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

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SUMMARY

The discussion focuses on solving a physics problem involving two masses, M1 and M2, connected by a frictionless pulley. Participants utilize Newton's Laws to derive the relationship between the tension (T), mass M1, and mass M2. The final equations established are M2g - T = M2(T/M1) and M2 = (M1 + M2)a/g, emphasizing the importance of isolating M2 to find its value in terms of T and M1.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with free body diagrams (FBD)
  • Basic knowledge of gravitational force calculations
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the application of Newton's Laws in pulley systems
  • Learn how to construct and analyze free body diagrams for complex systems
  • Explore the concept of tension in strings and its effects on connected masses
  • Practice solving problems involving multiple masses and forces in equilibrium
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Students in introductory physics courses, educators teaching mechanics, and anyone interested in understanding the dynamics of pulley systems and forces in motion.

soolights
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<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 :/
 
Last edited:
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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
 
Arman777 said:
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...?
 
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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 ?
 
Arman777 said:
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
 
soolights said:
we don't know which way it moves

Whats this means ?
 
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## ?
 
Arman777 said:
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.
 
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 ?
 
  • #10
Arman777 said:
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 :/
 
  • #11
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...?
 
  • #12
Deleted post .
 
  • #13
Arman777 said:
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..
 
  • #14
soolights said:
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?
 
  • #15
PetSounds said:
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?
 
  • #16
Sorry but that is not correct .

Draw a diagram and show what forces are acting on m1 and m2 .
 
  • #17
Nidum said:
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--
 

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  • #18
Nidum said:
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

there's two forces acting on M2
 
  • #19
Your diagram does not really show the forces acting .
help-jpg.112117.jpg
 
  • #20
Arman777 said:
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

there's two forces acting on M2

so, for tension and weight..?
 
  • #21
Nidum said:
Your diagram does not really show the forces acting .
help-jpg.112117.jpg
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?
 
  • #22
soolights said:
so, for tension and weight..?
yes,try to do what I did for M1
 
  • #23
Write down the equation then you solved
 
  • #24
Arman777 said:
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?
 
  • #25
yep correct

Now we have two equations
T=M1a
M2g-T=m2a so...?
 
  • #26
Arman777 said:
yep correct
so m2a = w-t, how to get rid of w?
 
  • #27
Whats w ohh I assume you know that.Thats the weight of the m2
 
  • #28
Arman777 said:
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)
 
  • #29
The questions ask you to find m2 in terms of T and m1.You have 2 equations as I said.Get rid of a.
 
  • #30
Arman777 said:
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)
 

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