Pulley system, find the acceleration and tension

[Orodruin]

It would be more accurate to say it would accelerate to the right due to the pull of the hanging mass. Of course, if the system starts at rest this will lead to it moving to the right.

 

[Davidllerenav]

It would be more accurate to say it would accelerate to the right due to the pull of the hanging mass. Of course, if the system starts at rest this will lead to it moving to the right.

I see. Ok, I wrote the sum of forces like this:

• Mobile with mass M: $T_1-F_r=M\cdot a_1\Rightarrow T_1-\mu N = M\cdot a_1 \Rightarrow T_1=\mu Mg+M\cdot a_1\Rightarrow T_1=M(\mu g+a_1)$
• Hanging body with mass m: Since it moves down, the force of gravity is bigger than the tension, thus $F_g-T_2=m\cdot a_2\Rightarrow mg-T_2=m\cdot a_2\Rightarrow T_2=m(g-a_2)$

Is it correct?

 

[Orodruin]

No, you mixed up $T_1$ and $T_2$ from one step to the next. It may seem petty, but you have done this several times in this thread alone and you need to take better care with this or you will make mistakes that are relatively easy to avoid.

 

[Davidllerenav]

No, you mixed up $T_1$ and $T_2$ from one step to the next. It may seem petty, but you have done this several times in this thread alone and you need to take better care with this or you will make mistakes that are relatively easy to avoid.

Sorry, I fixed it now. I'll try to be como careful. Now that it is fixed, is it correct?

 

[Orodruin]

Yes, but you need to find a relation between $T_1$ and $T_2$ to completely solve your system. Can you imagine where you can get such a relation from?

 

[Davidllerenav]

Yes, but you need to find a relation between $T_1$ and $T_2$ to completely solve your system. Can you imagine where you can get such a relation from?

I guess I need to make both of them somehow equal. But I don't know where I can get that relation from. I guess that since there are two $T_1$ tensions on the right side of the middle pulley, then the second tension must be half $T_1$?

 

[Orodruin]

I guess I need to make both of them somehow equal. But I don't know where I can get that relation from. I guess that since there are two $T_1$ tensions on the right side of the middle pulley, then the second tension must be half $T_1$?

Yes, this is the FBD of the middle pulley. An ideal (massless pulley) has the force equation $F = ma = 0 = T_2 - 2T_1$.

 

[Davidllerenav]

Yes, this is the FBD of the middle pulley. An ideal (massless pulley) has the force equation $F = ma = 0 = T_2 - 2T_1$.

I didn't knew that. So everytime I have a massless pulley I can assume that?

 

[Orodruin]

ecwrytumw

I think you need to double check the placement of your fingers on the keyboard ...
But yes. Any time you have an ideal massless object, the forces acting on it need to cancel out.

 

[Davidllerenav]

I think you need to double check the placement of your fingers on the keyboard ...
But yes. Any time you have an ideal massless object, the forces acting on it need to cancel out.

Sorry, I'm from my cellphone and didn't notice. I meant "everytime". So the forces cancel and the tensions also have that relationship?

 

[haruspex]

and the tensions also have that relationship?

The forces and torques cancel, so they may have a relationship, but not necessarily that one, exactly.
The right-hand pulley is on a bracket, so the tensions balance the force from that. The left-hand pulley is only in contact with the strings, so the forces from the tensions must balance somehow. All the strings are parallel, so it's easy, but in another problem the string that runs around the pulley might go off at some angle.