Solve Box Tipping Problem: μk=0.2, Weight=50kg

[jonnyjonny]

Homework Statement

The forces needed to consider for the tipping of a box

Relevant Equations

Moments
Forces

I would like to know the considerations needed to find the tipping of a box. It is given μk=0.2 and the box is 50kg

There are actually 2 ways of solving the problem:

1. To consider the moment about the centre of mass of the box, given that the normal force on the box will be redistributed.
So, considering moment about center of mass,
clockwise moment= Ff(0.5) + F(0.3)= 50(9.81)(μk)(0.5)+(600)(0.3)=229.05Nm
anticlockwise= (50)(9.81)(x) =490.5x Nm [Normal force x distance, x from the centre of mass]
For net moment about center of mass to = 0,
490.5x=229.05
∴x=0.467< 0.5 (Therefore, box won't tip because the normal force does not leave the box)

2. However, we can also consider the moment about the bottom right corner of the box, where it will potentially tip at.
In this case,
anticlockwise moment= W(0.5)= 245 Nm
clockwise moment= N(0.5-x) +600(0.8) = 496.18 Nm
∴ clockwise moment will result.

There seems to be a contradiction with the 2 methods as shown above. Could someone point out any errors or concepts that could have been missed?

 

[TSny]

If the center of mass has acceleration, then the relation $\sum \tau = I \alpha$ will hold if you take the origin to be the center of mass, as you did in your first method. But the relation does not generally hold about other points chosen as origin, as in your second method.

Suppose the crate is in free space (no gravity, friction or normal force) and the 600 N force is applied as shown

Here, the line of action of the force passes through the center of mass of the crate as the crate accelerates to the right. Consider the net torque about (1) center of mass, (2) lower right corner.

If you go to a frame of reference accelerating with the center of mass of the crate, then you can apply the principle of torques about any point if you include a "fictitious" force due to being in an accelerating reference frame.

 

[kuruman]

I am not sure what you are calculating in method 1. The use of $\mu_k N$ for the force of kinetic friction implies that the block is already sliding. If that is the case, why should it tip and not keep on sliding? I would go with method 2, but for that to work, you must first ascertain that the box will not start sliding before it starts tipping. For that you will need the coefficient of static friction which apparently is not given.

 

[Cutter Ketch]

I think you are stuck on a false premise. We know $\Sigma \Gamma = I \alpha$. We know that in the case of statics, i.e. when you are assured nothing is moving in some inertial reference frame and $\alpha$ is definitely zero, then you can say $\Sigma \Gamma = 0$ about any point. However, in the general case, there is no reason to expect the torques about any two points to add up to the same value or even the same direction. The moment of inertia depends on the choice of axis, the value of acceleration depends upon axis, and so do the torques.

Further, even if the sum of the torques about some axis add up to 0, that is not sufficient to say the object has no angular acceleration. All you can say is it has no angular acceleration about that axis.

It is very easy to construct all kinds of scenarios where the torque is zero about one axis but not others. Put your friction and normal force at the corner of the box and make the weight 460N. Or, take for example two opposed and displaced forces putting a torque on a box. Now suppose they aren’t quite parallel. Somewhere in space they cross. The torque about that point is zero.

If you pick an axis, calculate the torques and forces and momentum’s, as well as the linear forces and accelerations, you will get the correct motion. Some are just easier to calculate than others.

So, to summarize: if an object isn’t moving in some inertial reference frame, the sum of the torques about any axis add up to zero. However, the opposite is not true. Even if the sum of the torques add up to zero about some particular axis that doesn’t mean the object isn’t being torqued about some other axis.

 

[Cutter Ketch]

If the center of mass has acceleration, then the relation ∑τ=Iα∑τ=Iα\sum \tau = I \alpha will hold if you take the origin to be the center of mass, as you did in your first method. But the relation does not generally hold about other points chosen as origin, as in your second method.

Newton’s laws don’t hold true in an inertial reference frame?? In the case of your example with the box, picking the lower right corner of the box (in an inertial reference frame, i.e. not accelerating with the box), it certainly does hold true. The linear acceleration shown IS an angular acceleration about that point. $\Sigma \Gamma = I \alpha$ will always hold true in an inertial reference frame. Of course you have to calculate I properly for the chosen axis.

 

[TSny]

Newton’s laws don’t hold true in an inertial reference frame?? In the case of your example with the box, picking the lower right corner of the box (in an inertial reference frame, i.e. not accelerating with the box), it certainly does hold true. The linear acceleration shown IS an angular acceleration about that point. $\Sigma \Gamma = I \alpha$ will always hold true in an inertial reference frame. Of course you have to calculate I properly for the chosen axis.

Newton's rotational law in its general form is $\vec {\tau}_{\rm net} = \frac{d \vec L}{dt}$. This law holds in any inertial frame about any fixed point in the frame. However, I don't believe that you can generally replace $\frac{d \vec L}{dt}$ by $I \vec{\alpha}$.

So, $\vec {\tau}_{\rm net} = I \vec{\alpha}$ is not generally true, even in an inertial frame. However, it can be shown to be true for a rigid body if the center of mass is taken as the origin: $\vec {\tau}_{\rm net, cm} = I_{\rm cm} \vec{\alpha}$

For example, consider a single particle acted on by a constant force $\vec F$ as shown

The torque about the origin $O$ is constant: $\vec \tau = Fb \, \hat k$. How would you define $I$ and $\alpha$ so that $\tau = I \alpha$ about the origin?

 

[Cutter Ketch]

Newton's rotational law in its general form is $\vec {\tau}_{\rm net} = \frac{d \vec L}{dt}$. This law holds in any inertial frame about any fixed point in the frame. However, I don't believe that you can generally replace $\frac{d \vec L}{dt}$ by $I \vec{\alpha}$.

So, $\vec {\tau}_{\rm net} = I \vec{\alpha}$ is not generally true, even in an inertial frame. However, it can be shown to be true for a rigid body if the center of mass is taken as the origin: $\vec {\tau}_{\rm net, cm} = I_{\rm cm} \vec{\alpha}$

For example, consider a single particle acted on by a constant force $\vec F$ as shown
View attachment 260714

The torque about the origin $O$ is constant: $\vec \tau = Fb \, \hat k$. How would you define $I$ and $\alpha$ so that $\tau = I \alpha$ about the origin?

Yes, agreed. $\frac {dL} {dt}$. Sorry.

 

[kuruman]

The condition for sliding is $P \geq \mu_smg$. Sliding starts at $P = \mu_smg$.
The condition for tipping is $Ph\geq mg\dfrac{w}{2}$ where $h$ is the height at which $P$ is applied and $w$ is the width of the box. Tipping starts at $P = mg\dfrac{w}{2h}$.

For the box to tip before it starts sliding, the sliding threshold must be higher than the tipping threshold: $$\mu_smg > mg\frac{w}{2h}~\rightarrow~\mu_s>\frac{w}{2h}.$$Here, $h=0.8$ m and $w=1.0$ m which results in $\mu_s>0.4$. No information about $\mu_s$ is provided, only that $\mu_k=0.2$. If we assume that $\mu_s=0.2$ because that's all that's given, the box will slide before it tips.

 

[TSny]

Homework Statement:: The forces needed to consider for the tipping of a box
View attachment 260691

I would like to know the considerations needed to find the tipping of a box. It is given μk=0.2 and the box is 50kg

Please give the complete wording of the statement of the problem.

 

[jonnyjonny]

@TSny
The problem is not very relevant to the question I had but here it is.
Thanks a lot for your solutions anyway... I think I understand it a lot better now.

 

[jonnyjonny]

The condition for sliding is $P \geq \mu_smg$. Sliding starts at $P = \mu_smg$.
The condition for tipping is $Ph\geq mg\dfrac{w}{2}$ where $h$ is the height at which $P$ is applied and $w$ is the width of the box. Tipping starts at $P = mg\dfrac{w}{2h}$.

For the box to tip before it starts sliding, the sliding threshold must be higher than the tipping threshold: $$\mu_smg > mg\frac{w}{2h}~\rightarrow~\mu_s>\frac{w}{2h}.$$Here, $h=0.8$ m and $w=1.0$ m which results in $\mu_s>0.4$. No information about $\mu_s$ is provided, only that $\mu_k=0.2$. If we assume that $\mu_s=0.2$ because that's all that's given, the box will slide before it tips.

Im not too sure about using this... because if we assume tipping starts at Ph = mg(w/2), for the example Ph= 480Nm clockwise and mg(w/2)=245.25 Nm anticlockwise. The moment provided by the force will always be greater... furthermore, given that the normal gradually gets redistributed and its line of action gradually shifts from the centre to the right edge... it would always be acting to reinforce the clockwise moment to tilt it...
So I think the analysis would not work...

 

[haruspex]

we can also consider the moment about the bottom right corner of the box,

One has to be careful what is meant by that.
If you mean the corner of the box, however that might move, then you have to allow for the fact that it accelerates, so you have a non-inertial frame.
Safe is to take the initial position of the corner as a fixed point on the ground, and include the linear acceleration of the box's mass centre, multiplied by its height, as part of the angular acceleration about the axis.

 

[haruspex]

the box will slide before it tips.

Yes, but it can tip too. Bear in mind that the applied horizontal force is given as a constant. If it slips, the force has to do correspondingly more work.