Sign change problem :(.

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1. May 18, 2017

Buffu

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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 ?

2. May 18, 2017

Staff: Mentor

3. May 18, 2017

haruspex

As a vector equation that is true, but not as scalars.
How do you get that? Where did the $\sqrt{2}$ come from?

4. May 19, 2017

Buffu

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

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

$\theta = 45^\circ$

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

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

5. May 19, 2017

haruspex

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.

6. May 19, 2017

Buffu

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

7. May 19, 2017

haruspex

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?

8. May 19, 2017

Buffu

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.

9. May 19, 2017

haruspex

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

10. May 20, 2017

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

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

11. May 20, 2017

haruspex

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 $N\cos \theta-mg$.

12. May 20, 2017

Buffu

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