# Two blocks, a fixed pulley and friction

• SakuRERE
In summary, the conversation discusses a scenario where two masses, m1 and m2, are connected by a string and released from rest at time t=0. The system is subject to static and kinetic friction coefficients, μs and μk respectively. The first question asks for the minimum mass, m2,min, needed for the system to start moving, while the second question asks for the final velocity of both blocks when m2 has fallen a height H. The solution involves finding the maximum static friction for m1 and using it to determine the mass of m2 needed for the system to start moving. For the second question, the same acceleration is applied to both blocks, so they will have the same final velocity.
SakuRERE

## Homework Statement

a mass m1 is attached to a second mass m2 by an acme (massless, unstretchable) string. m1 sits on a table with which it has coefficients of static and dynamic friction μs and μk respectively. m2 is hanging over the ends of a table, suspended by the taut string from an acme pulley. at time t=0 both masses are released.
1. what is the minimum mass m2,min such that the two masses begin to move?
2. If m2= 2m2,min, determine how fast the two blocks are moving when mass m2 has fallen a height H (assuming the m1 hasn't yet hit the pulley)?

Σf=ma
fsmax=μs*fn
vf2=vi2+2aΔx[/B]

## The Attempt at a Solution

for a:
I am not sure about how I understand the question, but what comes to my mind when reading ( Begin to move) is that still the system acceleration is zero and that they are asking for the Fsmax (the maximum static friction) just before the system starts to move and the friction becomes Kinetic friction of Fk.
so:
for block m1
Σƒ= m*a=0
fsmax=T
μs*Fn= T
μs*m1g=T
now for block m2:
Σƒ=m*a=0
m2g=T

and by this m2g=μs*m1g so m2=(μs*m1g)/g

section b)
I guess since the mass of m2 is doubled, that here we will deal with the kinetic friction rather than the static friction.
since both are linked by a rope with the same pulley than they must have the same acceleration, and we should find it.
and after finding it then we will use this equation:
vf2=vi2+2aΔx
where vi= 0
so vf=√(2aΔx).
am I right??

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SakuRERE said:
so m2=(μs*m1g)/g
Yes, but you can simplify a little.
SakuRERE said:
section b)
I guess since the mass of m2 is doubled, that here we will deal with the kinetic friction rather than the static friction.
since both are linked by a rope with the same pulley than they must have the same acceleration, and we should find it.
and after finding it then we will use this equation:
vf2=vi2+2aΔx
where vi= 0
so vf=√(2aΔx).
am I right??
Yes.

SakuRERE
haruspex said:
but what is a
yes, of course, I will be substituting a in term of the given variables but I just didn't because I wasn't sure from my solution for section a at the first place. but now after your response, I will continue the solution. and regarding the H and Δx, I will be substituting Δx with H simply, but I wanted to write the equation in its original form so that everyone easily identifies it.
I will be happy if you grant me another help and answered this question:
for a moment I wondered, will the final velocity be the same for both blocks, I know they have the same acceleration and H will be the same as well but doesn't the friction affect the velocity of that block in the incline?

SakuRERE said:
I will be substituting a in term of the given variables
SakuRERE said:
will the final velocity be the same for both blocks, I know they have the same acceleration
Same acceleration over the same period of time, so...

SakuRERE
haruspex said:
Same acceleration over the same period of time, so...
Thanks! I just wanted to make sure and to move away any doubts that are used to appear out of nowhere :)

## 1. What is the purpose of a fixed pulley in a two-block system?

The fixed pulley serves to change the direction of the force applied to one of the blocks, making it easier to lift or move the load.

## 2. How does friction affect the movement of the two blocks?

Friction acts as a resistive force that opposes the motion of the two blocks. It can make it harder to move the blocks, and can also cause them to slide against each other.

## 3. Is there a limit to how much weight the two-block system can lift?

Yes, the maximum weight that the system can lift is equal to the weight of the heavier block. This is due to the conservation of energy and the fact that the force applied to one block is equal to the force needed to lift the other block.

## 4. Can the position of the fixed pulley affect the amount of force needed to lift the blocks?

Yes, the position of the fixed pulley can affect the direction and magnitude of the force needed to lift the blocks. Placing the pulley closer to the load can make it easier to lift, while placing it further away can require more force.

## 5. How does the number of pulleys in a two-block system affect the force needed to lift the load?

The more pulleys there are in the system, the less force is needed to lift the load. This is due to the fact that each additional pulley reduces the amount of force needed by half, making the system more efficient.

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