Motion of box on inclined plane connected by spring to a wall

[songoku]

Homework Statement

See the picture below. Initially, the system is at rest. Neglecting friction and mass of both spring and string,
a) what is initial acceleration of M? Assume initially the spring is at natural length $L_1$
b) find the tension when the system moves for the first time
c) is total energy of system conserved?
d) after moving a distance $x_o$ towards the wall, M will stop. Find $x_o$
e) find the equilibrium point of M, measured from initial position of M, when M oscillates
f) find the period of oscillation of system

Relevant Equations

∑F = m.a

Restoring force: F = k.x

ω = 2π / T

a) When the system is in motion for the first time, the force causing $M$ to move is contact force with $m$ so:
$$\Sigma F=M.a$$
$$N \sin \alpha=M.a$$
$$mg \cos \alpha \sin \alpha =M.a$$
$$a=\frac{mg \cos \alpha \sin \alpha}{M}$$

Is that correct?

b) Is acceleration of $m$ the same as $M$ when the system moves for the first time?

Thanks

 

[haruspex]

Are you sure N=mg cos(α)? If so, in which direction will m accelerate in the ground frame? Which direction is that relative to M?

 

[songoku]

Are you sure N=mg cos(α)?

Based on free body diagram (taking x-axis parallel to the slope of inclined plane), I think so

If so, in which direction will m accelerate in the ground frame? Which direction is that relative to M?

With respect to ground frame, $m$ will accelerate down the plane and $M$ to the right. With respect to $M$, I think $m$ will also accelerate down the slope.

Or maybe my imagination is wrong? Actually with respect to ground, $m$ is accelerating vertically downwards and with respect to $M$, $m$ is accelerating down the plane?

Thanks

 

[haruspex]

Based on free body diagram (taking x-axis parallel to the slope of inclined plane), I think so

The FBD gives you the forces, but the equation is ΣF=ma. You are assuming m has no component of acceleration (in the ground frame) normal to the plane.

With respect to ground frame, m will accelerate down the plane and M to the right. With respect to M, I think m will also accelerate down the slope.

How can those three be true simultaneously?

 

[songoku]

The FBD gives you the forces, but the equation is ΣF=ma. You are assuming m has no component of acceleration (in the ground frame) normal to the plane.

How can those three be true simultaneously?

I am not sure. Is there a certain way to determine the direction of acceleration of $m$ with respect to ground and relative to $M$?

Thanks

 

[Lnewqban]

I am not sure. Is there a certain way to determine the direction of acceleration of $m$ with respect to ground and relative to $M$?

Thanks

How much further down the sloped roof could this worker go?

 

[haruspex]

I am not sure. Is there a certain way to determine the direction of acceleration of $m$ with respect to ground and relative to $M$?

Thanks

If the wedge has acceleration $\vec a_w$ and the mass has acceleration $\vec a_{mw}$ relative to the wedge, what is the mass's acceleration in the ground frame?

 

[songoku]

How much further down the sloped roof could this worker go?

View attachment 280370

All the way to the bottom of the slope? Sorry I don't really understand how to answer that question

If the wedge has acceleration $\vec a_w$ and the mass has acceleration $\vec a_{mw}$ relative to the wedge, what is the mass's acceleration in the ground frame?

$\vec a_m = \vec a_{mw} + \vec a_w$

 

[Lnewqban]

All the way to the bottom of the slope? Sorry I don't really understand how to answer that question

Yes, sorry.
I was just trying to tell you that there is no much freedom for the mass m to move away from the wall because the string, even if the pulley attached to mass M moves as far to the right as the spring allows M to move.

Why would M move against the resistance of the spring?

 

[haruspex]

$\vec a_m = \vec a_{mw} + \vec a_w$

Ok, so what equation can you write for the component of $\vec a_{mw}$ normal to the wedge?
What equation relates N and mg to the component of $\vec a_m$ normal to the wedge?

 

[songoku]

Why would M move against the resistance of the spring?

Because there is contact force with $m$ that pushes $M$ to the right

Ok, so what equation can you write for the component of $\vec a_{mw}$ normal to the wedge?
What equation relates N and mg to the component of $\vec a_m$ normal to the wedge?

Should I first need to know the direction of $\vec a_{mw}$ and $\vec a_m$ to be able to answer this question?

Thanks

 

[haruspex]

Because there is contact force with $m$ that pushes $M$ to the rightShould I first need to know the direction of $\vec a_{mw}$ and $\vec a_m$ to be able to answer this question?

Thanks

For my first question , does the mass lose contact with the wedge? Bury itself into the wedge? No? So what equation can you write for the component of $\vec a_{mw}$ normal to the wedge?

 

[songoku]

For my first question , does the mass lose contact with the wedge? Bury itself into the wedge? No? So what equation can you write for the component of $\vec a_{mw}$ normal to the wedge?

I am not sure. I imagine the direction of $\vec a_{mw}$ is downwards parallel to the slope so the component of $\vec a_{mw}$ normal to the wedge is zero

 

[haruspex]

I am not sure. I imagine the direction of $\vec a_{mw}$ is downwards parallel to the slope so the component of $\vec a_{mw}$ normal to the wedge is zero

Yes.

 

[songoku]

What equation relates N and mg to the component of $\vec a_m$ normal to the wedge?

Is the direction of $\vec a_m$ vertically downwards?

Thanks

 

[Lnewqban]

If you observe the geometry of the string, you could see that:
1) The length of the RH portion of the string is reduced in the same magnitude that M moves towards the right.
2) The length of the LH portion of the string is increased in the same magnitude that M moves towards the right (because of 1 above); therefore, m slides down the slope and away from the pulley respect to M.

I believe that, respect to the ground, m moves downwards and towards the left as the spring gets compressed.

There is a reaction at the axis of the pulley due to the tension of the string, but perhaps its horizontal component is not significant enough to include it in the solution of this problem.

 

[haruspex]

Is the direction of $\vec a_m$ vertically downwards?

Thanks

Only if M=0.

Let me start again, laying out what I hope is a simpler path, working in scalars.
Let the acceleration of M be a.
Find a simple kinematic equation relating a_mw (which we know is parallel to the slope) to a.
Let the horizontal and vertical accelerations of m in the ground frame be a_x, a_y.
Write out the ΣF=ma equations in those directions. These will involve N, T and mg.
Write the horizontal ΣF=ma equation for M.

We need two more equations. These express a_x, a_y in terms of a, a_mw.
See how far you can get.

 

[songoku]

I believe that, respect to the ground, m moves downwards and towards the left as the spring gets compressed.

You mean the direction of acceleration of m with respect to ground is parallel to the slope?

Let the acceleration of M be a.
Find a simple kinematic equation relating a_mw (which we know is parallel to the slope) to a.

You mean a_mw = a_m - a , where a_m is acceleration of m with respect to ground?

Thanks

 

[haruspex]

You mean a_mw = a_m - a , where a_m is acceleration of m with respect to ground?

Thanks

No, just involving those two accelerations. Assume the whole string length is constant.

 

[haruspex]

respect to the ground, m moves downwards and towards the left

Whether it is left or right depends on α.

 

[songoku]

No, just involving those two accelerations. Assume the whole string length is constant.

Is it:

a = - a_mw cos α ?

Thanks

 

[haruspex]

Is it:

a = - a_mw cos α ?

Thanks

I gave you the hint to think about string length. If the wedge moves a distance x towards the wall, what happens to the length of string from the pulley to the mass?

 

[songoku]

I gave you the hint to think about string length. If the wedge moves a distance x towards the wall, what happens to the length of string from the pulley to the mass?

I tried to use that hint and also what Lnewqban posted in #16 but apparently I did not use it correctly.

When the wedge moves a distance x towards the wall, the pulley will also move as far as x towards the wall hence the length of string from the pulley to the wall decreases by x, so the length of string from the pulley to the mass increases by x since the total length should stay constant.

From there, I thought since $a$ is to the right and $a_{mw}$ is down the slope and the distance traveled by the mass and the wedge is the same, the magnitude of $a$ and $a_{mw}$ would be the same and I tried to find the component of $a_{mw}$ which is parallel to $a$, hence $a_{mw} \cos \alpha$ directed to the left

Edit: Ah I see my mistake, $|a|$ is the same as $|a_{mw}|$ but is not the same as $|a_{mw} \cos \alpha|$ as $|a_{mw} \cos \alpha|$ will be less than $a$

So the equation will be $|a|=|a_{mw}|$ ?

Thanks

 

[haruspex]

So the equation will be |a|=|a_mw| ?

Yes.
So onto the next part...
Let the horizontal and vertical accelerations of m in the ground frame be a_x, a_y.
Write out the ΣF=ma equations in those directions. These will involve N, T and mg.
Write the horizontal ΣF=ma equation for M.

We need two more equations. These express a_x, a_y in terms of a, a_mw.

 

[songoku]

Yes.
So onto the next part...
Let the horizontal and vertical accelerations of m in the ground frame be a_x, a_y.
Write out the ΣF=ma equations in those directions. These will involve N, T and mg.

Since $a_m = a_{mw} + a$ and for horizontal direction $|a|> |a_{mw} \cos \alpha|$, I assume horizontally $m$ will move to the right with respect to the ground

∑F = m.a_x
T cos α - N sin α = m.a_x ...(1)

For vertical direction:
∑F = m.a_y
mg - N cos α = m.a_y ...(2)

Write the horizontal ΣF=ma equation for M.

∑F = M.a
N sin α = M.a ... (3)

We need two more equations. These express a_x, a_y in terms of a, a_mw.

a_x = a - a_mw ... (4)

a_y = a_mw sin α ... (5)

Is that correct?

Thanks

 

[haruspex]

Since $a_m = a_{mw} + a$ and for horizontal direction $|a|> |a_{mw} \cos \alpha|$, I assume horizontally $m$ will move to the right with respect to the ground

I don't think that follows. Remember that a_mw is down and left.

For vertical direction:
∑F = m.a_y
mg - N cos α = m.a_y ...(2)

What about tension?

a_x = a - a_mw ... (4)

a_mw has a vertical component.

 

[songoku]

What about tension?

For vertical direction:
ΣF = m.a_y
mg - N cos α - T sin α = m.a_y ... (2)

a_mw has a vertical component.

a_x = a - a_mw cos α ... (4)

After a lot of algebra, I find the acceleration of M:

$$a=\frac{mg \cos⁡ \alpha}{M(\cos \alpha \cot \alpha+ \sin⁡ \alpha)+m \sin \alpha}$$

For question (b), I just need to substitute a to previous equations

For question (c), the answer is yes because total energy is always conservedFor question (d), would it be like, for M:
$$\Sigma F=M.a$$
$$N \sin \alpha - k.x = M.\frac{v.dv}{dx}$$
$$\int_0^{x_0} (N \sin \alpha - k.x)dx = M \int_p^{q} v~dv$$

Is this correct? And I suppose the value of $q$ = 0 but what will be the value for $p$?

Thanks

 

[haruspex]

I agree with your answer for a, but it can be simplified. Multiply top and bottom as necessary to get rid of the cot.
d is quite easy if you just consider energy.

 

[songoku]

I agree with your answer for a, but it can be simplified. Multiply top and bottom as necessary to get rid of the cot.

I got:
$$a=\frac{\frac{1}{2}mg \sin 2 \alpha}{M+m (\sin \alpha)^2}$$

d is quite easy if you just consider energy.

The equation will be: loss in GPE of m + gain in KE of m = strain energy ?

Thanks

 

[TSny]

∑F = M.a
N sin α = M.a ... (3)

Be sure to take into account all of the forces acting on M. Anything in contact with M can exert a force on M. If you consider the pulley as part of M, then note that the string is in contact with M. (I think @Lnewqban hinted at this in post #16.)

 

[etotheipi]

@TSny pointed out your error, but I'd like to provide a little more context just because in my experience pulleys are notoriously poorly explained in school.

Although this problem tells us to neglect friction between the string and the pulley, it's more instructive to consider the general case where there is friction (which we will assume to be limiting), and then - if you want - to set $\mu = 0$ afterwards.

Suppose that the section of string in contact with the pulley subtends an angle $(\pi - 2\alpha) / 2$, and superimpose a co-ordinate system whose $y$-axis coincides with the axis of symmetry.

Now consider a small element at an angle $\theta$ to the $x$-axis.

Balance the forces acting on the element tangential to the surface:$$T \left(\theta + \frac{d\theta}{2} \right) \cos{ \frac{d\theta}{2}} - T \left(\theta - \frac{d\theta}{2} \right) \cos{\frac{d\theta}{2}} - \mu N(\theta) = 0$$Balance the forces on the element normal to the surface:$$N(\theta) - T \left(\theta + \frac{d\theta}{2} \right) \sin{\frac{d\theta}{2}} - T \left(\theta - \frac{d\theta}{2} \right) \sin{\frac{d\theta}{2}} = 0$$You can simplify this by writing $\cos{\frac{d\theta}{2}} \sim 1 - \frac{d\theta^2}{2}$ and $\sin{\frac{d\theta}{2}} \sim \frac{d\theta}{2}$, and further by noticing that

$$T \left(\theta + \epsilon \right) = T(\theta) + T'(\theta) \epsilon + \mathcal{O}(\epsilon^2)$$If you write out those expansions then you'll obtain that, to first order in $d\theta$,$$T'(\theta) d\theta = \mu N(\theta) \ \ \ (1)$$and$$N(\theta) = T(\theta) d\theta \ \ \ (2)$$don't be put off by the fact that what I called $N(\theta)$ is actually differential form! You can just combine to get the differential equation$$T'(\theta) = \mu T(\theta)$$and you may be aware that this has a solution $T(\theta) = T(0)e^{\mu \theta}$; i.e. in the limiting case, the tension increases exponentially around the pulley. [Although the string is not in contact with the pulley at $\theta = 0$, you can just view $T(0)$ as some constant fixing the solution].

We're not there yet, though, because we want to ultimately work out the force that the pulley exerts on the string - and then by NIII, the force that the string exerts on the pulley. For example, the $y$-component of the contact force the pulley exerts on the string may be computed via an integral$$
\begin{align*}

\int_{\alpha}^{\pi - \alpha} dF_y = \int_{\alpha}^{\pi - \alpha}\left[ N(\theta) \sin{\theta} - \mu N(\theta) \cos{\theta} \right] &= \int_{\alpha}^{\pi - \alpha} (\sin{\theta} - \mu \cos{\theta}) T(\theta) d\theta \\ \\

&= T(0) \cos{\alpha} \left[ e^{\mu(\pi - \alpha)} + e^{\mu \alpha} \right]

\end{align*}$$But $T(0)e^{\mu(\pi - \alpha)} = T(\pi - \alpha)$ and $T(0)e^{\mu \alpha} = T(\alpha)$, so you can see the y-component of force that the pulley exerts on the string is$$F_y = T(\alpha) \cos{\alpha} + T(\pi - \alpha) \cos{\alpha}$$and thus by NIII, the string exerts $$\tilde{F}_y = -T(\alpha) \cos{\alpha} - T(\pi - \alpha) \cos{\alpha}$$on the pulley, or in other words, exactly equivalent to if the pulley were acted upon by two forces $T(\alpha)$ and $T(\pi - \alpha)$ in the direction of the string, even though the tension forces don't actually act on the pulley (they only act on neighbouring bits of string!). You can [and should!] do a similar calculation for the $x$-direction.

Another, perhaps more intuitive, way to think about it is to consider the string in contact with the pulley as part of the "pulley+string-in-contact-with-pulley system". Then, the force this system due to the string is just that exerted by the external axial tension on the bit of the string in contact with the pulley.

 

[haruspex]

Be sure to take into account all of the forces acting on M. Anything in contact with M can exert a force on M. If you consider the pulley as part of M, then note that the string is in contact with M. (I think @Lnewqban hinted at this in post #16.)

Damn! I had that originally then mislaid the scrap of paper. Thanks for picking it up.

 

[TSny]

Damn! I had that originally then mislaid the scrap of paper.

I knew you were aware of the string's force on M as evidenced by your replies in similar threads.

 

[etotheipi]

Well, it's a well known fact that @haruspex never makes mistakes...

 

[haruspex]

Well, it's a well known fact that @haruspex never makes mistakes...

"What never?"
"No never!"
"What never??"
"Well, hardly ever!"
(HMS Pinafore)
I know what happened... I had meantime been on another thread with a block and a pulley on a moving wedge, but in that case the string was tied to the wall.