What is the mass of the box held by Guy 3 in a three-pulley system?

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Homework Statement

Three guys hold up three boxes motionless on the pulley apparatus shown below. Guy 1 has a box with a mass of 40 kg, Guy 2 has a box with a mass of 100 kg, and Guy 3 has a box with an unknown mass. The boxes are, again, motionless. At some point they let go of them, but Guy 2 remains motionless. What is the mass of the box Guy 3 has?

here's the FBD but i actually added the forces, the diagram was given without the forces. are the forces directed properly? is there both tensions on the second one?

Homework Equations

• Fnet = m*a
• Fnet = F1 + F2 + F3 etc.
• Fg = m*g (g = 9.8 m/s^2)

The Attempt at a Solution

Three assumptions:

• given that Guy 2 remains motionless when they release the boxes, I'm guessing the implication is that accleration is 0.
  
• guy 2 has both tensions
  
• also, given that guy 2 has a heavier mass I'm guessing he's holding the middle box because it's weighted down more.

if those are wrong then please tell me a) why and b) what alternative assumptions I could make or if I need those assumptions even.

okay, so with that said, i got the T1 by secluding the Fnet of just Guy 1 (Fnet = T1 + (-Fg1) = m*a = T1 - m*g) so it ended up being T1 = 343 N.

guy 2 has both tensions, so i secluded his Fnet (Fnet = T1 + T2 + (-Fg2) = m*a = 343 + T2 - m*g) which ended up being T2 = 588 N.

guy 3 has the unknown mass, i secluded his Fnet = T2 + (-Fg3) = m*a = 588 - m*g, his mass being 60 kg.

turns out that's the wrong answer, so obviously i went wrong somewhere, probably the assumptions part or the secluding thing, idk. help me down the right path, spell it out for me if you would.

 

[page123]

Any advice?

 

[ehild]

The tension is the same everywhere in the string if the pulleys are massless.

ehild

 

[ashishsinghal]

given that Guy 2 remains motionless when they release the boxes, I'm guessing the implication is that accleration is 0.

That is correct!

guy 2 has both tensions

That is also correct! Same as ehild, tension is same all along the string.

also, given that guy 2 has a heavier mass I'm guessing he's holding the middle box because it's weighted down more.

No, no one is holding anything. Only tension in the string holds it together.

I will give you one more hint: The length of the rope is constant. I know it is obvious but this will help you solve the question.

 

[page123]

That is correct!


That is also correct! Same as ehild, tension is same all along the string.



No, no one is holding anything. Only tension in the string holds it together.

I will give you one more hint: The length of the rope is constant. I know it is obvious but this will help you solve the question.

So the tension and acceleration are the same? In what problem would the tensions be different and there's three objects involved? Or even if there were like 3000 objects connected by rope to pulleys, the tensions would be the same between each objects even if the objects were different masses?

And, forgive me, but I am getting the implication that the rope length is constant? Yeah, so that means it's motionless right? What do you mean?

 

[ashishsinghal]

rope length is constant meant the magnitude of acceleration of block 1 and 3 are same but I suppose you have figured it out earlier.

Tension in the string will always remain constant as long as the string is massless. I am now proving it to you.
If the tension in the rope is different then the string is also pulled by different force (Newton's third law). Then there will be a net finite force and since mass of the rope is 0, the acceleration will be infinite (or undefined).

 

[ehild]

I guess the axes of both upper pulleys are fixed, otherwise the whole thing would fall.

In most cases when you get a problem like this with strings and pulleys it is assumed that both the strings and the pulleys are massless. This involves that the tension is the same along a string.

The acceleration of any piece of a string is the same along itself , but the acceleration of m₂ is only half of it. If it moves upward by Δy₂, both pieces of the string between the middle pulley and the upper ones get shorter by 2Δy₂. If the displacements of m1 and m2 are Δy₁ and Δy₃, respectively,

Δy₁ +2Δy₂+Δy₃=0

as the length of the string is unchanged and the same relation is true for the accelerations:

a₁+2a₂+a₃=0.

But you need not bother about a₂ as box2 does not move, a₂=0.

ehild

 

[page123]

I guess the axes of both upper pulleys are fixed, otherwise the whole thing would fall.

In most cases when you get a problem like this with strings and pulleys it is assumed that both the strings and the pulleys are massless. This involves that the tension is the same along a string.

The acceleration of any piece of a string is the same along itself , but the acceleration of m₂ is only half of it. If it moves upward by Δy₂, both pieces of the string between the middle pulley and the upper ones get shorter by 2Δy₂. If the displacements of m1 and m2 are Δy₁ and Δy₃, respectively,

Δy₁ +2Δy₂+Δy₃=0

as the length of the string is unchanged and the same relation is true for the accelerations:

a₁+2a₂+a₃=0.

But you need not bother about accelerations as this is an equilibrium problem. If box2 stays in rest, both other boxes are motionless. The sum of forces on all boxes must be zero.

ehild

So basically to do this problem would I do a Fnet for masses 1 and 2 with the acceleration and get the tensions for both of them? would those tensions be the same as in:

fnet for mass 1 = m*a = 40*0 = T - fg1 = T - 40*9.8 = 0

fnet for mass 2 = m*a = 100*0 = T - fg2 = T - 100*9.8 = 0


the tensions would then obviously not be the same, so what am i doing wrong?

 

[ehild]

Sorry, I was wrong, I corrected it since then. Only mass 2 is motionless, the other move with accelerations of opposite directions and equal magnitude. See the corrected post.

ehild

 

[ashishsinghal]

So basically to do this problem would I do a Fnet for masses 1 and 2 with the acceleration and get the tensions for both of them? would those tensions be the same as in:

fnet for mass 1 = m*a = 40*0 = T - fg1 = T - 40*9.8 = 0

fnet for mass 2 = m*a = 100*0 = T - fg2 = T - 100*9.8 = 0


the tensions would then obviously not be the same, so what am i doing wrong?

The accelerations are not zero, consider acceleration as a and then solve. The acceleration is zero only for fg2. Now you have three variables: fg3, a and T and three equations. Solve them and you will surely get the correct answer.