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

[songoku]

@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.

View attachment 280600

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

View attachment 280585

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.

I really appreciate you type all this explanation to me but my brain cell just can't process this knowledge. Reading this so many times for these past 2-3 days and I still can't understand. Maybe I will leave this for now before my already few brain cells get permanent damage. I am sorry, and thank you very much for you effort

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.)

Is it correct that the horizontal component of tension acting on M is T sin α so the equation will be:
ΣF = M.a
N sin α + T sin α = M.a

Thanks

 

[haruspex]

Is it correct that the horizontal component of tension acting on M is T sin α

The same magnitude tension acts on the pulley at two different angles.
What is the horizontal component of each?
What is the net horizontal force?

 

[songoku]

The same magnitude tension acts on the pulley at two different angles.
What is the horizontal component of each?
What is the net horizontal force?

Net horizontal force of tension on M is T - T cos α ?

So equation for M is:
ΣF = M.a
N sin α + T - T cos α = M.a ?

Thanks

 

[haruspex]

Net horizontal force of tension on M is T - T cos α ?

So equation for M is:
ΣF = M.a
N sin α + T - T cos α = M.a ?

Thanks

Yes.

 

[songoku]

For question (d)

Change in GPE - change in KE = strain energy

mg x₀ sin α - 1/2 . m . (2a x₀) = 1/2 . k . (x₀)²

Then solve for x₀

Is that correct?

Thanks

 

[haruspex]

For question (d)

Change in GPE - change in KE = strain energy

mg x₀ sin α - 1/2 . m . (2a x₀) = 1/2 . k . (x₀)²

Then solve for x₀

Is that correct?

Thanks

Change in KE between what two states of motion?

 

[songoku]

Change in KE between what two states of motion?

Change in KE of $m$ before it moves and after it moves down the plane as far as $x_0$

 

[haruspex]

Change in KE of $m$ before it moves and after it moves down the plane as far as $x_0$

And what is its KE at x₀?

 

[songoku]

And what is its KE at x₀?

Ah I used wrong acceleration for $m$. I have to use $a_m = \sqrt {(a_x)^2+(a_y)^2}$ then use $v^2 = u^2 + 2a_mx_0$ to find speed of $m$ at $x_0$ and putting $u=0$. Am I correct?

 

[haruspex]

Ah I used wrong acceleration for $m$.

It compresses a spring. Acceleration will not be constant.
Note how x₀ is defined:
"after moving a distance xo towards the wall, M will stop"

 

[songoku]

It compresses a spring. Acceleration will not be constant.
Note how x₀ is defined:
"after moving a distance xo towards the wall, M will stop"

Initially the system is at rest so no KE and taking the position where $m$ will stop moving after going down the slope as far as $x_0$ as ground level, $m$ will have initial GPE with height equal to $x_0 \sin \alpha$

After moving a distance $x_0$, $M$ and $m$ will stop and the spring will be compressed so at this state there is no KE so the loss in GPE of $m$ will be equal to strain energy stored by the spring.

$$m~g~x_0 \sin \alpha=\frac 1 2 k~(x_0)^2$$

Is this correct?

If yes, I have another question regarding acceleration of oscillation. I learn that the acceleration of simple harmonic motion will be maximum at maximum displacement but in this case it looks like the acceleration is zero when the spring is compressed by $x_0$ so it seems that the acceleration is minimum at maximum displacement of the oscillation. What am I missing?

Thanks

 

[haruspex]

Is this correct

Yes.

it looks like the acceleration is zero when the spring is compressed by x0

It will be zero velocity but max acceleration.

 

[songoku]

It will be zero velocity but max acceleration.

So the acceleration of $m$ and $M$ will be maximum at $x_0$? It means that when while $M$ is moving to the right, its acceleration increases but its velocity decreases? And also the initial acceleration of $M$ is to the right at the instant it start to moves but when the spring has compression, the acceleration of $M$ changes direction to the left?

Thanks

 

[haruspex]

So the acceleration of $m$ and $M$ will be maximum at $x_0$? It means that when while $M$ is moving to the right, its acceleration increases but its velocity decreases? And also the initial acceleration of $M$ is to the right at the instant it start to moves but when the spring has compression, the acceleration of $M$ changes direction to the left?

Thanks

The equilibrium point (for M) of the system will be somewhere between the initial position and the displacement x₀. Until M reaches that point it will accelerate to the right, with decreasing magnitude. At the equilibrium point acceleration will be zero. Thereafter, acceleration is to the left.

 

[songoku]

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

 

[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