# Acceleration of wedge due to normal force exerted by block?

1. Jan 2, 2016

### ajay.05

This isn't actually a problem. It is a doubt which I have about wedges according to Newton's Laws

Consider a friction less wedge, of mass 'M', inclined at angle θ, placed on a smooth horizontal surface, on which a block of mass 'm' is placed. The block exerts some normal force perpendicular to the surface of the wedge, which can be resolved into two axes such that one is parallel to the ground. Now, after resolving we will have an horizontal component of force F=mgsinθ, which makes the wedge to move in that direction with an acceleration of a=mgsinθ/M. And since the wedge is friction less, the block moves down too.
This is what I've been taught by my teacher.

And My doubt here is that, the normal force is actually due to gravity(in this case), which has a vertical component of mg, then if we resolve the normal force according to the above method, then we will only have null vector in the horizontal direction, that is mgsinθ=0. Then, how does the wedge move?(I can certainly understand that if we actually apply a force on the block, besides gravity, then we may induce a force with a horizontal component, in which case the block will move)

In a nutshell, if the normal force itself is caused by only gravity, which has only a vertical component, then how is it possible to have a horizontal component, and alas, if there isn't no horizontal component, why should the wedge move?

2. Jan 2, 2016

### PeroK

First, do you doubt that if you push directly down on a wedge, then it will move to one side? Alternatively, consider a ball on an incline. Gravity is straight down, yet the ball will roll down and along the incline. The normal force from the surface turns some of the gravitational PE released into horzizontal motion. Another example would be a ski jumper. Think of how gravity creates largely horizontal motion in that case.

Second, this is an example of the vector nature of force. A downward force can be resolved into two angled components, one down and to the right and one down and to the left. Both components have a horizontal element.

3. Jan 2, 2016

### ajay.05

Yes, Of course I could understand your reply, and I don't argue that why should the block move along the incline, all I doubt is, will the wedge move in the opposite direction, due to the motion of the ball(if gravity is the only force acting on the system)?

4. Jan 2, 2016

### PeroK

If the wedge is able to slide along the surface, then a ball sliding down the wedge would push the wedge to one side as it moves. Note that in such a case the total horizontal momentum is conserved (assuming no friction between the wedge and surface).

5. Jan 2, 2016

### ajay.05

Thubs Up...I got the point now:)

6. Jan 2, 2016

### Mister T

Gravity is not the cause of the normal force. The normal force is caused by the deformation of the surface.

In fact, what is happening is that there is a contact force which can be resolved into two components, one parallel to the surfaces and the other perpendicular. The parallel component is called the friction force, and the perpendicular component is called the normal force.

7. Jan 3, 2016

### ajay.05

Actually, what I thought was normal force was caused due to gravity(when no other force acts on the system),So, if we consider this set-up in space, will there be a normal force? Also in this space set-up the gravitational attraction between the two particles will be negligible...

8. Jan 3, 2016

### Staff: Mentor

That's not right. Two ideal objects whose surfaces cannot deform will still exhibit normal forces at their interface. In this instance the normal force is a result of a component of the weight of the object, via Newton's third law.
No. The weight vector of the upper object can be resolved into two components as you say, but the downslope component is not the friction force; it's simply the component of the weight that is directed downslope. Friction is a separate force altogether that depends upon the "texture" (and other factors) of the mating surfaces and its magnitude is proportional to that of the normal force. Furthermore its direction, while parallel to the surface, depends upon the relative direction of motion of the surfaces; friction always opposes the relative motion.

9. Jan 3, 2016

### ajay.05

Yup, Exactly what I had in my mind...

10. Jan 3, 2016

### Mister T

This is the limit of high elasticity, the amount of deformation approaches zero as the amount of force increases beyond all bounds. Think about the spring force. It's the same concept.

If you have a block on a table (or ramp) there are two Third Law pairs. Earth exerts a gravitational force on the block, the block exerts a gravitational force on Earth. The block exerts a contact force on the table, the table exerts a contact force on the block.

That contact force exerted on the block by the table can be resolved into two components. The component perpendicular to the surface is called the normal force, the component parallel to the surface is called the friction force.

It's a result of the contact between the surfaces, as is the normal force. The bit about the friction force being proportional to the normal force is an approximation, and not a very good one.

11. Jan 3, 2016

### Staff: Mentor

It's also irrelevant here. Here the normal force depends upon the weight of the block, and more specifically, the component of the weight that is perpendicular to the shared surfaces.
No. That is misinformation. (in fact in this case the surfaces are frictionless, yet clearly there is a force acting to move the block downslope).

Define "contact force" and how it is calculated. Show how a normal contact force can have a component parallel to the surface. How can this contact force differ for objects with the same mass and geometry but composed of different materials?

I think your concepts are not helping the OP.

12. Jan 3, 2016

### Mister T

There is no force acting in that direction. The only forces acting on the block are the downward force of gravity and the contact force. Neither of those are directed down the slope.

It's a force exerted on one object by another as a result of their contact. In this example it's equal to the normal force, which in turn is equal to $mg \cos \theta$.

The normal force is the perpendicular component of the contact force. By definition it can't have a "component" parallel to the surface.

If you replaced the block in question with one of the same mass and geometry but composed of a different material, or even if you didn't, and then pressed on it with your hand, the contact force exerted on the block by the surface would be different because the deformation of the surfaces would be different.

13. Jan 3, 2016

### Staff: Mentor

Then you are claiming that a mass can accelerate in a direction without a force component acting in that direction.
Which implies that it is due to a component of the force due to gravity, contrary to what you previously claimed.
True.
So once again, how do you calculate this "contact force". Is there a useful way to do so in general that would lend itself to the solution of the OP's problem? How does it differ from simply calculating the relevant components of the gravitational force and determining the friction via the normal component and the coefficient of friction?[/quote]

14. Jan 3, 2016

### Mister T

There's a component in that direction, but the force itself doesn't point in that direction.

No, it's equal to $mg \cos \theta$ but it is not due to gravity, per se. It's due to a deformation of the surface.

No, the calculation of the value is not any different. All I'm saying is that we usually think of the friction force and the normal force as separate forces. But if you look at the description of those forces you realize that they are in a sense the same. They are both forces exerted on an object by a surface. For example, a block on a surface of a wooden board. The surface of the board exerts a force on the book. That force does not, in general, act in a direction perpendicular to the surface. It is conventional to refer to the parallel component as a friction force and the perpendicular component as the normal force. I can supply references if you're interested in looking at them.

There are two points that were raised in this thread that I disagree with. I know they are a source of confusion for students and I believe they are a source of confusion for the OP in this particular case.

1. The normal force is not related to the weight of the object via Newton's Third Law. Newton's Third Law is about a relationship between a pair of forces that act on different objects. The normal force and the weight force in this example are both acting on the same object. They are not a Third Law pair (an action-reaction pair) of forces. Again, I can supply references if you're interested in looking at them.

2. The normal force is not caused by the weight. If the weight of an object is 10 N, and you place it undisturbed on the horizontal surface of a wooden board, that board exerts a force on the object of magnitude 10 N. If you orient that surface so that it's vertical, place the same object in contact with it, and push on the block with a force of 10 N towards the board, you will again have the surface pushing on the object with a force of magnitude 10 N. In both cases the cause of the force is the same. It's the deformation of the surface. Again, if you are interested, I can supply references.

I have been teaching introductory physics at the college level as a full-time profession at three different colleges in succession continuously from 1986 to the present. None of this stuff is new, it's been around that entire time and it's a documented source of confusion for students. One example is the original post in this thread.