FRICTIONLESS INCLINED PLANE.. (some guidance please)

[Utsav]

I have this problem that I'm not too sure about...
it states...

you have an inclined plane with a mass of (capital) M. The angle of the incline of the plane is alpha. there is also a mass on the plane that has a mass of (lower case) m.

All of the surfaces (the inclined plane, the surface the plane lies on and the mass *ARE ALL FRICTIONLESS*.

Fast must the inclined plane be moving in order for the mass to stay exactly where it is.

(Answer in terms of the variables. Consider gravity to be -10 m/s2 [taking down to be negative]).

I'd appreciate any help!
THANKS

 

[newcool]

Do you know how to start the problem? What have you done?

 

[Doc Al]

Welcome to PF!

Realize that the inclined plane must be accelerating in order to prevent the mass from sliding down. Start by identifying the forces acting on the mass (there are only two). Then apply Newton's 2nd law for the vertical and horizontal components. Hint: The mass must have a horizontal acceleration (which is what you need to find), but set the vertical acceleration to zero.

 

[futb0l]

Welcome to PF! Try and draw a Free Body Diagram.

 

[Utsav]

Thanks

i'll try it and ask again if i need help

 

[Utsav]

Still not getting it?

I know that much...i just don't know how I'm supposed to fiugure out the needed applied force on the inclined plane.....

 

[MathStudent]

show what youve got so far:
list the forces acting on the mass... set up a coordinate system
I recommend a coordinate system where the x-axis lies along the inclined plane.

 

[Utsav]

i added an attachment of the diagram i made showing the forces

the inclination is alpha and the areas are all frictionless (*just a reminder)

 

[Doc Al]

I know that much...i just don't know how I'm supposed to fiugure out the needed applied force on the inclined plane.....

For the moment, forget about any force applied to the inclined plane. Concentrate on the mass resting on the plane. Your attachment did not appear, but a free body diagram should show the two forces acting on the mass. Consider vertical and horizontal components. You'll get two equations, which will allow you to solve for the acceleration.

In your original post it's not clear what exactly you are asked to find. But if you need to find the force on the plane, first find the acceleration as I outline above and in my earlier post.

 

[Utsav]

What i have to solve is 'hpw much force is required to keep the mass on the plane eactly where it is.'

Thanks for helpin' me out... ; )

 

[Utsav]

*attachment Of Diagram*

 

[Utsav]

ok so i did a little analysis n here's what I've got.

the force of gravity on the mass is '10*m.' derived into their x and y compontents, Fg in the x is -10 * sin(alpha); Fg in the y is 10 * cos(alpha). i don't think there is any y force to worry about, since the second law gives it the opposite = force (normal force).
The main force to focus on is the Fg in the x. we need to figure out how much force is required to overcome that force using the inclined plane. t h a t 's w h e r e i 'm s t u c k. Is the force needed to overcome the gravitational force of the mass 10 * sin(alpha) * 'M' (<-the mass of the inlined plane.)

thats what I've got...please give some feedback!
THANKS in advance

 

[Doc Al]

Hint: There are two forces acting on the mass. The weight acting down (which is just mg) and the normal force of the plane. Call the normal force "N" (it's an unknown). (What are the vertical and horizontal components of the normal force?) Now go back to my earlier posts and write down the equations for Newton's 2nd law for vertical and horizontal components. (Do not use components parallel and perpendicular to the plane! That will just make it harder.)

 

[Utsav]

i know that in order to find the vert. and horiz. force, you have to make a reference triangle. but I'm not sure what the angles of the triangle are (i am guessing that one of the angles is alpha and the other is 90 - alpha).

 

[Doc Al]

Figure out the angle that the normal force makes with the horizontal axis by drawing a careful diagram. Then you can find its components (I presume).

 

[Utsav]

tried that...i used the trig functions... wether or not alpha is an angle in the normal force, I'm still un sure...

and I'm also worried about what my answer will look like (with alphas and cos and sin etc.)

 

[Doc Al]

tried that...i used the trig functions... wether or not alpha is an angle in the normal force, I'm still un sure...

I don't know what you mean when you say you "tried that". Did you find the angle that the normal force makes with the horizontal (or vertical)? What is it? (Of course, it will be in terms of alpha.) Draw a picture of the normal force and label the angles.

and I'm also worried about what my answer will look like (with alphas and cos and sin etc.)

You are thinking way too hard on this one. Just do it. (It's much easier than you think.)

 

[Utsav]

i think i got it. here's what i think the answer is (try not to lose me in my explination; drawing what i say might help.)

you have the object with mas 'm'. it rests on an inclined plane with mass 'M' and inclination alpha. (all surfaces are fictionless)

OBJECT
the force to gavity on the object is 10m. the normal force is 10m * cos (alpha). the acceleration of the object is 10m * sin(alpha).

Keeping the object in place.
now that we know the force on the object, we need to keep it in place. the acceleration of the object is 10m * sin(alpha); the mass of the inclined plane is 10M; i think that the force applied must be greater than 10m * sin(alpha) + 10M.

Fapp > 10m * sin(alpha) + 10M

(i hope that's it)

 

[Doc Al]

the force to gavity on the object is 10m.

Right: the weight of the object is mg acting downward.

the normal force is 10m * cos (alpha).

No. This would be the normal force if the plane were stationary. But it's not: it is accelerating.

the acceleration of the object is 10m * sin(alpha).

No. This doesn't even have the units of acceleration.

Call the normal force "N" and the acceleration "a". Then do what I suggested in posts 3, 9, and 13. (Do it!)

Find the acceleration, then we'll worry about the force needed to produce that acceleration.

 

[plusaf]

i may be all wet, but how about this approach...

Right: the weight of the object is mg acting downward.
No. This would be the normal force if the plane were stationary. But it's not: it is accelerating.
No. This doesn't even have the units of acceleration.
Call the normal force "N" and the acceleration "a". Then do what I suggested in posts 3, 9, and 13. (Do it!)
Find the acceleration, then we'll worry about the force needed to produce that acceleration.

1) if the incline mass, M, were at rest, how fast would the block, m, on it, accelerate down the slope? i.e., what is the acceleration of mass m down the ramp of mass M with slope = alpha.
2) in order to keep the mass m in place on the incline of mass M, assuming all surfaces are frictionless, mass M must be accelerating at a rate which would cancel the acceleration of gravity on mass m, right? looks to me if mass M were accelerating at the same rate in the same direction as mass m was "trying" to accelerate down the ramp, all would be in balance... no?
3) here's the gotcha nobody's mentioned yet... what force, F, is needed to accelerate mass M to get it up to that needed rate of acceleration, is the wrong answer. the force is determined by F=ma, but here, we got the "a" above, but F=(M+m)a! you got to accelerate the combination of the two masses at that rate of acceleration. don't miss that part!

+af

 

[VietDao29]

Hi,
I have a suggestion. (I am not a native English speaker... so please forgive me if I make any misspell or confusing words).
Call a is the acceleration of the M.
In the non-inertial system of reference attached to the inclined plane. There are 3 forces acting on the mass:
+ Gravity : P = mg
+ Normal Force : This force is not a constant.
+ The 3rd force is called inertial force. If the inclined plane is moving at a (m/s^2) in the positive direction, then that force will be negative (parallel with the ground, not the inclined plane) and has the magnitude of: F = ma.
Since the mass is not moving relative to the inclined plane. The sum of three forces must be zero. Find it yourself, it's easy. (Remember that the mass can only move diagonally relative to the inclined plane, it cannot move straight up to the sky or down to the ground).
Hope this help,
If you have any questions or it's not clear, post it here.
Viet Dao,

 

[Doc Al]

1) if the incline mass, M, were at rest, how fast would the block, m, on it, accelerate down the slope? i.e., what is the acceleration of mass m down the ramp of mass M with slope = alpha.
2) in order to keep the mass m in place on the incline of mass M, assuming all surfaces are frictionless, mass M must be accelerating at a rate which would cancel the acceleration of gravity on mass m, right? looks to me if mass M were accelerating at the same rate in the same direction as mass m was "trying" to accelerate down the ramp, all would be in balance... no?

I'm not sure what you mean. I'm sure you don't mean what you said, which is that the ramp must accelerate downward (at an angle)! The ramp (and mass) both accelerate horizontally.

3) here's the gotcha nobody's mentioned yet... what force, F, is needed to accelerate mass M to get it up to that needed rate of acceleration, is the wrong answer. the force is determined by F=ma, but here, we got the "a" above, but F=(M+m)a! you got to accelerate the combination of the two masses at that rate of acceleration. don't miss that part!

This is true. That's why I keep insisting that one first determine the acceleration, then worry about what force is needed to produce the acceleration. One step at a time.

 

[Doc Al]

In the non-inertial system of reference attached to the inclined plane. There are 3 forces acting on the mass:
+ Gravity : P = mg
+ Normal Force : This force is not a constant.
+ The 3rd force is called inertial force.

While you are certainly correct that you can solve this problem easily by using the accelerated frame of the plane, I would advise beginners to stick to the usual inertial frame of reference. It's a very easy problem; no need to introduce inertial forces.

 

[plusaf]

ok so i did a little analysis n here's what I've got.
the force of gravity on the mass is '10*m.' derived into their x and y compontents, Fg in the x is -10 * sin(alpha); Fg in the y is 10 * cos(alpha). i don't think there is any y force to worry about, since the second law gives it the opposite = force (normal force).
The main force to focus on is the Fg in the x. we need to figure out how much force is required to overcome that force using the inclined plane. t h a t 's w h e r e i 'm s t u c k. Is the force needed to overcome the gravitational force of the mass 10 * sin(alpha) * 'M' (<-the mass of the inlined plane.)
thats what I've got...please give some feedback!
THANKS in advance

so, looking first at the small mass "m" on the frictionless slope (angle alpha) of the ramp (larger mass = M)...
yes, the force pulling mass m down the ramp = g*sin(alpha) [from f = ma = m*g*sin(alpha)]. OR, the mass "m" (if the ramp were not moving at all) would accelerate DOWN the ramp with acceleration a= F/m or m*g*sin(alpha)/m, which = g*sin(alpha).

in order that the mass NOT move down the ramp, what must be the net forces acting on it? hint: (zero). net forces=0 means net acceleration (relative to the ramp) =0.

therefore, for the mass m to not slide down the ramp, there must be another force trying to accelerate it "UP" the ramp. this can only happen if the ramp itself is accelerating in the same direction that the mass "m" is trying to slide down the ramp.

so, how fast must the ramp, M, be accelerating in order to balance the acceleration that the mass m would experience if the ramp were not moving?

 

[Utsav]

been busy lately...i feel like i should have figured this out by now.
i just wish i had the answer :(

 

[Doc Al]

Have you tried doing what I've suggested throughout this thread?

 

[arildno]

Utsav:
Let's take this piecewise:
1. Forces on angled block:
a) Its own weight, b)The normal force from the ground on it, c) The normal force upon it from the sliding block on it.
2. Forces on the sliding block:
a)Its own weight, b) The normal force from the angled block upon it.
3) Acceleration and velocity constraints:
a) The angled block has purely horizontal acceleration (why?)
b) The normal component of the angled block's acceleration must equal the normal component of the sliding block's acceleration (why?)

Do you agree to these relations?
(They have been given by Doc Al earlier)

 

[plusaf]

how about a new picture?

Utsav:
Let's take this piecewise:
1. Forces on angled block:
a) Its own weight, b)The normal force from the ground on it, c) The normal force upon it from the sliding block on it.
2. Forces on the sliding block:
a)Its own weight, b) The normal force from the angled block upon it.
3) Acceleration and velocity constraints:
a) The angled block has purely horizontal acceleration (why?)
b) The normal component of the angled block's acceleration must equal the normal component of the sliding block's acceleration (why?)

Do you agree to these relations?
(They have been given by Doc Al earlier)

How about using this picture as another "start"?
Inclined block = mass M
small block on inclined block = mass m
acceleration of the big block = A
acceleration of the small block down the slope = a (((=0)))
and F still =ma

 

[arildno]

1.The acceleration of the big block, A, is in the opposite direction of your drawing.
2. There is a normal force reaction couple betweenthe blocks, not a single normal force.
3. Whereever have you gotten the notion that a=0??

 

[Doc Al]

1.The acceleration of the big block, A, is in the opposite direction of your drawing.

If I'm picturing the problem correctly, both blocks will have a horizontal acceleration to the right. (For the condition that m not to slide down M.) This is the direction that A shows in the drawing.

 

[arildno]

Oops; I didn't see the condition in the original problem that m should not slide relative to M..

 

[plusaf]

thank you!

Oops; I didn't see the condition in the original problem that m should not slide relative to M..

yes, it looked to me that, in the notes flying back and forth, that point was missed, and it's one key to a correct solution.

look at limit cases: if they're all frictionless surfaces, a low force (or acceleration) will let the small mass slide down the ramp. (best example: no acceleration of the large mass, at all!)
a very high acceleration (force) applied to the large mass will accelerate it very quickly, and the small mass will zip UP the ramp (and presumably, be left behind... )

next "easy-see" boundary conditions: if the ramp has zero slope, any acceleration at all to it will enable the small mass to go "up" the ramp. if the ramp has an angle of 90 degrees, it there is no acceleration that can possibly keep the smaller block from going "down the ramp".

the correct answer lies inbetween these corner cases.

calculate on, folks! i expect a correct answer to appear here, now, in a matter of days, if not minutes!

good thinking (and a good re-read of the original question, too!

 

[plusaf]

correct!

If I'm picturing the problem correctly, both blocks will have a horizontal acceleration to the right. (For the condition that m not to slide down M.) This is the direction that A shows in the drawing.

that's my take on it!

note, please, the phrase in the original question, which i think misled the original poster as well as many others: "(how?) Fast must the inclined plane be moving in order for the mass to stay exactly where it is."

at any constant speed, the mass must and will slide down the ramp! the only way to prevent that is to "accelerate it up the plane just as fast as it would have been accelerating down the plane" if the bigger block weren't moving! the two accelerations must cancel for the small block to not move.

 

[Doc Al]

calculate on, folks! i expect a correct answer to appear here, now, in a matter of days, if not minutes!

The problem remains as simple to solve as ever. As I pointed out in post #3 of this thread, the key is finding the acceleration of the sliding mass given the constraints of the forces on it. Once that's found, then you can find the force that needs to be applied to the incline to produce that acceleration. (I assume that the real problem is to find the applied force; certainly not the speed, as the original poster mistakenly wrote.)

Still waiting for the original poster to solve it for himself.