Why Does Changing Force Direction Affect Acceleration Calculations on a Wedge?

[Buffu]

Homework Statement

A 45 deg wedge is pushed along a table with a constant accelaration A. A block of mass m slides without friction on the wedge. Find its accelaration.

Homework Equations

The Attempt at a Solution

$(x_b - x_w)\tan \theta = h -y_b$

$\ddot x_b + \ddot y_b = A$

Now equating forces horizontal and vertical forces on the block,

$N\cos \theta - mg = m \ddot y_b$

$N\sin\theta = m \ddot x_b$

Substituting these two in constraint equation,

$\sqrt{2}\times N = mg - mA$

$N = \dfrac{m}{2}(g + A)$

Substituting this back in the equations of forces on the block,

$\ddot x_b = \dfrac{g + A}{2}$ and $\ddot y_b = \dfrac{A - g}{2}$.

This is correct but if I take $mg -N\cos \theta = m \ddot y_b$ I get a wrong answer. For me, since the block is moving downward , $mg -N\cos \theta$ feels more natural and correct than $-mg +N\cos \theta$.
Anyway it does not give correct answer it gives $A = g$ which is wrong. I want to know why ?

 

[Mark44]

Homework Statement

A 45 deg wedge is pushed along a table with a constant accelaration A. A block of mass m slides without friction on the wedge. Find its accelaration.

Homework Equations

The Attempt at a Solution

View attachment 203762

$(x_b - x_w)\tan \theta = h -y_b$

$\ddot x_b + \ddot y_b = A$

Now equating forces horizontal and vertical forces on the block,

$N\cos \theta - mg = m \ddot y_b$

$N\sin\theta = m \ddot x_b$

Substituting these two in constraint equation,

$\sqrt{2}\times N = mg - mA$

$N = \dfrac{m}{2}(g + A)$

Substituting this back in the equations of forces on the block,

$\ddot x_b = \dfrac{g + A}{2}$ and $\ddot y_b = \dfrac{A - g}{2}$.

This is correct but if I take $mg -N\cos \theta = m \ddot y_b$ I get a wrong answer. For me, since the block is moving downward , $mg -N\cos \theta$ feels more natural and correct than $-mg +N\cos \theta$.
Wouldn't the block be moving upward? If the wedge is accelerating to the right, the weight of the block is acting downward, but the vertical component of its acceleration is in the opposite direction -- upward.

Anyway it does not give correct answer it gives $A = g$ which is wrong. I want to know why ?

 

[haruspex]

$\ddot x_b + \ddot y_b = A$

As a vector equation that is true, but not as scalars.

$\sqrt{2}\times N = mg - mA$

How do you get that? Where did the $\sqrt{2}$ come from?

 

[Buffu]

As a vector equation that is true, but not as scalars.

How do you get that? Where did the $\sqrt{2}$ come from?

$m\ddot x_b + m\ddot y_b = mA$

$N\sin \theta + N\cos \theta - mg = mA$

$\theta = 45^\circ$

$2N/\sqrt{2} = mg + mA \iff N\sqrt{2} = mg + mA$

As a vector equation that is true, but not as scalars.

Yes I know but I don't know the direction of those vectors.

 

[haruspex]

θ=45∘

Ok, I did not notice you were told that. I only saw the θ in the diagram.

Yes I know but I don't know the direction of those vectors.

The two accelerations are at right angles. How do you find the magnitude of their vector sum? Hint: Greek gent, name starting with P.

 

[Buffu]

Ok, I did not notice you were told that. I only saw the θ in the diagram.

The two accelerations are at right angles. How do you find the magnitude of their vector sum? Hint: Greek gent, name starting with P.

Pythagoras theorem. $\vec a = \sqrt{|\ddot x_b |^2 + |\ddot y_b |^2 }$ but how is magnitude tell me about direction ?

 

[haruspex]

Pythagoras theorem. $\vec a = \sqrt{|\ddot x_b |^2 + |\ddot y_b |^2 }$ but how is magnitude tell me about direction ?

It doesn't, but it is clearly a correct equation instead of the one you had.
But... you seem to be mixing up the two accelerations. "A" is the given acceleration of the wedge, not the block.

Think about the acceleration of the block relative to the wedge ( X and Y components). What is the component of the normal to the wedge?

 

[Buffu]

It doesn't, but it is clearly a correct equation instead of the one you had.
But... you seem to be mixing up the two accelerations. "A" is the given acceleration of the wedge, not the block.

Think about the acceleration of the block relative to the wedge ( X and Y components). What is the component of the normal to the wedge?

Is $\ddot x_b + \ddot y_b = A$ incorrect ?

Why ? I differentiated $(x_b - x_w)\tan \theta = h -y_b$ twice to get it.

 

[haruspex]

Is $\ddot x_b + \ddot y_b = A$ incorrect ?

Why ? I differentiated $(x_b - x_w)\tan \theta = h -y_b$ twice to get it.

Then please post that working. I do not understand what happened to the tan θ.

 

[Buffu]

In the small triangle,

$\tan \beta = \dfrac{h - y_b}{x_b - x_w} \iff (x_b - x_w)\tan \beta = h - y_b \iff (\ddot x_b - \ddot x_w)\tan \beta = -\ddot y_b$

Since $\beta = \theta = 45^\circ$, $\tan \beta = 1$.

$\ddot x_b - \ddot x_w = -\ddot y_b$.

 

[haruspex]

View attachment 203868

In the small triangle,

$\tan \beta = \dfrac{h - y_b}{x_b - x_w} \iff (x_b - x_w)\tan \beta = h - y_b \iff (\ddot x_b - \ddot x_w)\tan \beta = -\ddot y_b$

Since $\beta = \theta = 45^\circ$, $\tan \beta = 1$.

$\ddot x_b - \ddot x_w = -\ddot y_b$.

Oh, ok.
Very sorry, but because I engage in many threads at once I sometimes forget some crucial fact about a problem. In this case I forgot you had already explained θ is 45°. So now I agree with all your equations and can only apologise for all the noise.

Moving on to your original question:

since the block is moving downward , $mg -N\cos \theta$ feels more natural and correct

You have defined y as positive upwards, so $\ddot y$ is positive upwards. In the equation $\Sigma F_y=m\ddot y$, the forces must also be written as positive upwards, so $N\cos \theta-mg$.

 

[Buffu]

Oh, ok.
Very sorry, but because I engage in many threads at once I sometimes forget some crucial fact about a problem. In this case I forgot you had already explained θ is 45°. So now I agree with all your equations and can only apologise for all the noise.

Moving on to your original question:

You have defined y as positive upwards, so $\ddot y$ is positive upwards. In the equation $\Sigma F_y=m\ddot y$, the forces must also be written as positive upwards, so $N\cos \theta-mg$.

Oh, right, that was basic. I thought I was missing something very basic. lol.