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

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The discussion focuses on solving a pulley system problem involving a 200N weight and a 40N weight, with specific equations and forces outlined for each block. The key equations include the relationship between forces, friction, and acceleration for both blocks, emphasizing the mechanical advantage provided by the pulley system. The downward force on block A is calculated by subtracting twice the frictional force on block B from its weight. The masses of both blocks are derived from their weights, and the acceleration of block A is determined using these masses. The final velocity is calculated by incorporating the initial velocity and acceleration over time, highlighting the importance of understanding the mechanics involved in the system.
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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!
 

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  • 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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