Show that the acceleration of the system is upward

[cremelatte]

Homework Statement

A boy wants to reach a durian in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley, he pulls on the loose end of the rope with such a force that the spring scale reads 250N. The boy and the chair have masses 32kg and 16kg respectively.
a) show that the acceleration of the system is upward
b) find the magnitude of the acceleration
c) calculate the force the boy exerts on the chair

Homework Equations

2T + N - N - Wb - Wc
2T - Wb - Wc = (mb + mc)a
T + N - Wb = mba
T- N - Wc = mc a

The Attempt at a Solution

b) a = 0.61m/s
c) N =83N

My school teacher has gone through this question and he explained a million times but i don't get the equations above that he was trying to explain. I don't get why there's two T and why the boy and the chair are kind of different and everything is pretty confusing. Yup can someone explain how the boy even pulls himself and all the forces ;0

 

[tiny-tim]

welcome to pf!

hi cremelatte! welcome to pf!

I don't get why there's two T and why the boy and the chair are kind of different and everything is pretty confusing.

2T - Wb - Wc = (mb + mc)a

this is good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" …

it has to be applied to all the external forces on a body

it applies to any body, so in this case the body is boy-and-chair

(internal forces are not used … so you can ignore N)

draw a circle round the boy-and-chair … two ropes come out of it …

(you're only interested in what's inside the circle, so you don't care and you don't know that the ropes are joined)

so far as you're concerned, they are two entirely separate ropes which happen to have the same T

T + N - Wb = mba

Newton's second law for the boy, so only use the external forces on the boy

T- N - Wc = mc a

Newton's second law for the chair, so only use the external forces on the chair

can someone explain how the boy even pulls himself …

he's making the rope shorter!

 

[cremelatte]

haha okay thanks a lot! but just wondering if 250N is the reading on the spring scale, then how does it pull the boy and chair up when they weigh 32kg and 16kg (W = 470.88N). why doesn't the reading include this weight thing and so what's the actual force the boy uses to pull himself? :@





And now there's this other question:
A force F is required to push a crate along a rough horizontal floor at a constant speed v with friction present. What force is needed to push this crate along the same floor at a constant speed 3v if friction is the same as before?

I know that the answer to this is F, and that because constant speed and constant friction and all F = f, but then what affects the speed?

and if WD = F x d = work done against friction + K.E
F x d = f x d + 1/2 mv 2
if it is 3v,
F x d = f x d + 9/2 mv 2

WHYYY?

thanks a lot a lot a lot a lot thanks a lot a lot a lot a lot thanks a lot a lot a lot a lot thanks a lot a lot a lot a lot thanks a lot a lot a lot a lot thanks a lot a lot a lot a lot :) :) :) :)

 

[tiny-tim]

hi cremelatte!

… if 250N is the reading on the spring scale, then how does it pull the boy and chair up when they weigh 32kg and 16kg (W = 470.88N).

what is your https://www.physicsforums.com/library.php?do=view_item&itemid=26" equation for the boy-and-chair?

I know that the answer to this is F, and that because constant speed and constant friction and all F = f, but then what affects the speed?

as you know, a = 0, so the speed stays the same

so how does it get to be v or 3v?

that would depend on what happened earlier, which you're not told ​

and if WD = F x d = work done against friction + K.E

no, this is wrong

net https://www.physicsforums.com/library.php?do=view_item&itemid=75" = change in KE (and in PE),

and that's zero!

 

[cremelatte]

OHHHHH I GET IT OMGOMGOMG THANKS A LOT A LOT A LOT, anyway, :x JUST wondering,

1) why when i find the force the boy exerts on the chair, i only use one T?
2) the force the boy exerts on the chair is downwards towards the chair right, i realized if i used:
a) T - N - Wb = mba,
i get a negative N, but when i use

b) T + N - Wb = mba
i get a positive N.

Isnt N supposed to be positive when in eqn a and negative in eqn b? cause i thought its the force exerted ON the chair. and if upwards is taken as the positive direction,

-T + N + Wb = - mba
yeahh i still get a negative N.

so why is F exerted by boy on chair upwards and not towards the chair? kinda confused

hmmm. and yeahh there's Newton third law is equal and opposite, but

BOY on chair isn't it positive if downwards is positive direction... yeahh haha i know this

doesnt really affect the answer, but why do i take N as an upwards force... if its into the

chair...



And there's one last qn :X

A person pushes on two boxes with a horizontal force 100N force on a frictionless floor, as shown below. Box A is heavier than B.

(its flipped ard yeahhhh the picture)

Which of the following is correct?

A: Box A pushes on Box B with a force of 100N, and Box B pushes on A with a force of 100N

B: Box A pushes on Box B harder than box B pushes on box A

C: Boxes A and B push on each other with equal forces of less than 100N

D: The boxes will not begin to move unless the total weight of the two boxes is less than 100N


Okay the answer is C, but just wondering:
1) why is D wrong? is it if mass of A + B>100 then the box won't move? or what then the box won't move? (since its frictionless) :X
2) Why do the boxes push on each other with equal forces of less than 100N? where has the other remaining force went to... why isn't it 100N push on A which push on B which push back on A.

okay very confusing! :X


And also errmm wondering if you shoot a bullet from a gun, the recoil comes from the velocity of the gun after firing.

according to principal of momentum,
m(gun)u(gun) + m(bullet)u(bullet) = m(gun)v(gun) + m(bullet)v(bullet)

in this case, u(gun) is 0 in the beginning right, and u bullet is for example X. (when it leaves the gun the speed is X?)

then so the recoil is v(gun) which = m(bullet)u(bullet) = X.

So just wondering why is V(bullet) = 0?? where has the momentum of the bullet gone to... or is the whole thing wrong and so how is the transfer of momentum like in hereee :) :) :)



anw seriously thanks thanks thanks a lot okay! thanks for the time taken to help and all! yeahhh you all are super awesome i'll try helping around someday when i know my basics too! :)

 

[tiny-tim]

hi cremelatte!

Isnt N supposed to be positive when in eqn a and negative in eqn b? cause i thought its the force exerted ON the chair. and if upwards is taken as the positive direction,…

a https://www.physicsforums.com/library.php?do=view_item&itemid=73" on a body from another body in contact with it is always from the other body

2) Why do the boxes push on each other with equal forces of less than 100N? where has the other remaining force went to... why isn't it 100N push on A which push on B which push back on A.

you need to do F = ma on A, and on A-and-B …

what do you get? ​

So just wondering why is V(bullet) = 0??

that looks wrong …

if u is before and v is after, then obviously u(bullet) = u(gun) = 0

 

[cremelatte]

Wowww thanks you're a genius!

just one more clarification,

why when i find the force the boy exerts on the chair, i only use one T?

:@ thanks tim! :)

 

[tiny-tim]

when you do F = ma for the chair,

draw a circle round the chair, including only the bottom of the boy …

there's only one rope inside that circle, isn't there? ​

 

[cremelatte]

haha guess so
thanks tim!