in the acceleration of the mass on the ramp?

Friction can act as a resisting force to the motion of the mass on the ramp, reducing its acceleration. The amount of friction is dependent on the coefficient of friction between the mass and the ramp, as well as the normal force between the two surfaces. Thus, a higher coefficient of friction or a higher normal force will result in a greater frictional force and a decrease in the acceleration of the mass on the ramp.

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