Calculating Rank Pulley Forces with Newton's Equations

[isukatphysics69]

1. Homework Statement
In picture

Homework Equations

f=ma

The Attempt at a Solution

i can use Newtons equations here, but i am confused about the frictional force.. so it doesn't specify whether it is a frictional or unfrictional surface.. is there something I'm missing here? does friction not matter in this case?

 

[TSny]

Friction would matter, in general. But in order to have a definite answer, I think you should assume that the surface is frictionless. The problem statement should have been clearer.

 

[ehild]

View attachment 223391 1. Homework Statement
In picture

Homework Equations

f=ma

The Attempt at a Solution

i can use Newtons equations here, but i am confused about the frictional force.. so it doesn't specify whether it is a frictional or unfrictional surface.. is there something I'm missing here? does friction not matter in this case?

You do not need to bother about frictional forces. The pulleys are ideal, the friction between pulleys and axles is negligible and the friction between pulleys and strings is enough so the strings do not slip on the pulleys. And as @TSny said, you can assume frictionless surface of the table.

 

[isukatphysics69]

tHANK you guys, does anybody know what on Earth is going on with the last picture?? is the big block connected to the pulley or something?

 

[ehild]

tHANK you guys, does anybody know what on Earth is going on with the last picture?? is the big block connected to the pulley or something?

The big block is connected to the axle of the moving pulley.

 

[isukatphysics69]

The big block is connected to the axle of the moving pulley.
View attachment 223543

So the problem says that both blocks are attatched to an ideal string.. so in this case i think the block attatched to the pulley but the pulley is pulling the string down still

 

[isukatphysics69]

i am almost done deriving all of my equations for this problem i am currently on that last one E

 

[isukatphysics69]

So netforcebigblock = T/2 - MG? since the tension force is split ?

 

[ehild]

So netforcebigblock = T/2 - MG? since the tension force is split ?

No, the tension force in the common string does not split. There is two pieces of the same string pulling the pulley up.

 

[isukatphysics69]

No, the tension force in the common string does not split. There is two pieces of the same string pulling the pulley up.

So since the bigger box is twice as massive as the other box that means the smaller box should be accelerating up?

 

[isukatphysics69]

i have netforcesmallbox = t - mg and netforcebigbox = 2t - MG

 

[ehild]

i have netforcesmallbox = t - mg and netforcebigbox = 2t - MG

What are the directions of the net forces (and accelerations)?

 

[isukatphysics69]

What are the directions of the net forces (and accelerations)?

small block is positive going up and big block is negative going down so netforcebigblock = -2T +MG ?

 

[ehild]

small block is positive going up and big block is negative going down so netforcebigblock = -2T +MG ?

How are the accelerations related? You know that the length of the string is constant.
And note that M=2m. How much is the acceleration then?

 

[isukatphysics69]

How are the accelerations related? You know that the length of the string is constant.
And note that M=2m. How much is the acceleration then?

So acceleration is (-2T+MG)/2

 

[isukatphysics69]

But i will find acceleration of both systems momenterily

 

[ehild]

(-2T+MG)/2 is force, not acceleration. And the accelerations are related. Find T in terms of g and the masses.

 

[isukatphysics69]

i am getting acceleration = (-2*massofsmallblock*G - massofbigblock*G) / (massbigblock +2*massSmallblock)

 

[isukatphysics69]

i don't know if i am messing up my math it is 6AM i have not slept yet

 

[isukatphysics69]

so acceleration is -9.8 for system and when i plug that into find tension i am getting 0 so T = massofsmallblock*G + massofsmallblock*acceleration where mass of smallblock = 1kg and mass of big block = 2kg

 

[isukatphysics69]

So my ranking is D > A = B = C > E

 

[isukatphysics69]

My ranking is wrong >=[

 

[ehild]

i am getting acceleration = (-2*massofsmallblock*G - massofbigblock*G) / (massbigblock +2*massSmallblock)

No, it is not correct.
If the big mass M goes down by x, both pieces of the string attached to the moving pulley increase length by x. So the left piece of string shortens by 2x. If the big mass accelerates downward by a, the small block with mass m accelerates upward by 2a. Net force = mass times acceleration. What equations do you get for both blocks?

 

[isukatphysics69]

No, it is not correct.
If the big mass M goes down by x, both pieces of the string attached to the moving pulley increase length by x. So the left piece of string shortens by 2x. If the big mass accelerates downward by a, the small block with mass m accelerates upward by 2a. Net force = mass times acceleration. What equations do you get for both blocks?

ok i will try again.. i feel like i didn't make any wrong algebraic steps tho.

 

[isukatphysics69]

ok i figured out E and my friend told me that A and B are not equal.. how is that possible.

 

[isukatphysics69]

Ok i finished the ranking of tensions now i have to rank the accelerations
The accelerations of the big box

 

[ehild]

Show your work in detail, please,
What is the tension in the string in case A? What is the acceleration of the big box on the table?
What are the equations of motion of both boxes in case B? Solve them, What are the tension T and a, the common acceleration of the boxes?
Do the same for C.

 

[isukatphysics69]

Show your work in detail, please,
What is the tension in the string in case A? What is the acceleration of the big box on the table?
What are the equations of motion of both boxes in case B? Solve them, What are the tension T and a, the common acceleration of the boxes?
Do the same for C.

Case A: a = 0 since the netforce of pulling is 0
Case B : a = massSmallBox*G/(massBigBox+massSmallBox)
Case C: a = massBigBox*G/(massBigBox+massSmallBox)
Case D: a = (massBigBox*G-massSmallBox*G)/(massBigBox+massSmallBox)
Case E: aBigBox = (massBigBox*G - 2*massSmallBox*G)/(4*massSmallBox+massBigBox) = 0

 

[isukatphysics69]

 

[ehild]

Case A: a = 0 since the netforce of pulling is 0

The string is pulled by the force F = mg. What is the tension? What is the acceleration of the big box of mass M attached to it?

Case B : a = massSmallBox*G/(massBigBox+massSmallBox)
Case C: a = massBigBox*G/(massBigBox+massSmallBox)
Case D: a = (massBigBox*G-massSmallBox*G)/(massBigBox+massSmallBox
Case E: aBigBox = (massBigBox*G - 2*massSmallBox*G)/(4*massSmallBox+massBigBox) = 0

These are correct. Evaluate the accelerations for the case M=2m.