Maximum acceleration given a breakable edge

[Combinatorics]

Homework Statement

A mass $$m=12 kg$$ is placed on a cube, as shown in the attached figure. The cube is made of a breakable material, that breaks when a force larger than $$F_{max}=150N$$ is exerted on it.
Someone pulls the cube upwards with an acceleration of $$a \, \frac{meter}{sec^2}$$. What is the maximal acceleration possible for which the cube edges will not break? ($$\theta = \frac{\pi}{6} $$)

Relevant Equations

Newton's laws,
g=9.8 m/s^2

Using Newton's second law,
$$
cos\left( \frac{\pi}{6} \right) m(a+g) = 150 \Leftrightarrow a = \frac{\frac{2\times 150 }{\sqrt{3}}-mg}{m} =4.633.
$$
Unfortunately, the possible answers are
A. 1.025, B. 0.625, C. 3.75, D. 2.75, E. 1.75, F. 0.

What am I getting wrong? Isn't the force exerted on the inclined part of the cube equal to $cos\left( \frac{\pi}{6} \right) m(a+g) $?

Thanks

 

[BvU]

Hi,

You can check the expression by taking an extreme case:

Isn't the force exerted on the inclined part of the cube equal to $$cos\left( \theta \right) m(a+g) \qquad ?$$

Does that go to zero for $\theta\downarrow 0\$ ?

 

[Combinatorics]

Hi,

You can check the expression by taking an extreme case:

Does that go to zero for $\theta\downarrow 0\$ ?

Hi, why should the force tend to zero when $\theta\downarrow 0\$ ? The force indeed tends to zero when $\theta\uparrow \frac{\pi}{2}\$ which seems reasonable to me (as this expression represents the force exerted on the lower, inclined part of the cube. What am I getting wrong here?
Thank yoU!

 

[BvU]

Hi, why should the force tend to zero when $\theta\downarrow 0\$ ?

What is supposed to break first is the edge to the right of the mass. At the dashed line (the problem statement says it's a point mass ?! ).
If $\theta=0$ the mass doesn't lean into that edge

 

[Combinatorics]

What is supposed to break first is the edge to the right of the mass. At the dashed line (the problem statement says it's a point mass ?! ).
If $\theta=0$ the mass doesn't lean into that edge

View attachment 276822

You are right. The problem statements says it's a mass, and not a point mass. Apologize for this.
If $\theta = 0$ there is a force exerted on the bottom edge, which would break if $a$ is large enough. Isn't this correct? In any case, isn't the force exerted on the bottom (inclined edge) equal to what I stated? Why there is a force exerted on the right edge if both $a$ and $g$ are perpendicular to it?

Thanks!

 

[BvU]

Why there is a force exerted on the right edge if both a and g are perpendicular to it?

Check the direction of the normal force on the mass

 

[haruspex]

What is supposed to break first is the edge to the right of the mass.

Are you sure? It doesn't actually say that, and assuming it I get 11.85, nowhere near any of the options.
@Combinatorics , the title says "breakable edge", but the problem statement does not mention edge. Have you quoted the problem statement exactly?

Maybe the cube is hollow, in which case there are three normal forces on it to consider. The only one which seems to match an option is the sloped surface.

$$
\cos\left( \frac{\pi}{6} \right) m(a+g) = 150 \Leftrightarrow a = \frac{\frac{2\times 150 }{\sqrt{3}}-mg}{m} =4.633.
$$

How do you get $\cos(\theta) m(a+g)$ for the normal force on the sloped surface? Which direction did you consider for the balance of forces?

 

[BvU]

Are you sure?

I'm never absolutely certain (comes with experience) but I give it the best shot I can. I did see my approach doesn't give any of the choices, but (also experience) don't always trust given answers (*).

And this exercise isn't all that trustworthy in the first place: the ' that breaks when a force larger than 150 N' is so vague, it's almost a disqualifier. Again: I would be happy to be proven wrong.

(*) and vice versa: I can't find 150 N from the given answers

 

[Steve4Physics]

Using Newton's second law,
$$cos\left( \frac{\pi}{6} \right) m(a+g) = 150 \text{...}$$

$cos\left( \frac{\pi}{6} \right) m(a+g) = 150$ is wrong.
The vertical forces on the ball are:
- the upwards component of the 150N force
- the downwards weight.
These produce the vertical acceleration.

[EDIT. If not clear, the 150N force is the normal reaction between ball and 'slope'. Also, the reaction force on the side of the ball is smaller, so does not determine the breaking-point (which is not hard to prove).]

 

[Combinatorics]

Hi guys,

I went over all the comments here, and understood that my force balance is wrong. Unfortunately, I have no idea where is my mistake exactly.
The way I obtained the equation $\cos\left( \frac{\pi}{6} \right) m(a+g) = 150$ is:
By calculating the forces acting on the sloped part of the cube, we have (from Newton's 2 law):
$N-mg \cos\left( \frac{\pi}{6} \right) = ma \cos\left( \frac{\pi}{6} \right)$, so that $N=mg \cos\left( \frac{\pi}{6} \right) + ma \cos\left( \frac{\pi}{6} \right)$, and this should be less than or equal to $150 N$. Will you help me understand what is exactly wrong here?

Thank you all!

 

[haruspex]

By calculating the forces acting on the sloped part of the cube, we have (from Newton's 2 law):
$N-mg \cos\left( \frac{\pi}{6} \right) = ma \cos\left( \frac{\pi}{6} \right)$,

No, there are three forces acting on the cube, and since we are not given a mass, gravity is not one of them.
I think you mean from analysing the forces on the mass. They are gravity and two normal forces. All three have a component parallel to the slope.

 

[Steve4Physics]

I think you have mis-identified which force is 150N when the cube breaks,

Refer to the free-body diagram of the ball (which you should already have drawn!). You will see there are 3 forces acting on the ball:
1) it's weight, mg
2) the normal reaction, R₁, from the sloping contact surface of the cube
3) the normal reaction, R₂, from the vertical contact surface of the cube.

This means the 2 forces acting on the cube which could break it are the Newton 3rd law partners of R₁and R₂. These, of course, have the same magnitudes as R₁and R₂.

It is not hard to show |R₁|>|R₂|. This means the cube breaks if R₁>150N. So you need to find the acceleration which makes R₁=150N.

[EDIT: There is a 3rd force acting on the cube - the external force which accelerates the whole system upwards. I'm discounting this as a force which causes the breakage as I believe the intent of the question is to find R₁.]

If you are not clear how to proceed, see post #9 for a hint.

 

[haruspex]

I think you have mis-identified which force is 150N when the cube breaks,

AS I read @Combinatorics' work, there is no disagreement over which normal force matters, only a miscalculation of its relationship to mg etc.

 

[Steve4Physics]

AS I read @Combinatorics' work, there is no disagreement over which normal force matters, only a miscalculation of its relationship to mg etc.

Agreed. I was wrong to say ‘mis-identified’. @Combinatorics has identified the correct force - the normal reaction on the sloping surface.

But $m(a+g)$ is the normal force exerted on a vertically accelerating horizontal surface (as in a simple elevator problem). $\cos\left( \frac{\pi}{6} \right) m(a+g)$ is its component 30º to the vertical.

In this problem, the surface is not horizontal, so the component of $m(a+g)$ is the wrong force. That’s the mistake.

Can I encourage @Combinatorics to draw a free-body diagram of the ball and apply Newton’s 2nd law?

 

[Steve4Physics]

AS I read @Combinatorics' work, there is no disagreement over which normal force matters, only a miscalculation of its relationship to mg etc.

Agreed. I was wrong to say ‘mis-identified’. @Combinatorics has identified the correct force - the normal reaction on the sloping surface.

But $m(a+g)$ is the normal force exerted on a vertically accelerating horizontal surface (as in a simple elevator problem). $\cos\left( \frac{\pi}{6} \right) m(a+g)$ is its component 30º to the vertical.

In this problem, the surface is not horizontal, so the component of $m(a+g)$ is the wrong force. That’s the mistake.

Can I encourage @Combinatorics to draw a free-body diagram of the ball and apply Newton’s 2nd law?

 

[Combinatorics]

Agreed. I was wrong to say ‘mis-identified’. @Combinatorics has identified the correct force - the normal reaction on the sloping surface.

But $m(a+g)$ is the normal force exerted on a vertically accelerating horizontal surface (as in a simple elevator problem). $\cos\left( \frac{\pi}{6} \right) m(a+g)$ is its component 30º to the vertical.

In this problem, the surface is not horizontal, so the component of $m(a+g)$ is the wrong force. That’s the mistake.

Can I encourage @Combinatorics to draw a free-body diagram of the ball and apply Newton’s 2nd law?

Hi,

Thank you! Here is my attempt:
I defined $x$ and $y$ axis as shown in the image. From this I have that
$$
\sum F_x = N_1 \sin \theta - N_2
$$
$$
\sum F_y = N_1 \cos \theta - mg
$$
Is this correct?

Thank you very much for your help so far.

 

[Steve4Physics]

Hi,

Thank you! Here is my attempt:
I defined $x$ and $y$ axis as shown in the image. From this I have that
$$
\sum F_x = N_1 \sin \theta - N_2
$$
$$
\sum F_y = N_1 \cos \theta - mg
$$
Is this correct?

Thank you very much for your help so far.

Yes. Good. What can you say about the magnitude of $N_1$? If you apply 'F=ma' in the y-direction, you're almost home.

[EDIT. Typo' corrected.]

 

[Combinatorics]

Yes. Good. What can you say about the magnitude if $N_1$? If you apply 'F=ma' in the y-direction, you're almost home.

Amazing! So,
$$
ma = N_1 \cos \theta - mg \Rightarrow a_{\text{max}} = \frac{ 150 \cos \theta - mg}{m} = 1.025
$$
Which is indeed one of the answers!

Thank you very much! This was very helpful!

 

[BvU]

Still wonder where the breakline would run

The cube is made of a breakable material, that breaks when a force larger than $F_{max}=150$ N is exerted on it.
Someone pulls the cube upwards with an acceleration of $a$ m/s².

and how you pull it (massless cube? + 12 kg ball) up without breaking anything...

But that's just moping old me because I misinterpreted the question

(but happy, according to post #8 )