# Understanding Forces and Acceleration on a Sliding Object on an Incline

• Murilo T
in summary, the two solutions you provided are both correct, but the normal acceleration is different depending on how you look at it.
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?

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Murilo T said:
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.
Murilo T said:
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}##.
Murilo T said:
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?

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Murilo T
etotheipi said:
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
Murilo T said:
Aaaah, yeees. I see it now! Thank you very very much :)

No problem!

## 1. What causes a plane to slide sideways?

A plane can slide sideways due to a number of factors, such as strong crosswinds, pilot error, or mechanical issues. Crosswinds, in particular, can cause the plane to drift off course and appear to be sliding sideways.

## 2. Is sliding sideways on a plane dangerous?

Sliding sideways on a plane can be dangerous if it is not properly controlled by the pilot. Crosswinds can make landing and takeoff difficult, and if the pilot is not able to compensate for them, it can lead to accidents. However, modern planes are equipped with advanced technology and trained pilots to handle crosswind situations safely.

## 3. How do pilots handle sliding sideways on a plane?

Pilots are trained to handle crosswinds and sliding sideways on a plane. They use a technique called "crabbing" where they angle the plane into the wind to counteract the sideways movement. They also use rudder and aileron controls to maintain the correct direction and alignment with the runway.

## 4. Can sliding sideways on a plane be felt by passengers?

In most cases, passengers will not feel the plane sliding sideways. Pilots are trained to make smooth and controlled adjustments to counteract crosswinds, so the movement is often not noticeable. However, in severe cases, passengers may feel a slight tilt or bump as the plane corrects its course.

## 5. Are there any safety measures in place to prevent sliding sideways on a plane?

Yes, there are several safety measures in place to prevent sliding sideways on a plane. Air traffic control monitors weather conditions and informs pilots of any crosswinds or other hazards. Pilots also have access to real-time weather information and can make decisions to avoid areas with strong crosswinds. Additionally, modern planes have advanced technology such as wind shear detection systems to alert pilots of any sudden changes in wind direction or speed.

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