Finding the Total Mass in the Accelerating Blocks Problem: A Scientific Analysis

[Dansuer]

Homework Statement

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What horizontal force $F$ must be constantly applied to $M$ so that $M_1$ and $M_2$ do not move relative to $M$? neglet friction

Homework Equations

The Attempt at a Solution

Since $M_2$ is not moving, the tension of the string is $T=M_2 * g$. Plugging this into the euation $T=M_1 * a$ we find

that the acceleration of $M_1$ is $a=(M_2 / M_1)*g$

Since $M_1$ is at rest with respect to $M$, $M$ have the same accelleration $a$

So, $F = (M + M_2) * a$. or $F = (M_2 / M_1)*(M+M_2)*g$that's what i found. But the problem solution states that $F = (M_2 / M_1)*(M+M_1+M_2)*g$

which i don't see why it should be like that. Intuitively if i push on the block $M$, i push both $M$ and $M_2$, that's why i

found $(M+M_2)$, but i don't push on $M_1$, since there is no friction. It's movements are due to the force of $M_2$

pulling down.

 

[nayfie]

Since M1 is at rest with respect to M, M have the same accelleration a.
So, F = (M + M2) * a. or F = (M2 / M1)*(M+M2)*g

There may be no friction, but there's still the weight of M1 on M.

This means that the mass of the total system is (M + M1 + M2).

Hence, $F_{total} = (M + M1 + M2)*(\frac{M2}{M1})*g$.

 

[Dansuer]

I don't think that's the case.

writing the force acting on $M$ $M_1$ and $M_2$ we have$M$ x : $F - F_{M_2M} = M*a$

$M$ y : $N = M*g + F_{M_2M}$$M_1$ x : $T = m*a$

$M_1$ y : $F_{MM_1} = P$$M_2$ x: $F_{MM_2} = M_2*a$

$M_2$ y: $T = M_2*g$from $M$ x and $M_2$ x and given that $F_{M_2M}=F_{MM_2}$

we can find that $F - M_2*a = M*a$, and we find that $F = (M+M_2)*a$we can't(or i can't) do the same thing for the $M_1$ block, since all the force that connect $M$ and $M_1$ are on the y

axis, and we are taking about the x axis. We also don't have the friction that could "links the axis".

 

[nayfie]

$F_{tension} = M_{2}g = M_{1}a$

$a = (\frac{M_{2}}{M_{1}})g$

$F_{system} = M_{total}a$

It looks to me like you're over-complicating the question?

 

[Dansuer]

I might be, but i do that because i can't see why the total mass of the system should be $M + M_1 + M_2$. and not only $M + M_2$.

 

[nayfie]

Well $M_{1}$ is connected to the system by the tension in the string. When a force is 'created' on $M$; it 'pushes' against $M_{2}$.

Because there is no friction between $M_{1}$ and $M$; $M_{1}$ is going to want to move to the left. However, the movement of $M_{2}$ to the right is balancing this.

I'm not sure if this explanation helps or not. I tried to simplify it as much as possible.

That's just my understanding though. I could be completely wrong!

 

[Dansuer]

you say that $M_1$ wants to go to the left (relative to $M$) because of the acceleration of $M$. But i don't see that. For me $M_1$ wants to stay still(relative to $M$).

 

[nayfie]

Newton's first law of motion: every body remains in a state of constant velocity unless acted upon by an external unbalanced force.

Imagine you had a box with a glass of water on top. If you kicked the box, the glass of water (that initially has a velocity of 0) will want to stay still. Hence, relative to the box, the glass of water is going to be moving backwards.

This is what is happening in this question. There is no friction between $M_{1}$ and $M$. Hence, when a force is applied to $M$, $M_{1}$ is going to move backwards relative to $M$

However, $M_{2}$ is connected to both $M$ and $M_{1}$, so when the force is applied, the tension in the string provides the same acceleration that the rest of the system has.

 

[Dansuer]

exactly, if $M_1$ wants to stay still, it means that I'm not pushing on it, nor is $M$. the only one that is exerting force on $M_1$ it's the string, that would be acting wth or without the force that I'm doing.

 

[ojs]

It is the string that makes it so that $M_1$ is connected to the other two masses, and that makes it so that the total applied force must include all the masses and not just two of them.

But this is a very subtle example of Newtons law's and not at all easy to understand so if your still confused, please carry on asking or better yet have a sit down with your teacher and get a better explanation from him (sometimes difficult explaining things over a forum like this).

 

[Dansuer]

I don't think that's the case.

writing the force acting on $M$ $M_1$ and $M_2$ we have$M$ x : $F - F_{M_2M} = M*a$

$M$ y : $N = M*g + F_{M_2M}$$M_1$ x : $T = m*a$

$M_1$ y : $F_{MM_1} = P$$M_2$ x: $F_{MM_2} = M_2*a$

$M_2$ y: $T = M_2*g$from $M$ x and $M_2$ x and given that $F_{M_2M}=F_{MM_2}$

we can find that $F - M_2*a = M*a$, and we find that $F = (M+M_2)*a$we can't(or i can't) do the same thing for the $M_1$ block, since all the force that connect $M$ and $M_1$ are on the y

axis, and we are taking about the x axis. We also don't have the friction that could "links the axis".

what would really help is telling me what wrong reasoning i have done here. so that i can find, starting from the force diagrams, that the total mass is the sums of all three blocks. kind of like a proof