Pulley question -- 2 pulleys and 2 weights

[kieyard]

THIS THREAD WAS INCORRECTLY PLACED IN THE CLASSICAL PHYSICS FORUM, SO THERE IS NO TEMPLATE

Hi, everyone.
my teacher stuck me with these questions and i am need of a little help answering question 2.

here's my answers to question 1 first.
1a) a=0, as system would be in equilibrium when F=0
b) F=5mgN, as for a=5m/s the acceleration of B has to equal 2.5m/s and as its in equilibrium F has just got to equal m*a which is 2.5*2mg=5mgN
c) F=mga

i think my answers are right but if I am making a simple mistake let me know.

now my attempts at 2.

2a) first, B has to x/2 above the ground for the full system to work so original potential energy is 2mgx/2=mgx assuming A starts on the ground. now its the following kenetic energy bits that are confusing me because there will be two different V's one for A and one for B. V_A = 2V_B so assuming no energy gets lost to heat via friction whatever all of the gravitational energy plus energy input from F has to turn into kinetic energy so I am thinking.

mgx + Fx = 1/2mV_A ² + mV_B ²

which kind of works if your given enough information. but part B doesn't i don't think.

2b) substituting values into my answer to 2a

98 + 196 = 1/2[ V_A ² + 2V_B ² ]
so V_A ² + 2V_B ² = 588N
and V_A = 2V_B
so with further manipulation i get V_A = 19.8m/s

but using logic from question 1. if F=19.6 and B has a mass of 2kg, the acceleration of B = 9.8m/s ² therefore acceleration of A = 19.6m/s ² and then using SUVAT.

s=10m u=0 v=? a=19.6 m/s ²
v ² = u ² +2as
v = 23.2m/s

which is different. iv tried multiple times and i can't figure out what I am doing wrong. I am pretty sure its the WEP part not the SUVAT part as I am better at that but i do not know.

please help.

 

[SteamKing]

It would be nice if you had posted the text of the questions as well.

As your post reads now, we have no idea if you are correct or not.

 

[kieyard]

It would be nice if you had posted the text of the questions as well.

As your post reads now, we have no idea if you are correct or not.

iv updated it now so the picture of the question is on. sorry about that. it quite late haha

 

[haruspex]

Your answer to 1b) is wrong.
Write out the full force/acceleration equations for the two masses, including T as a common unknown.
Taking shortcuts often leads to errors.

 

[kieyard]

Your answer to 1b) is wrong.
Write out the full force/acceleration equations for the two masses, including T as a common unknown.
Taking shortcuts often leads to errors.

okay, here i go.

For mass A: T-mg = 5m =X2=> 2T-2mg = 10m
For mass B: F+2mg-2T= -2.5(2m) = -5m

adding the two equations:

F=5m

is that correct?

 

[haruspex]

okay, here i go.

For mass A: T-mg = 5m =X2=> 2T-2mg = 10m
For mass B: F+2mg-2T= -2.5(2m) = -5m

adding the two equations:

F=5m

is that correct?

Check your signs in the mass B equation. Think which way the acceleration is.

 

[kieyard]

Check your signs in the mass B equation. Think which way the acceleration is.

i have taken up to be my positive direction so a is positive for A and negative for B.

is that not right?

 

[haruspex]

i have taken up to be my positive direction so a is positive for A and negative for B.

is that not right?

The left hand side of your mass B equation takes down to be positive.

 

[kieyard]

The left hand side of your mass B equation takes down to be positive.

okay then. would that them make F=15m ?

 

[haruspex]

okay then. would that them make F=15m ?

Yes. But you should include units since the 15 is not dimensionless.

 

[kieyard]

Does that then make the answer to 1c) F=3ma N ?

 

[haruspex]

Does that then make the answer to 1c) F=3ma N ?

Yes, but this time you should leave out the units. That's because m is a mass and a an aceleration in unspecified units, so it's just F=3ma.

 

[kieyard]

so was my WEP question right?

 

[haruspex]

B has to x/2 above the ground for the full system to work

Do you mean it has to start x/2 above the ground? It certainly needs to be at least that.

original potential energy is 2mgx/2=mgx

You can't say what the original PE is unless you define a zero level. The loss in PE by B will be mgx.

mgx + Fx = 1/2mV_A ² + mV_B ²

How far does F move?

 

[kieyard]

assume the zero lvl is at the base of mass A and B is exactly X/2 above the ground.
the PE of the entire system is then: 2mgx/2 = mgx
the energy by F is: Fx/2
KE of A afterwards is 1/2*m*v_a²
KE of B afterwards is 1/2*2m*(v_b/2)² = m*(v_b/2)²

as v_b = 1/2v_a

KE of B can also be written as: m*(v_a/4)² = mv_a²/16

so the equation should be:
mgx + Fx/2 = 1/2mv_a² + mv_a²/16

which can be simplified to:
mgx + Fx/2 = 9/16mv²

is that right or am i missing somthing?

 

[haruspex]

KE of B afterwards is 1/2*2m*(v_b/2)²

Why v_b/2?

 

[kieyard]

i was thinking as it is half V_a but then i did it again. does that then make the end, mgx + Fx/2 = 3/4mv² is that right?

 

[haruspex]

i was thinking as it is half V_a but then i did it again. does that then make the end, mgx + Fx/2 = 3/4mv² is that right?

Yes.

 

[kieyard]

for part 2b I am still getting different answers depending on which method i use.

if F=3ma, 19.6=3(1)a, a=6.53.
using v² = u²+2as
v² = 2(6.53)(10) = 130.67
so, v= 11.43 m/s

but if we use mgx + Fx/2 = 3/4mv²
1(9.8)(10) + 19.6(10/2) = 3/4(1)v²
196 = 3/4v²
v²=261.33
v = 16.17 m/s

whats wrong?

 

[haruspex]

for part 2b I am still getting different answers depending on which method i use.

if F=3ma, 19.6=3(1)a, a=6.53.
using v² = u²+2as
v² = 2(6.53)(10) = 130.67
so, v= 11.43 m/s

but if we use mgx + Fx/2 = 3/4mv²
1(9.8)(10) + 19.6(10/2) = 3/4(1)v²
196 = 3/4v²
v²=261.33
v = 16.17 m/s

whats wrong?

I missed another error in the OP.
Think again about the total change in PE.

 

[kieyard]

would mass A also gain PE
so mgx + Fx/2 = 3/4mv² + mgx
which would leave you with Fx/2 = 3/4mv²

19.6(10)/2 = 3/4(1)v²
98 = 3/4v²
v² = 130.67
v = 11.43

which is the same as the suvat and forces one. :) we did it

 

[haruspex]

would mass A also gain PE
so mgx + Fx/2 = 3/4mv² + mgx
which would leave you with Fx/2 = 3/4mv²

19.6(10)/2 = 3/4(1)v²
98 = 3/4v²
v² = 130.67
v = 11.43

which is the same as the suvat and forces one. :) we did it

Yes.

 

[kieyard]

it was right and i then received more questions

what iv noticed is that the questions are the same, so this is basically a check before i hand in my answers as hopefully ill get them right.
iv also noticed B will never be as low as A due to the ropes making a infinite series upto 1 or x but this time B can only move x/4 if A is on the ground.

okay question time.

3a) a = 0 as in equilibrium again

b) do c) first and put a = 5 gives F = 25m N

c) F+4mg-4T = 4ma/4 = ma
T-mg = ma ==> 4T-4mg = 4ma
so F = 5ma

4) 4mgx/4 + Fx/4 = 1/2(4m)(v/4)² + 1/2mv² + mgx
mgx + Fx/4 = mv²/8 + 1/2 mv² + mgx
Fx/4 = 5/8 mv²

4b) 19.6(10)/4 = 5/8(1)v²
49 = 5/8v²
v² = 78.4
v = 8.85 m/s

 

[haruspex]

it was right and i then received more questions
View attachment 86870
what iv noticed is that the questions are the same, so this is basically a check before i hand in my answers as hopefully ill get them right.
iv also noticed B will never be as low as A due to the ropes making a infinite series upto 1 or x but this time B can only move x/4 if A is on the ground.

okay question time.

3a) a = 0 as in equilibrium again

b) do c) first and put a = 5 gives F = 25m N

c) F+4mg-4T = 4ma/4 = ma
T-mg = ma ==> 4T-4mg = 4ma
so F = 5ma

4) 4mgx/4 + Fx/4 = 1/2(4m)(v/4)² + 1/2mv² + mgx
mgx + Fx/4 = mv²/8 + 1/2 mv² + mgx
Fx/4 = 5/8 mv²

4b) 19.6(10)/4 = 5/8(1)v²
49 = 5/8v²
v² = 78.4
v = 8.85 m/s

Looks right.

 

[kieyard]

he had more. just checking again for 5 and need help on 6.

5) we have an equation to find V when B is there.

Fx/4 = 5/8mv²

we can then work out an equation for when B isn't there:

Fx/4 = 1/2mv² + mgx

now F has to be greater than the weight of B as the question says so let F = 100N, m=1KG, and x = 10m.

with B:
100(10)/4 = 5/8(1)v²
250 = 5/8v²
v² = 400
v = 20 m/s

without B:
100(10)/4 = 1/2(1)v² + 1(9.8)(10)
152 = 1/2v²
v² = 304
v = 17.44 m/s

therefore with B is more efficient.

6) now as its the method with B i want to assume B changes as P increases so that the system remains in equalibrium without F as it did between questions 1-2 and 3-4.
if that's the case i worked out the mass of B = 2^p-1
it would also have less to travel, x/2^p-1 m i think?

so would it look something like:

2^p-1mgx/2^p-1 + Fx/2^p-1 = 1/2m(v/2^p-1)² + 1/2mv² + mgx

mgx + Fx/2^p-1 = 1/2mv²/2^2p-2 + 1/2mv² + mgx

Fx/2^p-1 = mv²/2^2p-1 + 1/2mv²

Fx/2^p-1 = (2+2^2p-1)(mv²)/2^2p

now that doesn't seem right at all so i would for you to check it over and correct me please haha

thanks.

 

[haruspex]

I'm not sure how to interpret 5. You have compared a given F, with and without B. But maybe they intend you compare some F in conjunction with B, for a total force of F+m_Bg, with just a force F'=F+m_Bg.

Does your general equation for part 6 produce the right answers for p=2, 3?

 

[kieyard]

i think its removing the weight of B aswell. and my equation gave me 12.52 m/s instead of 11.43 m/s for question 2 values. where am i going wrong?

 

[haruspex]

i think its removing the weight of B aswell. and my equation gave me 12.52 m/s instead of 11.43 m/s for question 2 values. where am i going wrong?

Take your working for part 6, substitute p=2 everywhere, and compare it with your previous working for the p=2 case.

 

[kieyard]

Fx/2^2-1 = (2+2^2(2)-1)mv²/2²⁽²⁾

Fx/2 = 10/16mv² = 5/8mv²

which isn't the same as

Fx/2 = 3/4mv²