Two boxes connected with a pulley, one hanging

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The discussion revolves around a physics problem involving two masses connected by a pulley system, with one mass on a table and the other hanging. Participants analyze the forces acting on the masses and the torque on the pulley, leading to equations for acceleration and angular acceleration. There is confusion regarding the signs of tension forces and the correct application of equations, particularly in relation to the direction of motion and friction. After clarifying the equations and correcting the signs, a participant arrives at values for acceleration and angular acceleration, indicating progress in solving the problem. The conversation highlights the importance of careful consideration of forces and directions in physics problems.
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Two boxes connected with a pulley, one hanging *updated*

Homework Statement


There is a Mass (m1) on a table. The coefficient of friction between the table and m1 is 0.18. The second mass (m2) is hanging, suspended by a string around a pulley (m3) attached to m1. The pulley has a radius (R) and is considered a solid cylinder.
m1=3.5[kg]
m2=2.8[kg]
m3=1.9[kg]
R=0.089[m]

Find:
a) alpha of the pulley
b.)Ft1
c.)Ft2
d.)omega @ (6.9)

Homework Equations


I for solid cylinder = (1/2)M2R2
Fnet=ma
Torquenet=I*alpha
alpha=a/R

The Attempt at a Solution


I drew a torque fbd for the pulley, I found I for the pulley, I drew fbd's for the boxes, got my equations like

m1-->
X) Ft1-Ff=m1a
Y) Fn-Fg=0

Fn=m1g
Ft1=m1a+Ff
Ft1=m1a+Mew*m1*gm2--->
Y) Ft2-Fg=m2(-a)
Ft2=-m2a+m2g
and finally

Torquenet=I*alpha so
Torque1=R*sin(90)*Ft1 = R*Ft1 (k)
Torque2=R*sin(90)*Ft2 = R*Ft2 (-k)
==>R*Ft1-R*Ft2=I*alpha

and alpha= a/R

So
R*(m1a+Mew*m1*g)-R*(-m2a+m2g)=((1/2)M3R2)*(a/R)

That equation just has one uknown --> a

For a i got 3.97[m/s2]
For alpha I am getting like 44 [rads/s2] and others are getting about about 16 [rad/s2]
Can someone spot what, if anything, I'm doing something wrong?EDIT: Sorry for the mistakes..
 
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It would be easier to spot an error if you showed your three primary equations (for m1, m2, and m3) and your solution for acceleration symbolically (without numerical values plugged in).
 
Which is which? Is m3 the pulley or the hanging mass?
 
Doc Al said:
Which is which? Is m3 the pulley or the hanging mass?

The question identifies the masses. m3 is the pulley
 


frozenguy said:

Homework Statement


There is a Mass (m1) on a table. The coefficient of friction between the table and m1 is 0.18. The second mass (m2) is hanging, suspended by a string around a pulley (m3) attached to m1. The pulley has a radius (R) and is considered a solid cylinder.
m1=3.5[kg]
m2=2.8[kg]
m3=1.9[kg]
R=0.089[m]

Find:
a) alpha of the pulley
b.)Ft1
c.)Ft2
d.)omega @ (6.9)

Homework Equations


I for solid cylinder = (1/2)M2R2
Fnet=ma
Torquenet=I*alpha
alpha=a/R


The Attempt at a Solution


I drew a torque fbd for the pulley, I found I for the pulley, I drew fbd's for the boxes, got my equations like

m1-->
X) Ft1-Ff=m1a
Y) Fn-Fg=0

Fn=m3g

It should be m_1 there
 
nrqed said:
The question identifies the masses. m3 is the pulley

unfortunatly in my following equations though, I treated m3 as the mass used in Force of tension 2, where that should have been m2..

I'm starting to screw myself up now, after solving for a from that last equation I wrote up there, I got a different value for a then I originally got..

Then I realized that Ft1 is negative on my fbd, so I tried changing that and I got a totally different number that no one else has.

All I have to do is find my three equations, and solve for a right?? Why is this proving to be so difficult for me..
 


frozenguy said:
Torquenet=I*alpha so
Torque1=R*sin(90)*Ft1 = R*Ft1 (k)
Torque2=R*sin(90)*Ft2 = R*Ft2 (-k)
==>R*Ft1-R*Ft2=I*alpha
To make my point more clear: the rotation is clockwise, so the angular rotation is
alpha (-k), not alpha k . This is why we end up with

R*Ft1-R*Ft2=- I*alpha

or

-R*Ft1+R*Ft2= I*alpha
 
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nrqed said:
You are on the right track! Don't get discouraged, you are almost done.

Looks to me like it should be -Ft1+R*Ft2= I alpha
since the mass m2 is falling, the torque on that side is larger than the torque from T1

Ok so with my original equation where I had Ft1 as a positive direction, the equation I got for a was this:

a=(g(m2-Mew*m1))/(m1+m2-(m3/2))

After changing the direction of Ft1 I get a is equal to

a= (-g(Mew*m1-m2))/((m3/2)-m1-m2)

They both result with a value of 3.97[m/s2] but the second equation results in a negative value.. doesn't that mean that one of my directions is off if the value is negative?
 


frozenguy said:
Ok so with my original equation where I had Ft1 as a positive direction, the equation I got for a was this:

a=(g(m2-Mew*m1))/(m1+m2-(m3/2))

After changing the direction of Ft1 I get a is equal to

a= (-g(Mew*m1-m2))/((m3/2)-m1-m2)

They both result with a value of 3.97[m/s2] but the second equation results in a negative value.. doesn't that mean that one of my directions is off if the value is negative?

There is something fishy going on, indeed. Give me a minute to double check everything.
Meanwhile, can you check carefully the question and make sure that it is exactly what you wrote in your post? Thanks
 
  • #10


frozenguy said:
Ok so with my original equation where I had Ft1 as a positive direction, the equation I got for a was this:

a=(g(m2-Mew*m1))/(m1+m2-(m3/2))

After changing the direction of Ft1 I get a is equal to
Wait...you have to change the sign of both Ft1 and Ft2 in your torque equation, not just one
 
  • #11


Using

R T_2 - R T_1 = \frac{1}{2} m_3 R^2 \frac{a}{R}

and using the same T2 and T1 you had, I find

a = \frac{m_3 g - \mu m_1 g}{m_3/2+m_1 + m_2 }
 
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  • #12


nrqed said:
There is something fishy going on, indeed. Give me a minute to double check everything.
Meanwhile, can you check carefully the question and make sure that it is exactly what you wrote in your post? Thanks

diagram2.jpg
 
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  • #13


nrqed said:
Wait...you have to change the sign of both Ft1 and Ft2 in your torque equation, not just one

Why do I have to change the sign for Ft2? Its positive in the fbd for m3 and the torque from Ft2 is still out of the board (-)
 
  • #14


frozenguy said:
diagram1.jpg


please note that I changed the pulley to m3 and the hanging mass to m2 when I posted; I forget why.

AHHH!
And where do they give the coefficient of friction?

In any case, the formula is

a = \frac{ m(hanging) g - \mu m(on \,table) g}{m(pulley)/2 + m(on \,table) + m(hanging)}
 
  • #15


frozenguy said:
Why do I have to change the sign for Ft2? Its positive in the fbd for m3 and the torque from Ft2 is still out of the board (-)

I mean that the correct equation is

Ft2 - Ft1 = I alpha
 
  • #16


nrqed said:
I mean that the correct equation is

Ft2 - Ft1 = I alpha

I don't get how Ft2 is positive now though. Isn't it (-k) direction (so negative in torque equation) and positive in the fbd for m3?
 
  • #17


frozenguy said:
I don't get how Ft2 is positive now though. Isn't it (-k) direction (so negative in torque equation) and positive in the fbd for m3?

This is what I explained in post number 7

The angular acceleration is clockwise. This means that the angular acceleration is alpha(-k)

So we get

T2 (-k) + T1 (k) = I alpha (-k)

therefore

-T2 + T1 = - I alpha

or

T2 - T1 = I alpha
 
  • #18


nrqed said:
This is what I explained in post number 7

The angular acceleration is clockwise. This means that the angular acceleration is alpha(-k)

So we get

T2 (-k) + T1 (k) = I alpha (-k)

therefore

-T2 + T1 = - I alpha

or

T2 - T1 = I alpha
Oh..

Sorry, I didn't see post 7 somehow. Ok so that makes sense. I thought alpha could be positive and negative, infact I believed it must be able to be +or- but a peer said no. lol.. so it kind of escaped me..

Thank you so much.. So Ft1 will be = -m1a-Mew*m1g

because Ft1 itself is negative in the fbd for m3..

so with the negative Ft1 in the torque equation, that will distribute out and make those two factors positive (in Ft1 itself)
 
  • #19


frozenguy said:
Oh..

Sorry, I didn't see post 7 somehow. Ok so that makes sense. I thought alpha could be positive and negative, infact I believed it must be able to be +or- but a peer said no. lol.. so it kind of escaped me..
well, one has to be very careful about the language used. To be exact, alpha itself is always positive (it's the magnitude of the angular acceleration). But the components of the angular acceleration vector may be negative. Note that when I write alpha (-k), the value of alpha itself is positive.
Thank you so much.. So Ft1 will be = -m1a-Mew*m1g

because Ft1 itself is negative in the fbd for m3..

so with the negative Ft1 in the torque equation, that will distribute out and make those two factors positive (in Ft1 itself)

not quite. T1 is positive. And it is given by T1 = m1 a + mu m1 g, as you had initially written.
 
  • #20


nrqed said:
well, one has to be very careful about the language used. To be exact, alpha itself is always positive (it's the magnitude of the angular acceleration). But the components of the angular acceleration vector may be negative. Note that when I write alpha (-k), the value of alpha itself is positive.


not quite. T1 is positive. And it is given by T1 = m1 a + mu m1 g, as you had initially written.

Ok I got myself all mixed up... I'm on track now lol

Again, thanks so much..

oh, and the picture for some reason doesn't include the Mew, I apologize, but the written part of the question does.

So I get a=2.93 [m/s2]
alpha=33[rad/s2]

thank you!
 

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