Solving a Tricky Physics Pulley Problem: Finding M2 with Tension and Mass M1

[soolights]

<Mentor's note: moved from a technical section, therefore no template>​

hey guys, i have a question about a tricky physics 1 problem our teacher gave us today, paraphrased as follows:

there is a mass M1 on a flat, frictionless horizontal table connected to a pulley, and on the other side of the pulley is another mass M2 the pulls downward (off the table). the system is frictionless. the tension of the string is T and the mass of the table mass is M1, so what is M2 in terms of M1 and T?

we have a test soon and i would really appreciate help :/

 

[Arman777]

You can use Newton's Laws to find an Solution.Draw a free body diagram for each object then equal them to the net force.I can try to help you If u give me something.You should try to do it yourself a bit at least

 

[soolights]

You can use Newton's Laws to find an Solution.Draw a free body diagram for each object then equal them to the net force.I can try to help you If u give me something and this looks like homework question.You should try to do it yourself a bit at least

i don't understand how to list these out with so few information. there's nothing given for acceleration, movement, or relationships between anything...?

 

[Arman777]

If you think there's no friction and both $M_1$ and $M_2$ are not moving..thats impossible I think.Does $M_1$ makes a circular motion ?

 

[soolights]

If u think there's no friction and both $M_1$ and $M_2$ are not moving..thats impossible I think.Does $M_1$ makes a circular motion ?

yes i made a mistake, we don't know which way it moves

 

[Arman777]

we don't know which way it moves

Whats this means ?

 

[Arman777]

Ok I think $M_1$ makes a circular motion.If there's no friction this is the seems only possible way.Now What forces acting on $M_1$ and what's the total force on acting on $M_1$ ?

 

[soolights]

Ok I think $M_1$ makes a circular motion.If there's no friction this is the seems only possible way.Now What forces acting on $M_1$ and what's the total force on acting on $M_1$ ?

no, it doesn't make a circular motion. we just don't know the acceleration of the system. everything is linear! (these problems should be solvable without a calculator)
the tension force is acting on M1 and pulling it toward the edge of the table, along with the gravitational weight force and normal force from the table. the net force for M1 is the tension force.

 

[Arman777]

Ok,I understand,Now the idea is the Net force acting on the object should be equal the total force.and

$\vec F_{total}=m\vec a$ can you find an equation for M1 using this ?

 

[soolights]

Ok,I understand,Now the idea is the Net force acting on the object should be equal the total force.and

$\vec F_{total}=m\vec a$ can u find an equation for M1 using this ?

fnet = ma, so m = fnet/a? i don't know if this will help or not because it doesn't provide acceleration :/

 

[Arman777]

As you said there's Tension acting on the Object, that's the net force , and this net force should be equal to Total force so...?

 

[Nidum]

Deleted post .

 

[soolights]

As you said there's Tension acting on the Object, that's the net force , and this net force should be equal to Total force so...?

so the only force is the tension...but how does that link to the mass of m2 hanging off the table? m2's only force is the tension force holding it up..

 

[PetSounds]

fnet = ma, so m = fnet/a? i don't know if this will help or not because it doesn't provide acceleration :/

If M2 is pulling downward off the table, don't you already know its acceleration?

 

[soolights]

If M2 is pulling downward off the table, don't you already know its acceleration?

oh gosh! since there's no friction and it pulls downward, it's just 9.81. fnet = t, so using fnet = ma, m2 = t/9.8?

 

[Nidum]

Sorry but that is not correct .

Draw a diagram and show what forces are acting on m1 and m2 .

 

[soolights]

Sorry but that is not correct .

Draw a diagram and show what forces are acting on m1 and m2 .

this is the FBD I've been using--

 

[Arman777]

Sorry but that is not correct .

Draw a diagram and show what forces are acting on m1 and m2 .

I am trying to that...M1 has T (thats the net force) and that's equal to M1a (total force) so,T=M1a.

You have to do this M2

Theres two forces acting on M2

 

[Nidum]

Your diagram does not really show the forces acting .

 

[soolights]

I am trying to that...M1 has T (thats the net force) and that's equal to M1a (total force) so,T=M1a.

You have to do this M2

Theres two forces acting on M2

so, for tension and weight..?

 

[soolights]

Your diagram does not really show the forces acting .

i'm not sure what else there is to put...normal, tension, gravitational forces are all there and i don't know of any other ones that are applicable?

 

[Arman777]

so, for tension and weight..?

yes,try to do what I did for M1

 

[Arman777]

Write down the equation then you solved

 

[soolights]

yes,try to do what I did for M1

okay, so force in the y direction is both tension and gravity, so fnet = w-t (is necessarily w>t?). so w-t = m2a?

 

[Arman777]

yep correct

Now we have two equations
T=M1a
M2g-T=m2a so...?

 

[soolights]

yep correct

so m2a = w-t, how to get rid of w?

 

[Arman777]

Whats w ohh I assume you know that.Thats the weight of the m2

 

[soolights]

Whats w ohh I assume you know that.Thats the weight of the m2

OK let me try again lol this isn't going too well for me is it
m2a = w-t
m2a = m2g-t
m2a-m2g=-t
m2(a-g)=-t
m2=-t/(a-g)

 

[Arman777]

The questions ask you to find m2 in terms of T and m1.You have 2 equations as I said.Get rid of a.

 

[soolights]

The questions ask you to find m2 in terms of T and m1.You have 2 equations as I said.Get rid of a.

oh, i missed that message, sorry
T=M1a , a = T/M1
M2g-T=m2a ,
M2g-T=M2(T/M1)

 

[Arman777]

Thats correct.Leave m2 alone and you find the answer

 

[Arman777]

For correction we know that m2g-T=m2a and T=m1a
so
m2g-m1a=m2a
m2g=(m1+m2)a

As we expected.

 

[soolights]

For correction we know that m2g-T=m2a and T=m1a
so
m2g-m1a=m2a
m2g=(m1+m2)a

As we expected.

therefore since we're looking for m2 and only m2 we can simplify to m2=(m1+m2)a /g?

 

[Arman777]

oh, i missed that message, sorry
T=M1a , a = T/M1
M2g-T=m2a ,
M2g-T=M2(T/M1)

This is true.I just did that for to check that what we did was right or wrong.

The right equation is M2g-T=M2(T/M1).Leave M2 alone in this equation.Thats the answer.

sorry that I confused you

 

[soolights]

This is true.I just did that for to check that what we did was right or wrong.

The right equation is M2g-T=M2(T/M1).Leave M2 alone in this equation.Thats the answer.

sorry that I confused you

no problem, but shouldn't you be right if we are solving for M2? all the m2 should be on the LHS right?