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Object B is on top of object A and there is no friction between the two. A horizontal force acts on object A, will the force also act on object B?

My intuition is that the force only acts on object A , and thus only object A will accelerate in the horizontal direction, and object B will remain in it's original position (not accelerated).

However, when approaching this question "show that tanθ=F/W where W is the combined weight of object A and B" we can treat the two objects as one and the force thus acts on both objects. The block B does not move relative the wedge. (Friction between the wedge and block or between the wedge and the ground is negligible).

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Can someone explain when can two objects be treated as one? In the question I proposed, I thought the two objects can't be treated as one when the force only acts on object A. Thanks for any help.

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  • $\begingroup$ While you are pushing the wedge, do you notice that the block is also sliding down the wedge due to gravity? $\endgroup$ – QuIcKmAtHs Aug 4 '18 at 13:48
  • $\begingroup$ Right the block does slide down due to earth's gravitational force on the block. And the block sliding down should be independent of the horizontal force F i believe (i.e F doesnt act on the block)? $\endgroup$ – Bøbby Leung Aug 4 '18 at 13:52
  • $\begingroup$ Suppose you are carrying a box. If you hold it normally, you're only touching the bottom half of the box, so when you move forward, why can you treat the box as one thing? Shouldn't the top half of the box just stay in its original position? The force is not acting on it. $\endgroup$ – knzhou Aug 4 '18 at 14:01
  • $\begingroup$ While we're at it, you're not really touching the bottom half of the box either. You're touching the atoms on the outside surface of the box. Shouldn't only those atoms move, while the entire rest of the box stays floating in place? $\endgroup$ – knzhou Aug 4 '18 at 14:01
  • $\begingroup$ Point is, treating "two objects" as one is completely normal and you do it constantly, every day. You can do it whenever the two objects move in the same way. $\endgroup$ – knzhou Aug 4 '18 at 14:02
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Rule-of-thumb: You can consider all objects as one, when they move together, or if their relative motions are not important or negligible.

  • All particles in your body are treated as one system (your "body), when details are not important.
  • All components and parts of a car, including passengers, are often treated as one system (the "car").
  • In your situation, if you know both blocks will move together with same accelerations, then they can be treated as one object.

Now, it is not always smart to treat them as one object. Sometimes it doesn't simplify your task, and other times it simplifies it too much. For example, if you need to find the friction between the blocks, then you can't treat both blocks as one system - then that friction force would be an internal force and would thus not appear in your Newton's equations.

In the question, you are correct when saying that a pushing force $F$ is only exerted on the incline and not on the box on top. That pushing force $F$ does not act on the box. But if you set up Newton's laws on the incline and on the box, then you will find, that the normal force between incline/box will have to take the same size (its horizontal component) as $F$ in order to push the top box along. So, although $F$ is only exerted on the incline, it propagates to the box by means of other forces. (In the same way you could say that $F$ actually only acts on the particles most to the left in the incline. Those particles then in turn push their neighbours, which push their neighbours and so on. The force $F$ propagates so that all particles feel the same force. And that is passed on to the box.

Now, if you knew before-hand that the box would not move together, then that would mean that the force propagates to the box is not equal to $F$. Similarly, if all particles would not move together in an object such as the incline, then that incline must be made of an elastic material (or is maybe cracked and broken into pieces or it is elastic). Then the force exerted from the left side is not propagated to the right side - like pushing on a soft pillow or a spring.

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  • $\begingroup$ Hi, thank you for the detailed answer. But the wedge and block scenario has no friction in between them to keep them together, so I'm not sure why does the force act on both the wedge and the block. $\endgroup$ – Bøbby Leung Aug 4 '18 at 15:42
  • $\begingroup$ @BøbbyLeung Oh, sorry for my simplified writing. Friction is only one force that could be present with a horizontal force-component. Another force that definitely is present is the normal force, which also has a horizontal component. $\endgroup$ – Steeven Aug 4 '18 at 15:57
  • $\begingroup$ Oh, so is it because the normal force keeps the two object together intact, so the two objects can be treated as one? $\endgroup$ – Bøbby Leung Aug 4 '18 at 16:01
  • $\begingroup$ Right, but how can the normal force help with propagating the applied force F like how friction would propagate? $\endgroup$ – Bøbby Leung Aug 4 '18 at 16:10
  • $\begingroup$ @BøbbyLeung The normal force has a horizontal component. Just like friction could have. This horizontal component is causing the horizontal acceleration of the top box, so that it too can accelerate like the bottom box. $\endgroup$ – Steeven Aug 4 '18 at 18:04
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When two or more objects are treated as one, we are concerned with the movement of the centre of mass of the closed system( i.e the objects being treated as one). As we do all the time like when we are doing calculations concerned with movement of cars. The car may comprise of a system of people and its components and to make life easier we consider the motion of the car as a whole .

In the case of the wedge, notice that if there was no force acting on the wedge, both objects would accelerate horizontally due to the horizontal components of their normal contact forces.

But this is not the case in this situation, a force is being applied on the wedge. This force will counteract the horizontal component of the normal contact force on it due to the block and produce a resultant forward movement. On top of that the horizontal forces now acting on the block are F and the horizontal component of the normal contact force on he block. Therefore the horizontal component of the velocity of the block is greater than that of wedge relative to the ground. I hope this was useful.

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in general you can treat two objects as one in the same matter as you would do in real life. So if they are glued or screwed together there are one, if they are just lie on each other there are not.

About the answer to your problem:

To begin with, your text and the drawing describe different situations. I think you mixed up A and B in the second sentence. Following I will discuss the case described in the picture: Square B is on wedge A wile a force $F$ is applied on the wedge A.

If the force F is applied on the wedge A, the wedge will apply a force on the square B. Since there is no friction this force can only be orthogonal to the surface between the two objects. This force is called the normal force $F_n$. Via a parallelogram of forces (PoF) $F_n$ can be split in two other forces, one perpendicular $F_{n\perp}$ and one parallel $F_{n\parallel}$ to the surface. Additional, a gravitational force $F_g$ perpendicular to the surface, on which the wedge lie's, is assumed.

In sum, there are now three forces which apply on the cube B. Two perpendicular to the surface ( $F_{n\perp}$ + $F_g$ which can be summed up) and one parallel $F_{n\parallel}$ to it.

There are three possible cases to come up with as a result:

  1. If the gravity is stronger then the upward force, the cube will slide of the wedge. (With additional PoF $F_g$ and $F_{n\parallel}$ can both be transformed in a normal force and a force parallel to the upside of the wedge to find the resulting relevant forces)
  2. If the gravity and the upward force are canceling each other out, the cube will stay were it is.(Here $F_{n\parallel}$ will accelerate the cube equally, as $F$ accelerates the wedge)
  3. If the normal force exceeds the gravitational force, the cube will slide up and fall eventually off the wedge. (The cube will still be accelerated in the direction of $F$. Not as much as the wedge though)

As you can see, there will be a resulting force on the cube B, but it may not be the same force F that has been applied to the wedge A.

I hope that helps and is not to much.

If there are any mistakes, please let me know.

Cheers

Kai

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  • $\begingroup$ Thanks for correcting my typo. But for the scenario 2 you mentioned, shouldn't it be "Fn∥+g∥ accelerate the cube equally, as F accelerates the wedge"? $\endgroup$ – Bøbby Leung Aug 4 '18 at 16:06
  • $\begingroup$ sorry I mean mg∥, where m is the mass of the block. $\endgroup$ – Bøbby Leung Aug 4 '18 at 16:15
  • $\begingroup$ In scenario 2 the entire gravitational force cancels out with $F_{n\perp}$, so that no force accelerate the cube perpendicular to the surface (therefore it doesn't move up or down). The only remaining force is $F_{n\parallel}$. ($mg_\parallel = 0$ should apply anyway, since the gravitational force is assumed to be perpendicular to the ground, unless you mean parallel in respect to the surface of the wedge.) $\endgroup$ – KaiK Aug 4 '18 at 18:05
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Based on your clarification that block B is not to move relative to block A, both blocks can be treated as one per Steeven’s rule of thumb. In this regard, the equation given describes the required relationship between F and the total weight W for there to be no relative motion between the blocks. To develop the formula consider the following.

The acceleration of Block B down the incline due to gravity is

$g sin \theta$

Now the force F that accelerates the combined masses along the horizontal surface (x direction) is given by

$F=(M_A+M_B) a_x$

giving an acceleration in the x direction of

$a_x=F/(M_A+M_B)$

The component of that acceleration down along the incline is give by

$ Fcos\theta/(M_A+M_B)$

For there to be no relative motion along the incline, the two accelerations down the incline have to be equal, thus

$Fcos\theta/(M_A+M_B) = g sin\theta$

$Fcos\theta=(M_A+M_B)g sin\theta$

$Fcos\theta=(W_A+W_B)sin\theta$

$F/(W_A+W_B)= tan\theta$

$F/W = tan\theta$

($W= W_A+W_B$)

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