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Forces with Pulleys

  • Thread starter Max0007
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  • #1
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Homework Statement


Note that the vector F is 50N.
I need to find Theta and the Mass of the Box(M).

http://i5.minus.com/ib08xMuKZxTSbc.png [Broken]

Homework Equations


What I have so far:

Pulley 2:
FY: T2cos(20) - T1sin(60) + Fsin(45) + T2sin(30) = 0
FX: T2sin(20) + T1cos(60) + Fcos(45) - T2cos(30) = 0

Pulley 1:
Fx= T1sin(Theta) + T1cos(60) = 0
Fy= - T1cos(Theta) - T1sin(60) = 0

Box (M);
Fx= T2sin(20) - T1sin(Theta) = 0
Fy= -Wm +T2cos(20) + T1cos(Theta) = 0

The Attempt at a Solution



N/A
 
Last edited by a moderator:

Answers and Replies

  • #2
907
88
You are using x's and y's along with the sin's and cos's. I think this is confusing because T2y = T2cos(20). Maybe re-write the equations without the x's and y's and solve for M and theta,
 
  • #3
gneill
Mentor
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At first, concentrate your efforts at pulley 2. You have two equations with two unknowns there!
 
  • #4
Stephen Tashi
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FY: T2ycos(20) - T1ysin(60) + Fysin(45) + T2ysin(30) = 0
Perhaps you mean:
[itex] T1 \sin(70 deg) + T1 \sin(30 deg) = |F| \sin(45 deg) + T2 \sin(60 deg) [/itex]
 
  • #5
66
1
Perhaps you mean:
[itex] T1 \sin(70 deg) + T1 \sin(30 deg) = |F| \sin(45 deg) + T2 \sin(60 deg) [/itex]
Where did the 70 deg come from?
 
  • #6
Stephen Tashi
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Where did the 70 deg come from?
I took the complementary angle for 20 deg. You could use cos(20 deg) if you prefer. I find it simpler to see only sin(..)'s for y components.
 
  • #7
66
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I took the complementary angle for 20 deg. You could use cos(20 deg) if you prefer. I find it simpler to see only sin(..)'s for y components.
I am little confused, you see the angle 30 deg, that force is going which way? right or left.
 
  • #8
Stephen Tashi
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I am little confused, you see the angle 30 deg, that force is going which way? right or left.
Do you mean the force on pulley 2 due to the tension in the cord that makes the 30 deg angle? It can be regarded as having two components. One component pulls down, the other component pulls to the right.
 
  • #9
66
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Do you mean the force on pulley 2 due to the tension in the cord that makes the 30 deg angle? It can be regarded as having two components. One component pulls down, the other component pulls to the right.
I mean the direction of the Vector T is it negative or positive in that case on the 30 deg one.

FY:+ T2sin(30) OR - T2sin(30)
FX: - T2cos(30) OR + T2cos(30)


which is the correct one?
 
Last edited:
  • #10
Stephen Tashi
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I mean the direction of the Vector T is it negative or positive in that case on the 30 deg one.
I don't know what you mean by "the vector T". The tension in a cord is not a force. It is a property of the cord that let's one deduce the magnitudes of various forces. Tension is a non-negative number. At a particular end of the cord the tension causes a force. A force in 2-D is neither "positive" or "negative". It is the individual components of the force that can be positive or negative. The x-component of the force on pulley 2 due to the tension of the cord is positive if you consider x axis to be horizontal with the positive direction to the right. The y-component of that force is negative if you consider the y-axis to be vertical and upward to be the positive direction. The equation for the y-components of the forces on pulley 2 can be written:

[itex] -T1 \sin(70 deg) - T1 \sin(30 deg) + |F| \sin(45 deg) + T2 \sin(60 deg) = 0 [/itex]
 
  • #11
66
1
I don't know what you mean by "the vector T". The tension in a cord is not a force. It is a property of the cord that let's one deduce the magnitudes of various forces. Tension is a non-negative number. At a particular end of the cord the tension causes a force. A force in 2-D is neither "positive" or "negative". It is the individual components of the force that can be positive or negative. The x-component of the force on pulley 2 due to the tension of the cord is positive if you consider x axis to be horizontal with the positive direction to the right. The y-component of that force is negative if you consider the y-axis to be vertical and upward to be the positive direction. The equation for the y-components of the forces on pulley 2 can be written:

[itex] -T1 \sin(70 deg) - T1 \sin(30 deg) + |F| \sin(45 deg) + T2 \sin(60 deg) = 0 [/itex]
Why is it negative T1 sin(70 deg) ? How I see it its positive? If I use T1*cos20 would it be positive ?
 
  • #12
66
1
Why is it negative T1 sin(70 deg) ? How I see it its positive? If I use T1*cos20 would it be positive ?
Also are my other equations good? if I missed a negative I will get wrong answers.
 
  • #13
Stephen Tashi
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Why is it negative T1 sin(70 deg) ? How I see it its positive? If I use T1*cos20 would it be positive ?
The y-component of force on pulley 2 due to the tension T1 is downward. So it is considered negative regardless of whether you use cos(20 deg) or sin(70 deg) to express it.

Also are my other equations good? if I missed a negative I will get wrong answers.
The notation in your equations doesn't make sense. Rewrite them using proper notation, as paisiello2 suggested.
 
  • #14
66
1
The y-component of force on pulley 2 due to the tension T1 is downward. So it is considered negative regardless of whether you use cos(20 deg) or sin(70 deg) to express it.



The notation in your equations doesn't make sense. Rewrite them using proper notation, as paisiello2 suggested.
I already removed the y/x from my equations just after he posted. It did not update for you?
 
  • #15
Stephen Tashi
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I already removed the y/x from my equations just after he posted. It did not update for you?
They did update.

On pulley 2, there are two components forces pulling up and two pulling down. There should be 2 terms with minus signs.
On pulley 2, there are two components of force pulling left and two pulling right, There should be 2 terms with minus signs.

On pulley 1, you didn't account for the forces exerted by the support that holds the axis of the pulley. (It isn't useful to write equations for pulley 1.)
 
  • #16
66
1
They did update.

On pulley 2, there are two components forces pulling up and two pulling down. There should be 2 terms with minus signs.
On pulley 2, there are two components of force pulling left and two pulling right, There should be 2 terms with minus signs.

On pulley 1, you didn't account for the forces exerted by the support that holds the axis of the pulley. (It isn't useful to write equations for pulley 1.)
Looks like I cannot edit my thread anymore so here I will update;

Pulley 2:
FY: -T2cos(20) - T1sin(60) + Fsin(45) + T2sin(30) = 0
FX: T2sin(20)+ Fcos(45) -T1cos(60) - T2cos(30) = 0

Box (M);
Fx= T2sin(20) - T1sin(Theta) = 0
Fy= -Wm +T2cos(20) + T1cos(Theta) = 0

I have the Calculator Ti - Nspire, I think with that I can easily use the solve function and find T2 and T1 in Pulley 2, then with T1 and T2 I will be able to find Wm and Theta is that right ?

Thank you so much for helping me.
 
  • #17
Stephen Tashi
Science Advisor
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You have sign mistakes in the equations for pulley 2.
 
  • #18
66
1
You have sign mistakes in the equations for pulley 2.
Pulley 2:
FY: -T2cos(20) + T1sin(60) + Fsin(45) - T2sin(30) = 0
FX: T2sin(20)+ Fcos(45) -T1cos(60) - T2cos(30) = 0

I think this is correct now.
 
  • #19
Stephen Tashi
Science Advisor
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There's still a sign error in the FX: equation.

Are you familiar with "free body diagrams" ?
 
  • #20
66
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There's still a sign error in the FX: equation.

Are you familiar with "free body diagrams" ?
FX: -T2sin(20)+ Fcos(45) -T1cos(60) + T2cos(30) = 0
Here we go. I am sleepy so I make stupid mistakes.
 
  • #21
Stephen Tashi
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FX: -T2sin(20)+ Fcos(45) -T1cos(60) + T2cos(30) = 0
Here we go. I am sleepy so I make stupid mistakes.
That's correct.-
 
  • #22
66
1
That's correct.-
Thank you so much for the help Stephen, you helped me a lot. I truly appreciate the time you took to help me and explaining me things.
 

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