# Doubt on this problem from David Morin's book - Intro to Classical Mech

Will someone please explain the foll. doubts in this solved problem from David Morin's book.

TIA
sree
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DOUBTS :
1. What is the principle of eqn. 3.78 ? If V is block speed and Vy is its component in y dirn., why should V + Vy be constant ?

2. In Why should final speed ofblock Vf and its comp. in y direction be equal ?

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• DOUBTS_Morin.doc
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AlephZero
Homework Helper
This is a horrible looking argument. I can follow it, but I'm taking Bertrand Russell's position here when he said "The fact that I can't see what is wrong with it does not prove it is right".

Anyhow, assuming for the sake of argument that it is right:

1. He says the force of gravitiy down the plane, and the magnitude of the friction force (whatever direction it acts in), are always equal. Therefore, the deceleration due to friction in the direction of motion, and the acceleration down the plane, are always equal. So in each small increment of time, the reduction in v (the deceleration) equals the increase in vf (the acceleration).

So v + vf is constant.

2. He is assuming that the final motion will be in a straight line down the plane. That makes sense, because any sideways component of velocity would be reduced by the friction force, so the sideways velocity will decrease to 0.

Because the coefficient of friction is related to the angle of the plane, you know that motion straight down the plane will be at constant speed.

If the mass is moving straight down the plane, then its speed down the plane, vf, is equal to its total speed, v. And from (1) you know that v + vf = C = V

Honestly, I think this explanation makes little or no sense.
Edit: I'm talking about the attached doc file

The problem is that if the small increment of v in a small increment of time (I always call it acceleration, btw) is a rotating vector.
It cannot be added algebraically to the gravity acceleration, but a vector addition must be followed.

It is questionable whether the block should gain any speed at all in the down direction, because every acceleration is balanced by friction.
At minimum you should solve a differential equation. Stating vf+vf=C, sounds me too simple.

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Dear AlephZero and Quinzio
I'm confused. My thoughts are as below.
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The block is moving initially with Vx = V (in x direction - which is in plane of slope), and Vy = 0 (y dirn is down slope). So one has to take sqrt (Vx 2 + Vy 2 ) = V. How can you add Vx + Vy and say it is constant ? What is the principle ?
(If consvn. of momentum principle is used then inila mom = mV, final = mVy, therefore Vy = V but this is not what is being done here)
In his explanation Vx final is not zero. He is calculating it using the reln in eqn 3.78.
How is Vf + Vf = 2 Vf obtained ? After along time x vel becomes zero and block begins moving down with Vf. How is 2 Vf obtained ?

TIA
sree

This may be a realistic disposition of the accelerations here.
Acceleration due to friction has a component is the gravity direction (opposite) and it should be taken in account.

The gravity force has a normal component to v, so it will make the speed vector rotate, but whether or not it will end up to V/2, i would like a more complete explanation.

[PLAIN]http://img844.imageshack.us/img844/4613/mgsina.jpg [Broken]

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A simple numerical simulation like the one in the code below shows that the speed really ends up to V/2.
But the explanation is something more complex than vf+vf=C.
The analytical solution is a diffrential equation.
In the textbook, they're trying to make a difficult topic look easy, as a sort of magic trick. A very bad way to teach.

Code:
Dim friction_x As Double
Dim friction_y As Double
Dim speed As Double = 10
Dim speed_x As Double
Dim speed_y As Double
Dim angle As Double = 0
Dim delta_t As Double = 0.01
Dim mgsina = 0.1
For index = 0 To 50000
friction_x = mgsina * Math.Cos(angle)
friction_y = mgsina * Math.Sin(angle)
speed_x = speed * Math.Cos(angle)
speed_y = speed * Math.Sin(angle)

speed_x -= (friction_x) * delta_t
speed_y -= (friction_y - mgsina) * delta_t

speed = Math.Sqrt(speed_x ^ 2 + speed_y ^ 2)
angle = Math.Atan2(speed_y, speed_x)
Next

AlephZero
Homework Helper
The block is moving initially with Vx = V (in x direction - which is in plane of slope), and Vy = 0 (y dirn is down slope). So one has to take sqrt (Vx 2 + Vy 2 ) = V. How can you add Vx + Vy and say it is constant ? What is the principle ?
No, he is not addng vx+vy. He is saying the total speed v (i.e. sqrt (vx2 + vy2) + vy is constant.

(If consvn. of momentum principle is used then inila mom = mV, final = mVy, therefore Vy = V but this is not what is being done here)
You can't use conservation of momentum unless you include the forces acting on the particle. But if you don''t know the direction of motion, you don't know the direction of the friction force, so this doesn't help much.

In his explanation Vx final is not zero. He is calculating it using the reln in eqn 3.78.
How is Vf + Vf = 2 Vf obtained ? After along time x vel becomes zero and block begins moving down with Vf. How is 2 Vf obtained ?
v in the equations is not the same as vx. v is the speed of the block, in whatever direction it is moving.

At the start, v = vx because the block is moving sideways across the slope. At the end, v = vy because it is moving down the slope. In between, v2 = vx2 + vy2.