Acceleration of a moving wedge with a falling block

[Yalanhar]

Homework Statement

In the figure, the masses m of the box and M of the wedge, as well as the angles A and B, are all known. All wires are ideal and friction is negligible. It is necessary to determine the acceleration acquired by the wedge. Gravity is equal to g

Relevant Equations

## F = ma##

When the box travels a ## X## distance, the wedge travels ## \frac{X}{2}##. So ##a = 2A##

Using the wedge as a non inertial frame:

I didn't use (4). Using (2) on (3) and then on (1) I got:
##2mA=mgsin\alpha +mAcos\alpha + \frac{-mgcos\alpha sin\alpha +mAsin^2\alpha +MA}{2cos\beta - sin\alpha}##

##4mAcos\beta - 2mAsin\alpha = 2mgsin\alpha cos\beta - mgsin^2\alpha +2mAcos\alpha cos\beta - 2mAcos\alpha sin\alpha - mgcos\alpha sin\alpha + mAsin^2\alpha +MA##

But I can't continue. The expression is monstrous and I don't know where to start the FactorizationThe answer: ##A = \frac{2mgsin\alpha cos\beta}{M+msin^2\alpha+m(2cos\beta -cos\alpha)^2}##

 

[kuruman]

Homework Statement:: In the figure, the masses m of the box and M of the wedge, as well as the angles A and B, are all known. All wires are ideal and friction is negligible. It is necessary to determine the acceleration acquired by the wedge. Gravity is equal to g
Homework Equations:: ## F = ma##

When the box travels a X distance, the wedge travels ## \frac{X}{2}##.

Why? Say the block travels distance ##ds## down the incline. This means that the string from the mass ##m## to the pulley gets longer by ##ds##. How far does the block travel in the negative x-direction relative to the wedge? The string on the right side of the wedge gets shorter by the same amount ##ds##. How far does the wedge travel in the x-direction relative to the floor? How far does the block travel relative to the floor?

 

[Yalanhar]

Why? Say the block travels distance ##ds## down the incline. This means that the string from the mass ##m## to the pulley gets longer by ##ds##. How far does the block travel in the negative x-direction relative to the wedge? The string on the right side of the wedge gets shorter by the same amount ##ds##. How far does the wedge travel in the x-direction relative to the floor? How far does the block travel relative to the floor?

I don't know the name in english. Geometric link?

As the wires are inextensible, using the "geometric link" we can relate the accelerations. In this image's case when A travels D, then B travels D/2. Because both wire travels D/2 + D/2 = D. Inextensible.

In the wedge case, when the box travels D, the the wires travels D/2. But now that I'm looking at, i think ##a = 2Acos\beta##. In any case I'm not sure

 

[Cutter Ketch]

i think a=2Acosβ

You’ve used “A” to mean a couple of different things in your problem statement and diagram, but here you mean A to be the magnitude of the acceleration of the wedge, yes? And a is the magnitude of the acceleration of the block, right?

Yes, that cosine is definitely necessary. However I think you have it in the wrong place. 2A should be less than a, correct? It’s just an algebra error. Try again.

 

[haruspex]

I don't know the name in english. Geometric link?
View attachment 255529
As the wires are inextensible, using the "geometric link" we can relate the accelerations. In this image's case when A travels D, then B travels D/2. Because both wire travels D/2 + D/2 = D. Inextensible.

In the wedge case, when the box travels D, the the wires travels D/2. But now that I'm looking at, i think ##a = 2Acos\beta##. In any case I'm not sure

The trick with the kinematics aspect of such questions is to think about components of motion in the directions of the strings.
E.g. look at where the string attaches to the bottom right of the wedge. If the wedge has acceleration A to the right, what is the component of that in the direction of the string attaching there? What does that tell you about how the length of that section of string changes (assuming we are starting from rest)?
Consider the changes in the lengths of the other sections in the same way and combine.

 

[TSny]

The problem statement does not mention whether or not the wedge and block are at rest at the instant shown. But I think they must be at rest in order to get the given answer. In general, the acceleration of the block relative to the wedge depends on the velocity and position of the wedge as well as the acceleration of the wedge.

Also, in equation (3) of the original post, check if you have the correct trig function for the first term on the right side.

 

[Chestermiller]

But now that I'm looking at, i think ##a = 2Acos\beta##. In any case I'm not sure

I've worked this problem out to completion, and get the answer given by your book. If, in the equation ##a = 2Acos\beta##, you mean that A represents the horizontal acceleration of the wedge and a represents the acceleration of the block relative to the wedge (in the direction tangent to the wedge surface), this is correct. But this does not represent the absolute acceleration of the block in the direction tangent to the wedge surface (i.e., as reckoned from the stationary laboratory frame of reference) because the wedge itself is accelerating (and its horizontal acceleration has a component parallel to the incline).

 

[Yalanhar]

I've worked this problem out to completion, and get the answer given by your book. If, in the equation ##a = 2Acos\beta##, you mean that A represents the horizontal acceleration of the wedge and a represents the acceleration of the block relative to the wedge (in the direction tangent to the wedge surface), this is correct. But this does not represent the absolute acceleration of the block in the direction tangent to the wedge surface (i.e., as reckoned from the stationary laboratory frame of reference) because the wedge itself is accelerating (and its horizontal acceleration has a component parallel to the incline).
[/QUOTE]

Yes, exactly.
But I can't get to the answers.
Are these equations correct?

For the block:
x -> ## 2mAcos\beta = mgsin\alpha + mAcos\alpha - T## (1)
y -> ##mgcos\alpha - mAsin\alpha = f## (2)

Wedge
x -> ##fsin\alpha + Tcos\beta = MA## (3)

When I try to solve I get ##A = \frac{mgsin\alpha (cos\alpha + cos\beta)}{M+msin^2\alpha+m(2cos^2\beta-cos\beta cos\alpha)}##

 

[haruspex]

Are these equations correct?

For the block:
x -> ## 2mAcos\beta = mgsin\alpha + mAcos\alpha - T## (1)
y -> ##mgcos\alpha - mAsin\alpha = f## (2)

Wedge
x -> ##fsin\alpha + Tcos\beta = MA## (3)

When I try to solve I get ##A = \frac{mgsin\alpha (cos\alpha + cos\beta)}{M+msin^2\alpha+m(2cos^2\beta-cos\beta cos\alpha)}##

Your forces on the wedge are incomplete. Think about the pulley.

 

[Chestermiller]

Yes, exactly.
But I can't get to the answers.
Are these equations correct?

For the block:
x -> ## 2mAcos\beta = mgsin\alpha + mAcos\alpha - T## (1)
y -> ##mgcos\alpha - mAsin\alpha = f## (2)

Wedge
x -> ##fsin\alpha + Tcos\beta = MA## (3)

When I try to solve I get ##A = \frac{mgsin\alpha (cos\alpha + cos\beta)}{M+msin^2\alpha+m(2cos^2\beta-cos\beta cos\alpha)}##

I confirm your equation for the tangential component of the force balance on the block. But, instead of writing a separate force balance on the wedge, write down the horizontal force balance for the combination of wedge and block (including the accelerations of both). The only external horizontal force acting on this combination is ##2T\cos{\beta}##.

 

[Yalanhar]

I confirm your equation for the tangential component of the force balance on the block. But, instead of writing a separate force balance on the wedge, write down the horizontal force balance for the combination of wedge and block (including the accelerations of both). The only external horizontal force acting on this combination is ##2T\cos{\beta}##.

Thanks! I got the answer!

Here my full resolution:

Block on X axis
##2mAcos\beta = mgsen\alpha +mAcos\alpha -T (1)##

Wedge + block X axis
##0=(m+M)A - 2Tcos\beta## Therefore ##T=\frac{(m+M)A}{2cos\beta}## (2)

(2) in (1)

##2mAcos\beta = mgsen\alpha +mAcos\alpha - \frac{mA+MA}{2cos\beta}##
##4mAcos^2\beta = 2mgsen\alpha cos\beta + 2mAcos\alpha cos\beta - mA - MA##

##A = \frac {2mgsen\alpha cos\beta}{4mcos^2\beta -2mcos\alpha cos\beta +M+m}##(3)

##m = m(sen^2\alpha + cos^2\alpha) = msen^2\alpha + mcos^2\alpha## (4)

(4) in (3)

##A = \frac {2mgsen\alpha cos\beta}{4mcos^2\beta -2mcos\alpha cos\beta +mcos^2\alpha +M+msen^2\alpha}##(3)
##A = \frac {2mgsen\alpha cos\beta}{M+msen^2\alpha + m(2cos\beta - cos\alpha)^2}##

 

[Chestermiller]

Thanks! I got the answer!

Here my full resolution:

Block on X axis
##2mAcos\beta = mgsen\alpha +mAcos\alpha -T (1)##

Wedge + block X axis
##0=(m+M)A - 2Tcos\beta## Therefore ##T=\frac{(m+M)A}{2cos\beta}## (2)

(2) in (1)

##2mAcos\beta = mgsen\alpha +mAcos\alpha - \frac{mA+MA}{2cos\beta}##
##4mAcos^2\beta = 2mgsen\alpha cos\beta + 2mAcos\alpha cos\beta - mA - MA##

##A = \frac {2mgsen\alpha cos\beta}{4mcos^2\beta -2mcos\alpha cos\beta +M+m}##(3)

##m = m(sen^2\alpha + cos^2\alpha) = msen^2\alpha + mcos^2\alpha## (4)

(4) in (3)

##A = \frac {2mgsen\alpha cos\beta}{4mcos^2\beta -2mcos\alpha cos\beta +mcos^2\alpha +M+msen^2\alpha}##(3)
##A = \frac {2mgsen\alpha cos\beta}{M+msen^2\alpha + m(2cos\beta - cos\alpha)^2}##

The horizontal acceleration of m is not A. It is ##A-2A\cos{\beta}\cos{\alpha}##

 

[TSny]

##A = \frac {2mgsen\alpha cos\beta}{4mcos^2\beta -2mcos\alpha cos\beta +mcos^2\alpha +M+msen^2\alpha}##(3)
##A = \frac {2mgsen\alpha cos\beta}{M+msen^2\alpha + m(2cos\beta - cos\alpha)^2}##

The last equation does not follow from the equation before it. Note that
##m(2cos\beta - cos\alpha)^2 = 4mcos^2\beta -##4##mcos\alpha cos\beta +mcos^2\alpha ##

@Chestermiller is pointing out that there is an error in your equation (2) where you treated both masses as having the same horizontal acceleration ##A##.

 

[Yalanhar]

Wedge + block X axis
##0=(m+M)A - 2Tcos\beta## Therefore ##T=\frac{(m+M)A}{2cos\beta}## (2)

The last equation does not follow from the equation before it. Note that
##m(2cos\beta - cos\alpha)^2 = 4mcos^2\beta -##4##mcos\alpha cos\beta +mcos^2\alpha ##

@Chestermiller is pointing out that there is an error in your equation (2) where you treated both masses as having the same horizontal acceleration ##A##.

yeah true

 

[Yalanhar]

The horizontal acceleration of m is not A. It is ##A-2A\cos{\beta}\cos{\alpha}##

But that is on a inertial reference frame right? If yes, then my equation 1 is wrong.
I'm trying to solve on the wedge reference of frame, that's why A is perpendicular to g, as show on the upper right corner. But now I'm stuck again

 

[Yalanhar]

@Chestermiller is pointing out that there is an error in your equation (2) where you treated both masses as having the same horizontal acceleration ##A##.
[/QUOTE]
If I want to stick to the non inertial frame then I can't use block+wedge system, I'll try on Earth's reference

 

[Yalanhar]

I confirm your equation for the tangential component of the force balance on the block. But, instead of writing a separate force balance on the wedge, write down the horizontal force balance for the combination of wedge and block (including the accelerations of both). The only external horizontal force acting on this combination is ##2T\cos{\beta}##.

I don't understand how to balance the forces of block and wedge while they have diferent horizontal accelerations.
m is ##A - 2Acos\beta cos\alpha##
M is ##A##

 

[Chestermiller]

If you write separate force balance equations on the block and wedge, and then add the two equations together, all the internal forces cancel out, and you are left with: $$2T\cos{\beta}=MA+m(A-2A\cos{\beta}\cos{\alpha})$$

 

[TSny]

If I want to stick to the non inertial frame then I can't use block+wedge system, I'll try on Earth's reference

You can stick to the non-inertial frame and still set up the equation for the block+wedge system. You just need to be sure to include the fictitious forces as external forces acting on the system.

For the inertial frame, @Chestermiller has given the equation for the system:

$$2T\cos{\beta}=MA+m(A-2A\cos{\beta}\cos{\alpha})$$

For the non-inertial frame, you get an equivalent equation where the ##MA## and ##mA## terms appear on the left side as fictitious forces instead of on the right side.

 

[Yalanhar]

If you write separate force balance equations on the block and wedge, and then add the two equations together, all the internal forces cancel out, and you are left with: $$2T\cos{\beta}=MA+m(A-2A\cos{\beta}\cos{\alpha})$$

For the block
##m(A-2Acos\beta cos\alpha) = Tcos\alpha - fsen\alpha)## (1)

For the wedge
##MA = 2Tcos\beta +fsen\alpha -Tcos\alpha## (2)

(2) + (1)

##MA + m(A-2Acos\beta cos\alpha) = 2Tcos\beta## (3)

Using non inertial frame, and the force balance of the block on tangente
##2mAcos\beta = mgsen\alpha + mAcos\alpha - T## (4)

(3) in (4):
##2mAcos\beta = mgsen\alpha - mAcos\alpha - \frac{MA+mA-2mAcos\beta cos\alpha}{2cos\beta}##

##4mAcos^2\beta = 2mgsen\alpha cos\beta + 2mAcos\alpha cos\beta - MA - mA +2mAcos\alpha cos\beta##

##A(4mcos^2\beta - 2mAcos\alpha cos\beta + M + m - 2mAcos\alpha cos\beta) = 2mgsen\alpha cos\beta####m = m(sen^2x + cos^2x)####A(4mcos^2\beta - 4mAcos\alpha cos\beta + M + msen^2\alpha + mcos^2\beta) = 2mgsen\alpha cos\beta##

##A[m(2cos\beta - cos\alpha)^2 +M +msen^2\alpha] = 2mgsen\alpha cos\beta##

##A = \frac{2mgsen\alpha cos\beta}{[m(2cos\beta - cos\alpha)^2 +M +msen^2\alpha]}##I think it's all sorted out.