# B How can gravity have a horizontal component?

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1. Apr 28, 2016

### EddiePhys

In the above case, mgcos(theta) is the component of gravity perpendicular to the surface of the wedge. Considering all surfaces to be frictionless, the horizontal component of the normal force would cause the wedge to accelerate towards the right, and the horizontal component mgcos(theta)-the horizontal component of the normal force on the block would be the net horizontal force on the block causing the block to move along with the wedge to the right. How is this possible? How can gravity cause a net horizontal acceleration?

2. Apr 28, 2016

### nasu

It does not have one.
You are missing some components in your diagram. When you consider the missing ones, you will see that gravity has no horizontal component.
The normal force (at the interface bloch-wedge) does have a non-zero horizontal component but this is a different force, is not gravity.

3. Apr 28, 2016

### EddiePhys

But isn't the normal force in this case caused due to gravity? And the horizontal component mgcos(theta)-the horizontal component of the normal force on the block would be the net horizontal force on the block causing the block to move along with the wedge to the right. In this case wouldn't it be gravity that's causing a net horizontal acceleration?

4. Apr 28, 2016

### rcgldr

The normal force opposes a component of gravity, m g cos(θ). If the blocks slides, the horizontal component of acceleration is due to the horizontal component of the normal force, gravity only exerts a vertical and downwards force on the block.

5. Apr 29, 2016

### EddiePhys

Alright, that explains why the wedge is moving to the right. But what causes the block to move along with the wedge to the right? My speculation was that horizontal component mgcos(theta)-the horizontal component of the normal force on the block would be the net horizontal force on the block causing the block to move along with the wedge to the right. If this is true, then wouldn't gravity be responsible for causing a net rightward acceleration?

6. Apr 29, 2016

### rcgldr

Why do you assume the block moves to the right? Linear momentum is conserved, even with the external force of gravity, since the force is perpendicular to the direction of motion. The wedge moves to the right, the block moves to the left, and the center of mass of wedge and block only moves downwards, not horizontally.

7. Apr 29, 2016

### mandyscott

This is a complicated question. I also need the answer.

8. Apr 29, 2016

### FactChecker

The force of gravity is only down. Simple machines, like a wedge or a lever, can redirect the energy. When gravity pulls one end of a see-saw down, the other end goes up. That does not mean that the gravity has an upward component. As others have said, the simple machines are supplying forces that were not components of gravitational force.

9. Apr 29, 2016

### PeroK

Gravity, like any force, is a vector. You can resolve a vector into components any way you like. For example:

$(0, -1) = (1, 0) + (-1, -1)$

That resolves a "downward" vector into a "horizontal" vector plus a "45-degree" vector. You can view gravity as having a horizontal component, but it will have a second component with an equal and opposite horizontal force.

This might sound odd until you think of what you do with the block and wedge problem: you resolve gravity into a component normal to the surface and a component tangential to the surface.

You could equally argue that gravity doesn't have either of these components, both of which include some force in the horizontal direction. Gravity is not pushing tangentially down the slope: it's always pushing vertically.

But, I think, this is an insight into the vector nature of force: you really can resolve gravity into these two components. One of the components (the normal force) is cancelled out, leaving only the unopposed tangential component.

Now, you could view the horizontal force here as really originating from the wedge (not from gravity). If you look at the force vectors differently, you could have gravity as a single vertical force, resolve the normal force from the wedge into a vertical component and a horizontal component: this leaves a nett (reduced) vertical component due to gravity and a horizontal component due to the wedge. That would avoid resolving the gravity vector. The block moves down due to what's left of gravity and across due to the horizontal component of the normal force from the wedge.

But, I think this misses the point about the true vector nature of forces. Gravity really is acting in the normal and tangential directions simultaneously; it really is a vector with vector properties.

10. Apr 29, 2016

### EddiePhys

Ah, I see. Makes sense now.

11. Apr 29, 2016

### EddiePhys

12. Apr 29, 2016

### Gordon Mutten

Its also worth noting that you need to know the mass of the wedge to calculate the horizontal (and vertical) accelerations of the wedge and block. Mass times acceleration of one is equal and opposite to mass times acceleration of the other.

13. Apr 29, 2016

### EddiePhys

Very interesting, yet a little confusing how this works. Do you know of any book/online resource that discusses this in depth?

14. Apr 29, 2016

### PeroK

Any book on vectors will cover what you need. And physics books will use the properties of vectors where needed. Vectors are just about the most useful thing in applied maths and physics.

It's not really an "in-depth" thing. This simple example of resolving gravity into a component normal to a slope and a component down a slope is, I would say, just about the whole subject.

15. Apr 29, 2016

### Merlin3189

Glad you got it sorted. Just like to add a slightly dissident view.
First, that the horizontal forces arise from the block pushing the wedge and the wedge pushing the block.
Second, I don't think you can resolve a vertical force into a partially horizontal force. You can resolve it into two or more forces so long as the net horizontal force is zero and the net vertical force is equal to the original force. So here you can resolve gravity into the sum of a force along the slope and a force perpendicular to the slope (or in any other directions you like), but both must be present and their horizontal effects must be equal and opposite.

The original diagram failed because it showed a net horizontal force due to gravity.