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Gravity Pulling Blocks

  • Thread starter anastasiaw
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Homework Statement


M1 and M2 are two masses connected as shown. The pulley is light (massless) and frictionless. Find the mass M1, given that M2 (3.5 kg) accelerates downwards at 3.35 m/s^2, θ is 35o, and μk is 0.25.

http://img525.imageshack.us/img525/7076/prob31fricpullplaneyi0.gif [Broken]


Homework Equations





The Attempt at a Solution


The total force (which is the total mass times the acceleration) is equal to the sum of the forces.

Forces:
Gravity = M1g
Normal = M1g(sinθ)
Applied = M1a
Kinetic Friction = μkM1a

g=9.81m/s^2

a(M1 + M2) = -M1g +M1g(sinθ) +M1a -μkM1a
M1a +M2a = M1g((sinθ)-1) +M1a -μkM1a
M2a = M1(g((sinθ)-1) -μka)
M1 = M2a/(g((sinθ)-1) -μka)
M1 = (3.5*3.35)/(9.81((sin35)-1) -0.25*3.35)
M1 = 11.725/(9.81(-.426)-0.8375)
M1 = -2.34 kg

A negative answer isn't even possible, but I tried putting in "2.34 kg" is an answer and that didn't work. I don't know what I'm doing wrong.
 
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Answers and Replies

  • #2
rock.freak667
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I get M1=2.058kg.
If that is correct then I got it by simply using the tension as a force in the system, then found the two resultant forces acting. Then solved.
 
  • #3
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I get M1=2.058kg.
If that is correct then I got it by simply using the tension as a force in the system, then found the two resultant forces acting. Then solved.
Yeah it's 2.056 kg. Can you show me the math for that?
 
  • #4
rock.freak667
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The tension(T) in the strings are also forces in the system. (T is up the plane in the string at m1 and vertically opposite to the weight of m2)

The components of the weight of m1 is m1gsin35(down the plane) and m1gcos35(perpendicular to the plane).

The motion is downward so that the resultant force up the plane is

[itex]m_1a=T-m_1gsin35-\mu m_1gcos35...(*)[/itex]

the resultant motion of m2 is

[itex]m_2a=m_2g-T...(**)[/itex]

Now you have the value of a(the downward acceleration),g,m2.
 
  • #5
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The tension(T) in the strings are also forces in the system. (T is up the plane in the string at m1 and vertically opposite to the weight of m2)

The components of the weight of m1 is m1gsin35(down the plane) and m1gcos35(perpendicular to the plane).

The motion is downward so that the resultant force up the plane is

[itex]m_1a=T-m_1gsin35-\mu m_1gcos35...(*)[/itex]

the resultant motion of m2 is

[itex]m_2a=m_2g-T...(**)[/itex]

Now you have the value of a(the downward acceleration),g,m2.
Thanks a lot! This is a big help. I understand it now.
 
  • #6
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this is a silly question but how do you factor out the m1 so that you solve for m1. my problem looks like this

M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (7.00 kg) accelerates downwards at 2.19 m/s2, that theta is 20.0°, and that muk is 0.500.

i have m1*2.19 = 53.27N-m1*3.35m/s/s-m1*4.60m/s/s, but i do not know how to get it to a m1= ____ kg.
 
Last edited:
  • #7
rock.freak667
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Put everything with m1 on one side,then factor it out.
 
  • #8
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i will try that thanks
 
  • #9
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thank you soo much it clicked with me you are my favorite people in the world right now!
 
  • #10
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I was just wondering if you could help me with the same problem. My prof is useless. He's completely conceptual.
M1=?
M2=7.0 kg
/mu_k=.38
A=3.05 m/s^2
Theta=25

If possible can you show the problem as you worked it? I'm completely lost and need to know how to do this for a test at the end of the week.
 
Last edited:
  • #11
rock.freak667
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I was just wondering if you could help me with the same problem. My prof is useless. He's completely conceptual.
M1=?
M2=7.0 kg
/mu_k=.38
A=3.05 m/s^2

If possible can you show the problem as you worked it? I'm completely lost and need to know how to do this for a test at the end of the week.
Just follow the way I showed in post #4.
 
  • #12
13
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I have tried, but I'm unsure as to what I am getting from the equations. I don't know what I am supposed to do with the values acquired...
 
  • #13
rock.freak667
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I have tried, but I'm unsure as to what I am getting from the equations. I don't know what I am supposed to do with the values acquired...
Show the equations you got and we will see from there.
 
  • #14
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That is just it, I have no clue where to start.
 
  • #15
rock.freak667
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That is just it, I have no clue where to start.
Start with the free body diagram shown in post #1 and then write the components of the weight of the mass M1. (Were you also given the angle the plane is at?)
 
  • #16
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FBD for m1 would consist of the contact force pushing up (parallel to the ramp), the tension force pulling m1 up the ramp, and gravity pulling down. But how can I solve for any of that if I don't have the mass?
 
  • #17
rock.freak667
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FBD for m1 would consist of the contact force pushing up (parallel to the ramp), the tension force pulling m1 up the ramp, and gravity pulling down. But how can I solve for any of that if I don't have the mass?
I think you mean pulling down. Assuming you have the angle θ, and assuming M2 moves downwards. Consider the two systems separately. Knowing that the sum of the forces in any direction is equal to ma,

For T and M2, can you write an expression for the resultant force? Do the same for T and M1

(If you still aren't too sure how to do it, look back at the equations in post #4)
 
  • #18
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Post #4 confuses me. My prof is absolutely worthless, and I am unsure of what each equation you've stated, is solving for.
 
  • #19
rock.freak667
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Post #4 confuses me. My prof is absolutely worthless, and I am unsure of what each equation you've stated, is solving for.
Ok let's just consider everything to the left of the pulley, parallel to the plane. What are the forces acting? (state whether up or down the plane as well)
 
  • #20
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The horizontal component of gravity pulling on the block(so down the plane), tension pulling the block up the plane. Then of course the contact force of the plane pushing up on m1.
 
  • #21
rock.freak667
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The horizontal component of gravity pulling on the block(so down the plane), tension pulling the block up the plane. Then of course the contact force of the plane pushing up on m1.
You also have friction acting against the motion. Now if we assume M2 moves downwards, then what in what direction does M1 move and hence in which direction does the frictional force act?
 
  • #22
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So friction acts against the motion, so down the plane correct? I understand that much, but don't know where to start plugging in variables and such.
 
  • #23
rock.freak667
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So friction acts against the motion, so down the plane correct? I understand that much, but don't know where to start plugging in variables and such.
Right, so T is up, component of the weight is down, friction is down. So for this piece of the system can you form an equation for the resultant force M1a ?
 
  • #24
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m1a=T-weight-friction?
 
  • #25
rock.freak667
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m1a=T-weight-friction?
Right. But write the weight and the friction in terms of M1.


Now considering the mass M2, what are the forces acting? Find the resultant force M2a now.
 

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