Deformation of a spring being accelerated

[Like Tony Stark]

Homework Statement

The picture shows a block of $0.5 kg$ that interacts with a spring of constant $40 \frac{N}{m}$. The body lies on a wedge and they are static with respect to the room that accelerates with $a=0.5 \frac{m}{s^2}$ to the right. The horizontal and the wedge form an angle of $40°$. Determine the deformation of the spring, the normal force acting on $A$ and the acceleration needed to make the deformation equal to 0.

Relevant Equations

Newton's equations

I wrote Newton's equations for the block seen from the non inertial frame. The axis are inclined.
$x) Fe+W_x-f*cos(\alpha)=0$
$y) N-f*sin(\alpha)-W_y=0$
Where $f*$ is the pseudo-force and $Fe$ is the elastic force. I set the acceleration as 0 because they are in equilibrium.
The thing is that when I solve the equations I got that the deformation is negative, but it should be positive because the spring is being stretched.

 

[haruspex]

Homework Statement: The picture shows a block of $0.5 kg$ that interacts with a spring of constant $40 \frac{N}{m}$. The body lies on a wedge and they are static with respect to the room that accelerates with $a=0.5 \frac{m}{s^2}$ to the right. The horizontal and the wedge form an angle of $40°$. Determine the deformation of the spring, the normal force acting on $A$ and the acceleration needed to make the deformation equal to 0.
Homework Equations: Newton's equations

I wrote Newton's equations for the block seen from the non inertial frame. The axis are inclined.
$x) Fe+W_x-f*cos(\alpha)=0$
$y) N-f*sin(\alpha)-W_y=0$
Where $f*$ is the pseudo-force and $Fe$ is the elastic force. I set the acceleration as 0 because they are in equilibrium.
The thing is that when I solve the equations I got that the deformation is negative, but it should be positive because the spring is being stretched.

Check all your signs. That's the commonest error in force analysis, and using tilted axes and noninertial frames makes it no easier.
The equations you posted aren't actually wrong - they don't become wrong until you make the substitutions - but they certainly look as though you are about to go wrong.

An approach I prefer is to make up and right positivefor forces and accelerations (or such tilted to match the plane) and write the force sums all with + signs. When substituting the known values (like mg) apply the appropriate signs.

 

[Like Tony Stark]

Check all your signs. That's the commonest error in force analysis, and using tilted axes and noninertial frames makes it no easier.
The equations you posted aren't actually wrong - they don't become wrong until you make the substitutions - but they certainly look as though you are about to go wrong.

An approach I prefer is to make up and right positivefor forces and accelerations (or such tilted to match the plane) and write the force sums all with + signs. When substituting the known values (like mg) apply the appropriate signs.

I've done it but changing the tilted axes for common axes.
So my equations are
$x)-f*+Fecos(\alpha)+Nsin(\alpha)=0$
$y) Ncos(\alpha)-mg-Fesin(\alpha)=0$

$f*=m.a=0.5×0.5=0.25 N$
$\alpha =40°$
$Fe=k \Delta x=40 \Delta x$ (I put the sign depending on the free body diagram
$mg=0.5×10=5 N$
$a_x=a_y=0$ because they're at rest in the non inertial frame.

But I'm still getting a negative value for $\Delta x$

 

[haruspex]

I've done it but changing the tilted axes for common axes.
So my equations are
$x)-f*+Fecos(\alpha)+Nsin(\alpha)=0$
$y) Ncos(\alpha)-mg-Fesin(\alpha)=0$

$f*=m.a=0.5×0.5=0.25 N$
$\alpha =40°$
$Fe=k \Delta x=40 \Delta x$ (I put the sign depending on the free body diagram
$mg=0.5×10=5 N$
$a_x=a_y=0$ because they're at rest in the non inertial frame.

But I'm still getting a negative value for $\Delta x$

Well, you did not do what I suggested wrt signs. In my approach you would write e.g. $f_*+F_e\cos(\alpha)+N\sin(\alpha)=0$, where $F_e$ will be negative because it acts to the left, and the angle is negative because it is tilted clockwise.
No matter... I will work with your approach.

f* and N both act to the right, but you have given them opposite signs in (x).
In (y), N and Fe both act up, but they have opposite signs.

 

[Like Tony Stark]

Well, you did not do what I suggested wrt signs. In my approach you would write e.g. $f_*+F_e\cos(\alpha)+N\sin(\alpha)=0$, where $F_e$ will be negative because it acts to the left, and the angle is negative because it is tilted clockwise.
No matter... I will work with your approach.

f* and N both act to the right, but you have given them opposite signs in (x).
In (y), N and Fe both act up, but they have opposite signs.

Why the normal an elastic force should have the same sign? The normal points upwards and the elastic force tries to put the block where is was before since it is stretching.
Then why do you say that $f*$ is positive? The acceleration is to the right so the force will be to the left

 

[haruspex]

Why the normal an elastic force should have the same sign? The normal points upwards and the elastic force tries to put the block where is was before since it is stretching.

We are both right and both wrong. At this stage, you do not know which way the force acts. You have effectively defined downslope as positive for the spring force, so if the extension comes out negative it just means the spring is actually under compression.
Bear in mind that if the room were not accelerating then the spring wouid definitely be under compression, so it comes to whether the acceleration is greater or less than some threshold.

Then why do you say that f∗f∗f* is positive?

On that, you are right and I was wrong.

 

[Like Tony Stark]

We are both right and both wrong. At this stage, you do not know which way the force acts. You have effectively defined downslope as positive for the spring force, so if the extension comes out negative it just means the spring is actually under compression.
Bear in mind that if the room were not accelerating then the spring wouid definitely be under compression, so it comes to whether the acceleration is greater or less than some threshold.

On that, you are right and I was wrong.

Thanks!