How do you calculate the acceleration of a block sliding down a plane?

[tryingtolearn1]

Homework Statement

A block starts at rest and slides down a frictionless plane inclined at an angle $\theta$. What should $\theta$ be so that the block travels a given horizontal distance in the minimum amount of time?

Relevant Equations

$$F=ma$$

When drawing a diagram of the forces acting on the block, I have the following forces: $$\sum F_x =mg\sin\theta = ma$$ $$g\sin\theta = a$$ however the solution has $$F_x = ax = (g \sin\theta) \cos \theta$$ but I am not sure how they got that? I know the normal force is $$N=mg\cos\theta$$ but the normal acts on the y-axis.

 

[haruspex]

Homework Statement:: A block starts at rest and slides down a frictionless plane inclined at an angle $\theta$. What should $\theta$ be so that the block travels a given horizontal distance in the minimum amount of time?
Relevant Equations:: $$F=ma$$

When drawing a diagram of the forces acting on the block, I have the following forces: $$\sum F_x =mg\sin\theta = ma$$ $$g\sin\theta = a$$ however the solution has $$F_x = ax = (g \sin\theta) \cos \theta$$ but I am not sure how they got that? I know the normal force is $$N=mg\cos\theta$$ but the normal acts on the y-axis.

Are you using the same XY axes as in the solution?

 

[tryingtolearn1]

Are you using the same XY axes as in the solution?

yes, I am using the x-y plane to determine the acceleration. The book states the following:

The component of gravity along the plane is $$g \sin\theta$$. The acceleration in the horizontal direction is therefore $$a_x = (g \sin\theta) \cos\theta$$

 

[haruspex]

yes, I am using the x-y plane to determine the acceleration. The book states the following:

That's not what I asked. My question is whether your XY axes are the same as in the official solution.
Looks to me that the book takes X as horizontal and Y as vertical , whereas you are taking X as parallel to the plane and Y as normal to the plane. That would explain the differences in the algebra.

 

[tryingtolearn1]

That's not what I asked. My question is whether your XY axes are the same as in the official solution.
Looks to me that the book takes X as horizontal and Y as vertical , whereas you are taking X as parallel to the plane and Y as normal to the plane. That would explain the differences in the algebra.

The solution doesn't provide a diagram so I am not sure what XY axis my books takes. Here is the diagram I drew that I assumed is what the book claimed:

 

[haruspex]

The solution doesn't provide a diagram so I am not sure what XY axis my books takes. Here is the diagram I drew that I assumed is what the book claimed:

So rework it using X as horizontal and Y as vertical and see whether you get the same as the book.

 

[tryingtolearn1]

So rework it using X as horizontal and Y as vertical and see whether you get the same as the book.

Hm, I don't understand? Isn't this problem exactly similar to finding the acceleration of a block sliding down on a frictionless inclined plane thus the acceleration is $$a=g\sin\theta?$$
Which can be found here
http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html

 

[Hamiltonian]

Homework Statement:: A block starts at rest and slides down a frictionless plane inclined at an angle $\theta$. What should $\theta$ be so that the block travels a given horizontal distance in the minimum amount of time?
Relevant Equations:: $$F=ma$$

When drawing a diagram of the forces acting on the block, I have the following forces: $$\sum F_x =mg\sin\theta = ma$$ $$g\sin\theta = a$$ however the solution has $$F_x = ax = (g \sin\theta) \cos \theta$$ but I am not sure how they got that? I know the normal force is $$N=mg\cos\theta$$ but the normal acts on the y-axis.

gsin$\theta$ is the acceleration along the incline.
then what will be the acceleration of the block in the horizontal direction(parallel to the base of the triangle formed by the inclined plane)?

 

[tryingtolearn1]

gsin$\theta$ is the acceleration along the incline.
then what will be the acceleration of the block in the horizontal direction(parallel to the base of the triangle formed by the inclined plane)?

Oh, hmm I kind of understand now but what does it mean to travel a horizontal direction when the plane is inclined?

 

[Hamiltonian]

Oh, hmm I kind of understand now but what does it mean to travel a horizontal direction when the plane is inclined?

just imagine the motion of the block in the x-direction irrespective of its motion in the y .

 

[haruspex]

Oh, hmm I kind of understand now but what does it mean to travel a horizontal direction when the plane is inclined?

JUst as you can resolve a force into horizontal and vertical components, you can do the same with accelerations, velocities and displacements.

 

[tryingtolearn1]

JUst as you can resolve a force into horizontal and vertical components, you can do the same with accelerations, velocities and displacements.

Yes, that I understand but what does it mean physically for the block to travel a horizontal direction? The block is inclined so the only direction it can physically travel is in the direction of the incline so why is the question asking to find the horizontal direction acceleration when the incline acceleration provides a better approximation?

 

[Hamiltonian]

Yes, that I understand but what does it mean physically for the block to travel a horizontal direction? The block is inclined so the only direction it can physically travel is in the direction of the incline so why is the question asking to find the horizontal direction acceleration when the incline acceleration provides a better approximation?

because the question is only concerned with the distance traveled by the block in the horizontal direction(which is not along the incline).

 

[haruspex]

Yes, that I understand but what does it mean physically for the block to travel a horizontal direction? The block is inclined so the only direction it can physically travel is in the direction of the incline so why is the question asking to find the horizontal direction acceleration when the incline acceleration provides a better approximation?

If you move distance s up a slope at θ to the horizontal then you have moved s cos(θ) in the horizontal direction and s sin(θ) in the vertical direction. Your total movement is the sum of the two.

 

[tryingtolearn1]

because the question is only concerned with the distance traveled by the block in the horizontal direction(which is not along the incline).

Oh I see. I will take the question as is then.