Understanding Forces and Acceleration on a Sliding Object on an Incline

[Murilo T]

Homework Statement

A block is placed on a plane inclined at an angle θ. The coefficient of friction between the block and the plane is µ = tan θ. The block initially moves horizontally along the plane at a speed V . In the long-time limit, what is the speed of the block?

Relevant Equations

ms˙ϕ˙ = mg sin θ cos ϕ

This question is from the David Morin ( Classical Mechanics ) - problem 3.7. I spent some time trying to figure it out the solution by myself, but since I couldn't, I looked into the solution in the book, but I got even more lost. So I searched for an online solution that could help me at least visualize the problem, and I found two solutions here, one here, and some one that already asked the same question here. There isn't any pictures of the problem anywhere, so this is my interpretation:

I get that the forces down the plane cancels out, and the equation for the acceleration in the tangential direction: m(ds^2/dt^2) = mg sinθ (sinϕ − 1).
But in the equation for the acceleration in the normal direction: m(ds/dt)(dϕ/dt) = mg sinθcosϕ, why the normal acceleration is (ds/dt)(dϕ/dt)?
And why the acceleration in the y-axis equals the negative acceleration in the tangetial direction?

 

[etotheipi]

I get that the forces down the plane cancels out

Remember that this occurs only in the large-$t$ regime; at the beginning, the $y$-components of the two forces do not cancel out, otherwise, $a_y$ would always be zero!

Or perhaps you mean the forces perpendicular to the plane? Those do indeed always cancel out.

But in the equation for the acceleration in the normal direction: m(ds/dt)(dϕ/dt) = mg sinθcosϕ, why the normal acceleration is (ds/dt)(dϕ/dt)?

Notice that$$\frac{d\mathbf{x}}{dt} = \frac{ds}{dt} \frac{d\mathbf{x}}{ds} = \dot{s} \frac{d\mathbf{x}}{ds}$$and hence that$$\frac{d^2 \mathbf{x}}{dt^2} = \frac{d^2 s}{dt^2} \frac{d\mathbf{x}}{ds} + \frac{ds}{dt} \frac{d^2 \mathbf{x}}{ds^2} \frac{ds}{dt} = \ddot{s} \frac{d\mathbf{x}}{ds} + \dot{s}^2 \frac{d^2 \mathbf{x}}{ds^2}$$but since the tangential unit vector is $\mathbf{e}_t = \frac{d\mathbf{x}}{ds}$ and the normal unit vector satisfies $\kappa \mathbf{e}_n = \frac{d\mathbf{e}_t}{ds} = \frac{d^2 \mathbf{x}}{ds^2}$, it follows from the chain rule that, since $\kappa := d\varphi / ds$,$$\frac{d^2 \mathbf{x}}{dt^2} = \frac{d^2 s}{dt^2} \mathbf{e}_t + \left(\frac{ds}{dt}\right)^2 \frac{d\varphi}{ds} \mathbf{e}_n = \ddot{s} \mathbf{e}_t + \dot{s} \dot{\varphi} \mathbf{e}_n$$which tells you that the tangential component of acceleration is $\ddot{s}$ and the normal component is $\dot{s} \dot{\varphi}$.

And why the acceleration in the y-axis equals the negative acceleration in the tangetial direction?

Since the components of force perpendicular to the plane cancel for all time [i.e. the motion is strictly rectilinear], you've essentially just got a particle acted upon by a component of weight $mg\sin{(\theta)} \mathbf{e}_y$ acting parallel to and down the plane, and a friction force $-mg\sin{(\theta)} \mathbf{e}_t$ acting in the negative tangential direction.

If you draw a diagram of this, looking directly down on the plane [i.e. in the direction of the plane normal], can you see why $a_y = -a_t$ when you draw the forces?

 

[Murilo T]

Remember that this occurs only in the large-$t$ regime; at the beginning, the $y$-components of the two forces do not cancel out, otherwise, $a_y$ would always be zero!

Or perhaps you mean the forces perpendicular to the plane? Those do indeed always cancel out.

Notice that$$\frac{d\mathbf{x}}{dt} = \frac{ds}{dt} \frac{d\mathbf{x}}{ds} = \dot{s} \frac{d\mathbf{x}}{ds}$$and hence that$$\frac{d^2 \mathbf{x}}{dt^2} = \frac{d^2 s}{dt^2} \frac{d\mathbf{x}}{ds} + \frac{ds}{dt} \frac{d^2 \mathbf{x}}{ds^2} \frac{ds}{dt} = \ddot{s} \frac{d\mathbf{x}}{ds} + \dot{s}^2 \frac{d^2 \mathbf{x}}{ds^2}$$but since the tangential unit vector is $\mathbf{e}_t = \frac{d\mathbf{x}}{ds}$ and the normal unit vector satisfies $\kappa \mathbf{e}_n = \frac{d\mathbf{e}_t}{ds} = \frac{d^2 \mathbf{x}}{ds^2}$, it follows from the chain rule that, since $\kappa := d\varphi / ds$,$$\frac{d^2 \mathbf{x}}{dt^2} = \frac{d^2 s}{dt^2} \mathbf{e}_t + \left(\frac{ds}{dt}\right)^2 \frac{d\varphi}{ds} \mathbf{e}_n = \ddot{s} \mathbf{e}_t + \dot{s} \dot{\varphi} \mathbf{e}_n$$which tells you that the tangential component of acceleration is $\ddot{s}$ and the normal component is $\dot{s} \dot{\varphi}$.

Since the components of force perpendicular to the plane cancel for all time [i.e. the motion is strictly rectilinear], you've essentially just got a particle acted upon by a component of weight $mg\sin{(\theta)} \mathbf{e}_y$ acting parallel to and down the plane, and a friction force $-mg\sin{(\theta)} \mathbf{e}_t$ acting in the negative tangential direction.

If you draw a diagram of this, looking directly down on the plane [i.e. in the direction of the plane normal], can you see why $a_y = -a_t$ when you draw the forces?

Aaaah, yeees. I see it now! Thank you very very much :)

 

[etotheipi]

Aaaah, yeees. I see it now! Thank you very very much :)

No problem!