Newton's Third Law on a Inclined Plane

[Zalajbeg]

Hello everyone. I think I have almost understood the third one of Newton's motion laws but I have some doubts:

Let us think an inclined plane block located on a surface, there is a box on it and assume there is no friction neither between the surface and block nor between the block and the box. The mass of the box is m, the mass of the inclined plane is M and the acceleration of gravity is g and the angle of the inclined plane is a.

And elementary vectors:

[j]: parallel to the gravity.
: perpendicular to [j]

[x]: parallel to the upper surface of the block
[y]: perpendicular to [y]

I have three scenarios but all of them have a paradox.

1)

The forces asserted to the block:

By box: m.g.[j]
By gravity: M.g.[j]
By the surface: -(M+m).g.[j]

The forces asserted to the box

By the block: -m.g.[j]
By the gravity: m.g.[j]

The weakness of this scenario is that; in this case the block shouldn't accelerate but we know it accelerates because there is no friction.

2)

The forces asserted to the block:

By box: m.g.[j]
By gravity: M.g.[j]
By the surface: -(M+m).g.[j]

The forces asserted to the box

By the block: -m.g.cosa.[y]
By the gravity: m.g.[j]= m.g.cosa.[y]+m.g.sina.[x]

The weakness of this scenario: This one explains the acceleration of the box but the forces asserted by the block and the box to each other don't have the same magnitude and it is against Newton's third law.

3)

The forces asserted to the block:

By box: m.g.cosa.[y]
By gravity: M.g.[j]
By the surface: -M.g.[j] + (-m.g.cosa.[y])

The forces asserted to the box

By the block: -m.g.cosa.[y]
By the gravity: m.g.cosa.[j]= m.g.cosa.[y]+m.g.sina.[x]

The weakness of the scenario: If we think the box and the block as a one body we can see the force asserted to the block by the surface is -(M+m).g.[j]Then what am I missing?

 

[tiny-tim]

Welcome to PF!

Hello Zalajbeg! Welcome to PF!

(btw, forces are "exerted on", not "asserted to" )

Your 3) is the closest, so I'll only comment on 3) …

3)

The forces asserted to the block:

By gravity: M.g.[j]

Yes.

By the surface: -M.g.[j] + (-m.g.cosa.[y])

No, the reaction force from the surface (which I assume is horizontal) is perpendicular to the surface (because the friction is zero) … it cannot have any component.

The only way to find any reaction force is by using Newton's second law …

in this case, the reaction force is vertical, and the vertical acceleration of the block is zero, so the reaction force must be minus the sum of the other two forces on the block.

The forces asserted to the box

By the block: -m.g.cosa.[y]

That would be correct if the block was not accelerating. But it is accelerating, so the acceleration of the box is not parallel to [x], and does not have zero [y] component.

By the gravity: m.g.cosa.[j]= m.g.cosa.[y]+m.g.sina.[x]

No, gravity is always mg …

why did you think cosa should be included?

 

[Zalajbeg]

I put the cosa wrongly to my last statement while copy-pasting Sorry.

Then I wonder this. What is the force exerted on the block by box. Isn't it m.g.[j] ? And according to Newton's third law, shouldn't the force exerted on the box by the block be -m.g.[j]. However in this case acceleration of the box would be zero. I don't know what I am missing.

 

[rcgldr]

I don't know how to solve this problem but to get things started:

Define:

m = mass of box
M = mass of plane
a = acceleration of box (downwards and right considered positive)
A = acceleration of plane (left considered negative)

Assume plane slopes down and to the right, with angle θ from horizontal.

Using coordinates that are perpendicular (x axis) and parallel (y-axis) to gravity

There is a Newton third law pair of horizontal forces that result in horizontal accelerations of plane and box:

M A_x = -m a_x

The vertical acceleration of the box depends on how fast the plane and box accelerate away from each other horizontally times sin(θ):

a_y = sin(θ) (a_x - A_x)

The vertical force the box applies to the plane = m (g - a_y) and is part of a Newton third law pair of forces downwards from box onto plane, upwards from plane onto box.

The total vertical force on the box = F_y = m g - m (g - a_y) = m a_y.

 

[tiny-tim]

Hi Zalajbeg!

What is the force exerted on the block by box. Isn't it m.g.[j] ? And according to Newton's third law, shouldn't the force exerted on the box by the block be -m.g.[j] …

Weight is always mg. Kinetic friction is always µN. And some applied forces are given in the question.

But the only way to find any reaction force between two surfaces is by using Newton's second law …

add up all the other forces, and multiply by -1.​

 

[Doc Al]

The forces asserted to the block:

By box: m.g.[j]

The force mg acts on the box, not on the block. It is the gravitational force of the Earth acting on the box. You cannot assume that the force the box exerts on the block equals the weight of the box.

 

[Zalajbeg]

You cannot assume that the force the box exerts on the block equals the weight of the box.

I think I can. Because the block does not have an acceleration on the y coordinates (parallel to gravity) Then the reaction force in this direction, exerted on the block by the surface is (-M+m).g.[j].

The other forces in this direction exerted on the block are the weight and the force exerted by the box (let us call it Fb)

Then

(-M+m).g.[j]+M.g.[j]+Fb=0

Hence;

Fb=m.g.[j]

But if you have an opposite idea I surely would love to read it.

 

[Zalajbeg]

I don't know how to solve this problem but to get things started:

Define:

m = mass of box
M = mass of plane
a = acceleration of box (downwards and right considered positive)
A = acceleration of plane (left considered negative)

Assume plane slopes down and to the right, with angle θ from horizontal.

Using coordinates that are perpendicular (x axis) and parallel (y-axis) to gravity

There is a Newton third law pair of horizontal forces that result in horizontal accelerations of plane and box:

M A_x = -m a_x

The vertical acceleration of the box depends on how fast the plane and box accelerate away from each other horizontally times sin(θ):

a_y = sin(θ) (a_x - A_x)

The vertical force the box applies to the plane = m (g - a_y) and is part of a Newton third law pair of forces downwards from box onto plane, upwards from plane onto box.

The total vertical force on the box = F_y = m g - m (g - a_y) = m a_y.

Then what is the vertical force applied on the block? M.g+m (g - a_y) ?

 

[rcgldr]

Then what is the vertical force applied on the block? M.g+m (g - a_y) ?

zero, any downwards force on the block is equally opposed by an upwards force from the surface the block slides on. Only the box has a net vertical force and a vertical component of acceleration (a_y).

I've looked at previous threads here, and it seems that no one has actually solved this problem.

 

[Doc Al]

I think I can. Because the block does not have an acceleration on the y coordinates (parallel to gravity)

While the block has no vertical component of acceleration, the box does. All you can conclude from this is that the net vertical force on the block must be zero.

Then the reaction force in this direction, exerted on the block by the surface is (-M+m).g.[j].

This is an invalid conclusion.

The other forces in this direction exerted on the block are the weight and the force exerted by the box (let us call it Fb)

Then

(-M+m).g.[j]+M.g.[j]+Fb=0

Hence;

Fb=m.g.[j]

But if you have an opposite idea I surely would love to read it.

Two points:
(1) If the force exerted by the box equaled the weight of the box, the box wouldn't slide down the incline--it would be in equilibrium.
(2) Since the contact between box and block is frictionless, the only force they can exert on each other must be normal to the surface.

 

[rcgldr]

Since the contact between box and block is frictionless, the only force they can exert on each other must be normal to the surface.

True, but this problem is complicated by the fact that the block/plane is accelerating. In the extreme case that the block has much lower mass then the box, then almost all of the horizontal acceleration occurs at the block and the path of the box is much steeper than the angle of the plane. The path of the box is always steeper than the angle of the plane unless the plane isn't moving.

I did a web search, but still didn't find a solution to this problem.

 

[Doc Al]

This problem has been discussed on PF many times.

Here's how I like to approach it. Call the acceleration of the wedge a_w and the acceleration of the block with respect to the wedge a. (a_w will be horizontal and a will be parallel to the wedge.)

Now apply Newton's 2nd law to the block in the accelerating frame of the wedge. That gives you one equation.

Then take advantage of the fact that the only horizontal forces on each are due to the normal force between them, so Ma_w = ma_h. (a_h is the horizontal component of the block's acceleration with respect to the ground, which is easily expressed in terms of a_w and a.) That's your second equation.

That's all you need. (That's not the only way to solve it, but it's one of the easier ways.)

 

[Zalajbeg]

While the block has no vertical component of acceleration, the box does. All you can conclude from this is that the net vertical force on the block must be zero.

This is an invalid conclusion.


Two points:
(1) If the force exerted by the box equaled the weight of the box, the box wouldn't slide down the incline--it would be in equilibrium.
(2) Since the contact between box and block is frictionless, the only force they can exert on each other must be normal to the surface.

Yes. I thought on this matter and I think I have finally got it. I missed the fact that the acceleration of the system (block+box) is not zero.

Thanks to everyone who has joined the discussion. Now I am capable of drawing free body diagrams for both the box and the block.

 

[rcgldr]

This problem has been discussed on PF many times.

Here's how I like to approach it. Call the acceleration of the wedge a_w and the acceleration of the block with respect to the wedge a. (a_w will be horizontal and a will be parallel to the wedge.)

Now apply Newton's 2nd law to the block in the accelerating frame of the wedge. That gives you one equation.

Then take advantage of the fact that the only horizontal forces on each are due to the normal force between them, so Ma_w = ma_h.

The issue with this approach is you have a ficticious force on the box since it's in an accelerating frame of reference. For example, if the upper surface of the block were horizontal, then there can be no horiztonal force between the frictionless block and box, but the box experiences a ficticious horizontal force due to the accelerating frame of reference.

 

[Doc Al]

The issue with this approach is you have a ficticious force on the box since it's in an accelerating frame of reference. For example, if the upper surface of the block were horizontal, then there can be no horiztonal force between the frictionless block and box, but the box experiences a ficticious horizontal force due to the accelerating frame of reference.

Sure, whenever you use an accelerating frame you'll have to add inertial forces. So?

And if the wedge surface were horizontal instead of angled, its acceleration would be zero so the fictitious force would end up being zero.

(That's not the only way of solving this, of course. You can always just stick to the inertial frame, but then you'll have to add the constraint that the block maintains contact with the wedge.)

 

[rcgldr]

Sure, whenever you use an accelerating frame you'll have to add inertial forces. And if the wedge surface were horizontal instead of angled, its acceleration would be zero so the fictitious force would end up being zero.

I was considering other scenarios that go along with this class of problems, such as what acceleration of the block is required to prevent any vertical acceleration of the box. Anyway, I've seen attempts to solve this problem, but none of them seemed to have final solutions. I think I could follow the approach I used above and end up with expression for box accelerations a_x and a_y in terms of the block's upper surface angle and ratio of mass box versus block, but I'm assuming the OP is supposed to figure this out.