Solving inclined plane in different axis.

In summary, the conversation discusses how to calculate the acceleration of a block on an inclined plane using different coordinate axes. The person mentions making a mistake when using horizontal and vertical axes and asks for help understanding why the vertical forces do not add up to zero. The response explains that the acceleration of the block is parallel to the incline, making it impossible for the vertical forces to sum to zero. It is suggested that using coordinates parallel and perpendicular to the surface is easier, but it is possible to solve the problem using vertical and horizontal axes with the added constraint of acceleration parallel to the surface. The conversation concludes with the person successfully solving the problem using this method.
  • #1
tousif1995
2
0
I can work out the acceleration of a block that will take along the plane when the axes are the ones along and perpendicular to the inclined plane. But, when I choose a horizontal and vertical axis I end up with the same mistake as this guy in this thread did:

https://www.physicsforums.com/showthread.php?t=238180

In that thread, it is said that the vertical forces don't add up to zero. I don't get it and hence I cannot solve the problem when the axis is horizontal and vertical. Can anyone help?
 
Physics news on Phys.org
  • #2
tousif1995 said:
In that thread, it is said that the vertical forces don't add up to zero. I don't get it and hence I cannot solve the problem when the axis is horizontal and vertical. Can anyone help?
The acceleration of the block is parallel to the incline, so it will have a vertical component. So the vertical forces on the block cannot sum to zero.

Obviously, it's much easier to solve the problem using coordinates aligned parallel and perpendicular to the surface. But if you insist, you can solve it using vertical and horizontal axes. It just takes a bit more work, since you must add the constraint that the acceleration is parallel to the surface.
 
  • #3
Thank you!

masinθ = -(Rcosθ-mg)
Rsin = macosθ

Just solved it! feels good :cool:
 

Related to Solving inclined plane in different axis.

1. How do you calculate the force needed to move an object up an inclined plane?

The force needed to move an object up an inclined plane can be calculated using the formula F = mg(sinθ + μcosθ), where F is the force, m is the mass of the object, g is the acceleration due to gravity, θ is the angle of the inclined plane, and μ is the coefficient of friction.

2. What is the advantage of using an inclined plane in different axis?

Using an inclined plane in different axis allows for the force needed to move an object to be spread out over a longer distance, making it easier to move the object. This is particularly useful when moving heavy objects.

3. How does the angle of the inclined plane affect the force needed to move an object?

The steeper the angle of the inclined plane, the greater the force needed to move an object. This is because the component of the force acting against gravity increases as the angle increases.

4. What is the relationship between the length of the inclined plane and the force needed to move an object?

The longer the inclined plane, the less force is needed to move an object. This is because the force is spread out over a longer distance, making it easier to move the object.

5. How does friction impact the movement of an object on an inclined plane?

Friction acts in the opposite direction of motion, making it harder for an object to move up an inclined plane. The coefficient of friction between the object and the surface of the inclined plane will determine the amount of force needed to overcome friction and move the object up the incline.

Similar threads

Replies
40
Views
3K
Replies
13
Views
2K
Replies
9
Views
1K
Replies
1
Views
1K
Replies
10
Views
2K
Replies
3
Views
2K
Replies
13
Views
1K
  • Classical Physics
Replies
21
Views
2K
Back
Top