How to avoid the use of a pseudo force

[timetraveller123]

Homework Statement

[/B]

Homework Equations

f = ma

The Attempt at a Solution

t₁ tension in string connecting M_c
t₁ tension in string connecting M_a and M_b
$m_c:\\ m_c g - T_1 = m_c a_c\\$
i considered pseudo for acting on masses b and a from a non inertial frame of reference
hence in that frame of reference mass b will go down with a_observed and mass a will go up with a _observed
$m_a,m_b:\\ a_{observed} = \frac{(m_b - m_a)a_c}{m_b + m_a}\\ a_b = a_c - \frac{(m_b - m_a)a_c}{m_b + m_a}\\ a_a = a_c + \frac{(m_b - m_a)a_c}{m_b + m_a}\\ T_2 = m_a a_a = m_b a_b\\ 2 T_2 = T_1\\$
is this correct i have feeling something has gone wrong

but how could i solve without pseudo forces i feel very uncomfortable having to switch frames of references

 

[ehild]

Homework Statement

View attachment 211246[/B]

View attachment 211245

Homework Equations

f = ma

The Attempt at a Solution

t₁ tension in string connecting M_c
t₁ tension in string connecting M_a and M_b
$m_c:\\ m_c g - T_1 = m_c a_c\\$
i considered pseudo for acting on masses b and a from a non inertial frame of reference
hence in that frame of reference mass b will go down with a_observed and mass a will go up with a _observed
$m_a,m_b:\\ a_{observed} = \frac{(m_b - m_a)a_c}{m_b + m_a}\\ a_b = a_c - \frac{(m_b - m_a)a_c}{m_b + m_a}\\ a_a = a_c + \frac{(m_b - m_a)a_c}{m_b + m_a}\\ T_2 = m_a a_a = m_b a_b\\ 2 T_2 = T_1\\$
is this correct i have feeling something has gone wrong

but how could i solve without pseudo forces i feel very uncomfortable having to switch frames of references

You do not need pseudo force. The "observed acceleration" in the accelerating frame of reference is the relative acceleration with respect to the table (in rest), equal in magnitude and opposite in direction for a and b. The acceleration a_a and a_b are a_c+a_relative.
Your work is correct so far, go ahead.

 

[timetraveller123]

no but i used the concept of pseudo force to derive the result for observed acceleration
is there a way to do away with it

 

[timetraveller123]

as observed in accelerated frame of reference

 

[ehild]

View attachment 211251
as observed in accelerated frame of reference

Forget accelerating frame of reference. Even then, if the acceleration of block A is a_rel with respect to the horizontal pulley, that accelerates with a_c , the relative acceleration of block B is -a_rel and a_a=a_c+a_rel and a_b=a_c-a_rel. Draw the FBD for each block, write the ΣF=ma equations in the rest frame of reference, eliminate the tensions and solve.

 

[timetraveller123]

Forget accelerating frame of reference. Even then, if the acceleration of block A is a_rel with respect to the horizontal pulley, that accelerates with a_c , the relative acceleration of block B is -a_rel and a_a=a_c+a_rel and a_b=a_c-a_rel. Draw the FBD for each block, write the ΣF=ma equations in the rest frame of reference, eliminate the tensions and solve.

wow i got the same result without having to use pseudo forces had been having a lot of problems when acceleration was involved but this a_rel concept cleared it up a lot thanks the rest is just solving the equation i have a few general questions :

lets say there is a massless string in space and two astronauts tug at it one with greater force then there is net force on the string how is that possible ?

 

[CWatters]

lets say there is a massless string in space and two astronauts tug at it one with greater force then there is net force on the string how is that possible ?

It's not.

Review Newton's laws.

 

[timetraveller123]

what i am saying is that the force applied by the astronauts can't be controlled by Newtons law ? and are you referring to the third law if so i don't see how that plays a part here please help

 

[ehild]

lets say there is a massless string in space and two astronauts tug at it one with greater force then there is net force on the string how is that possible ?

Can not one astronaut pull the rope with different force than the other.

 

[timetraveller123]

why not it is up to the will of the astronaut
i just can't visualize what does it physically look like

 

[ehild]

why not it is up to the will of the astronaut
i just can't visualize what does it physically look like

How do you exert force on the rope in empty space? It is not enough to want it.

 

[timetraveller123]

is it that i will start moving opposite direction?

 

[ehild]

is it that i will start moving opposite direction?

Yes, but how will you start to move when in empty space?

 

[vela]

What I am saying is that the force applied by the astronauts can't be controlled by Newton's law. And are you referring to the third law? If so, I don't see how that plays a part here. Please help.

Why not? It is up to the will of the astronaut. I just can't visualize what it physically looks like.

You can't picture it because, as CWatters said, it's impossible. Anything the astronaut can possibly do has to be consistent with Newton's laws. According to your reasoning, you can exert any force on a lightweight object as it's just a matter of will. The third law therefore implies that the object will exert the same magnitude force back on you. Now try to imagine yourself throwing a wad of paper with a 10-N force, about the weight of a kilogram. The paper must exert a 10-N force on your hand. Does that in fact happen?

 

[timetraveller123]

You can't picture it because, as CWatters said, it's impossible. Anything the astronaut can possibly do has to be consistent with Newton's laws. According to your reasoning, you can exert any force on a lightweight object as it's just a matter of will. The third law therefore implies that the object will exert the same magnitude force back on you. Now try to imagine yourself throwing a wad of paper with a 10-N force, about the weight of a kilogram. The paper must exert a 10-N force on your hand. Does that in fact happen?

yes that happens and i will accelerate backwards but how can i that apply to this situation if the astronaut tugs at the rope with force f backwards then he will accelerated forward same with the other astronaut then what about the other astronaut what is stopping him from applying a different force he would just accelerate with a different acceleration
i am sorry if am not getting what you are saying but it would if you could make it more clear to me

 

[vela]

yes that happens and i will accelerate backwards.

I guess I should have been clearer. Can you personally actually throw a wad of paper and exert 10 N of force on it? To do so, the wad of paper should push back on your hand with a force that feels like the force you'd need to hold a kilogram mass up. If you can't (or find it very hard to do), why do you suppose this is the case?