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

[haruspex]

For (e), I tried to use conservation of energy again but got stuck

You should start by thinking about what defines the equilibrium point. What would happen if you brought the system to rest at that point and released it?

 

[songoku]

You should start by thinking about what defines the equilibrium point. What would happen if you brought the system tonrest at that point and released it?

I guess $M$ will move to the left due to restoring force of spring and $m$ will move up the slope, starting with maximum velocity and minimum acceleration (zero)

 

[haruspex]

I guess M will move

Then it was not at equilibrium.

 

[songoku]

Then it was not at equilibrium.

By equilibrium point, do you mean equilibrium point of simple harmonic motion? What I know about equilibrium point of simple harmonic motion is it is the mid point of the oscillating motion where the velocity will be maximum and acceleration will be zero.

That's why I thought if I brought the system to rest at equilibrium point and then released it, it woukd execute such motion

Thanks

 

[haruspex]

By equilibrium point, do you mean equilibrium point of simple harmonic motion? What I know about equilibrium point of simple harmonic motion is it is the mid point of the oscillating motion where the velocity will be maximum and acceleration will be zero.

That's why I thought if I brought the system to rest at equilibrium point and then released it, it woukd execute such motion

Thanks

An equilibrium state of a system is a stable state. I.e a state in which it will remain unless perturbed.
Generally speaking, that can include dynamic equilibrium, such as a wheel rolling. But here it means a state in which there is no motion.

 

[songoku]

An equilibrium state of a system is a stable state. I.e a state in which it will remain unless perturbed.
Generally speaking, that can include dynamic equilibrium, such as a wheel rolling. But here it means a state in which there is no motion.

I think I imagine the motion of the system wrongly.

The equation of motion for $M$ is:
$$\Sigma F = M.a$$
$$N \sin \alpha +T-T \cos \alpha-kx=M.a$$

Since acceleration is not directly proportional to displacement, the system does not perform simple harmonic motion.

But the question states that $M$ will oscillate so it will move continuously left and right, just not in simple harmonic motion? If so, will there be a point which there is no motion? Since there is no friction, I think it will be continuously in motion.

Or by equilibrium point you mean the position where the speed is zero (hence no motion)? Won't it be the same as $x_0$?

Thanks

 

[haruspex]

Since acceleration is not directly proportional to displacement

How do you know it isn"t? Thar equation has T and N in it, which might also be proportional to x.

it will move continuously left and right, just not in simple harmonic motion?

It does not need to be SHM for this part of the question, but probably is.

by equilibrium point you mean the position where the speed is zero

No, it is the position it would stay in if you were to stop it there and then let go.
For a pendulum, it would be hanging vertically.
For a mass hanging from a spring, it would be when the force in the spring is mg.
In SHM it is the mid point of the oscillation.

 

[songoku]

No, it is the position it would stay in if you were to stop it there and then let go.
For a pendulum, it would be hanging vertically.
For a mass hanging from a spring, it would be when the force in the spring is mg.
In SHM it is the mid point of the oscillation.

In this case, it will be the mid point of oscillation? If I bring the system to stop at that point and then let go, the system will stay at rest, maybe because the resultant force is zero ($N \sin \alpha + T-T \cos \alpha=kx$). Am I correct?

Thanks

 

[haruspex]

In this case, it will be the mid point of oscillation?

Maybe, but you don't need to assume that.

because the resultant force is zero ($N \sin \alpha + T-T \cos \alpha=kx$).

Yes.

 

[songoku]

Yes.

If I use $\Sigma F = 0$ to find equilibrium position:

$$N \sin \alpha + T - T \cos \alpha=kx$$

I don't know what to do with $N$ and $T$ and I think I can not use $N$ and $T$ that I found in part (a) because both $N$ and $T$ depend on acceleration and initial acceleration when the system moves for the first time will not be same as acceleration at equilibrium point.

Or maybe there is other way to find the equilibrium position?

Thanks

 

[haruspex]

If I use $\Sigma F = 0$ to find equilibrium position:

$$N \sin \alpha + T - T \cos \alpha=kx$$

I don't know what to do with $N$ and $T$ and I think I can not use $N$ and $T$ that I found in part (a) because both $N$ and $T$ depend on acceleration and initial acceleration when the system moves for the first time will not be same as acceleration at equilibrium point.

Or maybe there is other way to find the equilibrium position?

Thanks

We are now faced with a standard 2D statics problem. TYpically, there are three balance equations available for each object that can move independently. Here you have two objects. Some equations would be unhelpful because they bring extra unknowns, e.g. the vertical force balance on the wedge, which would involve the normal force from the ground.
Which two additional equations will be useful?

 

[songoku]

We are now faced with a standard 2D statics problem. TYpically, there are three balance equations available for each object that can move independently. Here you have two objects. Some equations would be unhelpful because they bring extra unknowns, e.g. the vertical force balance on the wedge, which would involve the normal force from the ground.
Which two additional equations will be useful?

Equation for $m$ in x-direction: $T \cos \alpha = N \sin \alpha$

Equation for $m$ in y-direction: $mg = N \cos \alpha$

Combining those three equations, I get:
$$x=\frac{mg \tan \alpha}{k \cos \alpha}$$

For question (f), should we check first if the motion is SHM or not? Or there is other method?

Thanks

 

[haruspex]

Equation for $m$ in y-direction: $mg = N \cos \alpha$

Are those the only forces acting on m with vertical components?

 

[songoku]

Are those the only forces acting on m with vertical components?

I am so pissed I keep forgetting about tension acting on $m$. I think I can do question (e), adding $T \sin \alpha$ in equation for vertical component of $m$ and just find $x$ using algebra

How to do question (f)?

Thanks

 

[haruspex]

I can do question (e), adding Tsin⁡α in equation for vertical component of m and just find x using algebra

Actually it is easier than that. Treat the mass and wedge as a single unit.

How to do question (f)?

One way is to write out all the force and acceleration equations again, but this time for the general case of the wedge being at some displacement x from its equilibrium position and moving with velocity $\dot x$.
Another is to write out the sum of KE and GPE at that general position.
I think the second is less work, but you do need to be careful to get the KE of m correct.

 

[vcsharp2003]

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

View attachment 280353

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

I think these types of problems where there is a motion within another motion or multiple motions relative to ground unfolding in a complex manner, it's best to use the accelerating object as a frame of reference i.e. a non-inertial frame of reference to simplify things and then add a pseudo force on the other object being observed from an accelerating frame of reference.

In this situation, let's have an observer on accelerating wedge. This observer will see the block m as stationary since it's tied to the wall with a fixed length string. If a is the initial acceleration of wedge then apply a force ma in opposite direction on block m. We are focusing on the instant t=0 since we need the acceleration of wedge at t=0.

Below are the FBDs that I get for block m relative to accelerating wedge and for wedge relative to ground. Note that at t=0 there is no spring force since the problem states that the spring is not stretched initially. The second FBD is relative to ground, while first is relative to wedge.

Block observed from wedge

 

[haruspex]

I think these types of problems where there is a motion within another motion or multiple motions relative to ground unfolding in a complex manner, it's best to use the accelerating object as a frame of reference i.e. a non-inertial frame of reference to simplify things and then add a pseudo force on the other object being observed from an accelerating frame of reference.

In this situation, let's have an observer on accelerating wedge. This observer will see the block m as stationary since it's tied to the wall with a fixed length string. If a is the initial acceleration of wedge then apply a force ma in opposite direction on block m. We are focusing on the instant t=0 since we need the acceleration of wedge at t=0.

Below are the FBDs that I get for block m relative to accelerating wedge and for wedge relative to ground. Note that at t=0 there is no spring force since the problem states that the spring is not stretched initially. The second FBD is relative to ground, while first is relative to wedge.

Block observed from wedge
View attachment 281133
View attachment 281134

That was done in posts #25 and #27, except that horizontal and vertical coordinates were used.

 

[songoku]

I am really sorry for late reply

Actually it is easier than that. Treat the mass and wedge as a single unit.

Treating $m$ and $M$ as one unit, I get equation: $T=kx$ so combining this with horizontal and vertical equation for $m$, I get:
$$x=\frac{mg \sin \alpha}{k}$$

One way is to write out all the force and acceleration equations again, but this time for the general case of the wedge being at some displacement x from its equilibrium position and moving with velocity $\dot x$.
Another is to write out the sum of KE and GPE at that general position.
I think the second is less work, but you do need to be careful to get the KE of m correct.

Let $M$ moves a distance $d$ to the right from initial position. Equation of conservation of energy at this position:
$$mgd \sin \alpha=\frac 1 2 M {v_M}^2+\frac 1 2 m {v_m}^2 +\frac 1 2 k d^2$$

I don't know how to change $v_M$ and $v_m$. I can't use kinematics equation since the acceleration is not constant and I also don't know whether the motion is simple harmonic or not. I don't how this equation will lead to period of the motion.

I also tried to write down equation of Newton's 2nd law:
Equation for horizontal direction of $m$:
$$T \cos \alpha -N \sin \alpha=m.(a-a \cos \alpha)$$

Equation for vertical direction of $m$:
$$mg - N \cos \alpha-T \sin \alpha=m.a \sin \alpha$$

Equation for $M$:
$$T+N \sin \alpha-T \cos \alpha-kx=M.a$$

I also don't know how to change $a$

Thanks

 

[haruspex]

I am really sorry for late reply

Treating $m$ and $M$ as one unit, I get equation: $T=kx$ so combining this with horizontal and vertical equation for $m$, I get:
$$x=\frac{mg \sin \alpha}{k}$$Let $M$ moves a distance $d$ to the right from initial position. Equation of conservation of energy at this position:
$$mgd \sin \alpha=\frac 1 2 M {v_M}^2+\frac 1 2 m {v_m}^2 +\frac 1 2 k d^2$$

I don't know how to change $v_M$ and $v_m$. I can't use kinematics equation since the acceleration is not constant and I also don't know whether the motion is simple harmonic or not. I don't how this equation will lead to period of the motion.

I also tried to write down equation of Newton's 2nd law:
Equation for horizontal direction of $m$:
$$T \cos \alpha -N \sin \alpha=m.(a-a \cos \alpha)$$

Equation for vertical direction of $M$:
$$mg - N \cos \alpha-T \sin \alpha=m.a \sin \alpha$$

Equation for $M$:
$$T+N \sin \alpha-T \cos \alpha-kx=M.a$$

I also don't know how to change $a$

Thanks

I agree with your answer for e.
For f, your equations look good.
Use them to eliminate N and T. This should leave you with an equation relating a and x.

 

[songoku]

Use them to eliminate N and T. This should leave you with an equation relating a and x.

I get:
$$T=mg \sin \alpha + ma \cos \alpha - ma$$

and
$$a=\frac{mg \sin \alpha -kx}{M + 2m(1-\cos \alpha)}$$

How to proceed to find period? Thanks

 

[haruspex]

I get:
$$T=mg \sin \alpha + ma \cos \alpha - ma$$

and
$$a=\frac{mg \sin \alpha -kx}{M + 2m(1-\cos \alpha)}$$

How to proceed to find period? Thanks

Remember that a is $\ddot x$. The standard form of the SHM ODE is $\ddot x+\omega^2x=0$, but that is taking x as offset from the equilibrium position. Because your x is not measured from there the form you have is $\ddot x+\omega^2x=C$.
You could use the equilibrium position you found to adjust your x to be an offset from there. If you have done everything correctly c should disappear. Or you could just trust that it will disappear and ignore it.

 

[songoku]

Remember that a is $\ddot x$. The standard form of the SHM ODE is $\ddot x+\omega^2x=0$, but that is taking x as offset from the equilibrium position. Because your x is not measured from there the form you have is $\ddot x+\omega^2x=C$.
You could use the equilibrium position you found to adjust your x to be an offset from there. If you have done everything correctly c should disappear. Or you could just trust that it will disappear and ignore it.

Oh, so $\omega^2$ will be $\frac{k}{M+2m(1-\cos \alpha)}$ then using $\omega=\frac{2 \pi}{T}$ we can find the period

Sorry I have several questions:

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

Why can't I say $m$ will move to the right with respect to the ground? My reasoning is since the horizontal acceleration of $m$ is $a-a \cos \alpha$ and $|a|>|a \cos \alpha|$ and direction of $a$ is to the right, so horizontal acceleration of $m$ is to the right.

Or do you mean the direction of motion of $m$ with respect to the ground should be the resultant of horizontal and vertical direction?

For 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

Is this valid approach for (d)? If yes, what would be the limit for the RHS integration?

For (e), I tried to use conservation of energy again but got stuck

Let:
$d$ = distance move by $M$ from initial position to reach equilibrium point
$v_m$ = speed of $m$ after moving down the slope as far as $d$
$v_M$ = speed of $M$ after moving a horizontal distance $d$ to the right
Ground level is the position of $m$ after moving down a distance $d$ down the slope

I compare the initial condition where the system is at rest and when $M$ reaches equilibrium point

$$mg~d \sin \alpha = \frac1 2 m (v_m)^2 + \frac 1 2 M (v_M)^2 + \frac 1 2 k d^2$$

$v_m$ and $v_M$ will be the maximum speed of the oscillation so $v_m = v_M = \omega A$ where $A$ is the amplitude of oscillation and $A=x_0-d$

So:
$$mg~d \sin \alpha = \frac1 2 m (v_m)^2 + \frac 1 2 M (v_M)^2 + \frac 1 2 k d^2$$
$$mg~d \sin \alpha=\frac 1 2 m (\omega (x_0-d))^2 + \frac 1 2 M (\omega (x_0-d))^2 + \frac 1 2 kd^2$$

How to get rid $\omega$ from the equation? Thanks

Can I change $\omega$ in the last equation with $\omega = \sqrt{\frac{k}{M+m}}$

I think these types of problems where there is a motion within another motion or multiple motions relative to ground unfolding in a complex manner, it's best to use the accelerating object as a frame of reference i.e. a non-inertial frame of reference to simplify things and then add a pseudo force on the other object being observed from an accelerating frame of reference.

In this situation, let's have an observer on accelerating wedge. This observer will see the block m as stationary since it's tied to the wall with a fixed length string. If a is the initial acceleration of wedge then apply a force ma in opposite direction on block m. We are focusing on the instant t=0 since we need the acceleration of wedge at t=0.

Below are the FBDs that I get for block m relative to accelerating wedge and for wedge relative to ground. Note that at t=0 there is no spring force since the problem states that the spring is not stretched initially. The second FBD is relative to ground, while first is relative to wedge.

Block observed from wedge
View attachment 281133
View attachment 281134

Where can I learn about non-inertial frame in more detail?

Thanks

 

[haruspex]

Why can't I say m will move to the right with respect to the ground?

You can. I was handling two threads at once which had remarkably similar problems. m would have moved left in the other one.

Is this valid approach for (d)?

Yes, but you forgot the force from the string on the pulley again.
p and q would be the initial and final speeds, which are both zero here.
You should get the same equation you got in post #46.

 

[haruspex]

Can I change $\omega$ in the last equation with $\omega = \sqrt{\frac{k}{M+m}}$

No. Using KE was harder than I thought. Stick with your force approach.

 

[songoku]

Thank you very much for all the help and explanation haruspex, Lnewqban, Tsny, etotheipi, vcsharp2003

 

[vcsharp2003]

Where can I learn about non-inertial frame in more detail?

Its really very simple as explained below. You can look up https://en.wikipedia.org/wiki/Fictitious_force for more details.

If you put an observer in an accelerating frame of reference then the object being observed will have all "real" forces acting on it like gravity, tension, friction, normal reaction etc. PLUS a pseudo force that equals ma in magnitude having a direction opposite to the acceleration of frame of reference where a is acceleration magnitude of reference frame and m is mass of object being observed . We can then apply the Newtons Second Law of Motion just as we normally do in a non-inertial frame of reference. This is a standard practice when solving complex problems where one of the object is moving in a complex and hard-to-predict manner; the pseudo force simplifies our analysis of the problem greatly. In my view, the problem that you posted under this thread is an ideal candidate for such an analysis since it makes it very simple as the smaller mass is in equilibrium with respect to the wedge.

A common area of application for pseudo force is in uniform circular motion where the object in circular motion has a radial acceleration towards the center of the circle. If we now put an observer in a frame of reference that is having the same circular motion as the object then we have our observer in an accelerating frame of reference and the object will appear at rest relative to the frame of reference. We apply a force ma = mv²/r in a radially outward direction. This radial outward force is called centrifugal force and it's an example of a pseudo force.

Pseudo force is a fictitious force which has no real existence, but it helps us to apply Newton's Second Law of Motion from an accelerating frame of reference. As a rule, Newton's Second Law of Motion can only be applied when observer is in an inertial frame of reference (i.e. frame is at rest or moving at a constant velocity relative to ground).

 

[songoku]

Thank you again vcsharp2003 for al the explanation