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

  • Thread starter Thread starter SteliosVas
  • Start date Start date
  • Tags Tags
    Pulleys
AI Thread Summary
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.
SteliosVas
Messages
70
Reaction score
0

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
    26.2 KB · Views: 593
Physics news on Phys.org
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 :(
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top