What is the solution for a pulley system with 200N weight and 40N weight?

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Homework Help Overview

The discussion revolves around a pulley system involving two weights: a 200N weight and a 40N weight. Participants are analyzing the forces acting on each block and attempting to derive equations of motion based on the given parameters, including friction and acceleration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on each block, including the effects of friction and mechanical advantage. Some question the original poster's equations and suggest rechecking the forces involved. There is mention of needing three equations to solve the problem, focusing on the equations of forces for both blocks and their relationship in terms of acceleration.

Discussion Status

The discussion is active, with participants providing hints and corrections regarding the setup of the problem. While some equations have been presented, there is no explicit consensus on the solution, and the original poster expresses confusion about their approach.

Contextual Notes

Participants note the importance of correctly identifying the blocks and the mechanical advantage due to the pulley system. There is also a mention of the need to include friction in the calculations, which was initially overlooked.

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Homework Statement



Pulley system, attachment below. Would be impossible to explain without a diagram.



Homework Equations



Wa = 200N
Va0=2m/s
t= 2 seconds
Wb= 40N
μK=0.2
g=9.81m/s^2

The Attempt at a Solution



Okay my attempted solution was this:

For block A:

40N + Fn+Tension =m1a

T - 40N*0.2=40N/9.81

Block 2

200 + Fn + Tension = m2a

200N + T = 200N/9.81

So lost to be honest!
 

Attachments

  • Screen Shot 2014-08-25 at 2.32.45 pm.png
    Screen Shot 2014-08-25 at 2.32.45 pm.png
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Recheck the forces on each of the blocks (I didn't see you include the force of friction anywhere??)

Also, please use the correct names of the blocks ("A" and "B") to avoid confusion :-p (you said "Block A" and "Block 2")
Edit:
Hint: (in case I go to sleep before you reply) it only takes 3 equations to solve this problem:
The "equation of forces" on block A
The "equation of forces" on block B
And an equation which relates the acceleration of block A to the acceleration of block B
 
Last edited:
Ans:
final velocity= v(initial) + g(Wa-2μWb)/(4Wb+Wa)t
 
Maharshi Roy said:
Ans:
final velocity= v(initial) + g(Wa-2μWb)/(4Wb+Wa)t

That is correct, but Stelios already knows the answer. He doesn't just want the answer, he wants to understand it.
 
The downward force on block A is its weight minus twice the frictional force on B (twice because of the mechanical advantage of the two-fold pulley). This will be constant because (IIRC) sliding frictional force does not vary with speed.

That is ##F_A=W_A-2\mu_k W_B##

The mass of A is ##m_A=W_A/g##

The mass of B is ##m_B=W_B/g##

The acceleration of A is ##a_A=F_A/(m_A+2m_B)## The coefficient of 2 for ##m_B## is there because of the reverse mechanical advantage of the pulley: a downward force on the pulley translates to half that force on the single rope leading to B.

The speed at time ##t## will be ##v_{A0}+a_At##
 
Last edited:
Thanks Andrew, I forgot to include note the fact that their is mechanical advantage as a result of two pulleys in the system.

And as for block a and block 2 I think it was just a typo :(
 

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