HEEEELP!! How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart.

- anonymous

- schrodinger

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- anonymous

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- anonymous

How do I calculate the acceleration of the cart shown, given the following assumption: A frictional resistance of magnitude 2.0 N is acting on the cart. It first asks what the acceleration will be with no friction is acting on a cart and my formula is:
T = m1a
m2a = m2g - T
T = m2g - m2a
m1a+ m2a = m2g
a = m2g / (m1 + m2)

- anonymous

and the answer for my current problem stated above is 2.9 m/s^2... and I don't know how to solve it D;

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## More answers

- anonymous

Someone? Help? :(

- anonymous

a=F/m
F=driving force-resistant force
what's the driving force in this case?

- anonymous

driving force? It.. It doesn't say.

- anonymous

well i'm just asking you to think about it. it will help you in solving problems in the future. what force is causing the system to move?

- anonymous

ohh.. Tension force.

- anonymous

is this moving to the left?

- anonymous

to the right.

- anonymous

ok, so no it's not the tension force. it's the force from the hanging mass

- anonymous

really? ohh.. okay.

- anonymous

gravity causes the masses to move

- anonymous

ohh, i see it. 'Cause when the hanging mass falls the cart moves.

- anonymous

yup.

- anonymous

so \[driving force=m _{hanging}*g\]

- anonymous

alright.

- anonymous

now the resistant force is already given in the problem

- anonymous

how?
And is your equation the same as:
a = m2g / (m1 + m2)?

- anonymous

what are you using for m2?

- anonymous

2.0 kg

- anonymous

and you said how the resistant force is already given in the problem-- how?

- anonymous

ok then yes, it will end up as that equation you have but i will break it down
a=F/m
resistant force = 2N
F=2g-2N
=19.6N-2N
back to acceleration...
a=F/mass of the system = (19.6N-2N)/(2kg+4kg)

- anonymous

how did you get 2N as the resistant force? o.O
'cause the answer in the back of the text is 2.9 m/s^2

- anonymous

oops, actually different than yours but i'm confident in what i put up there

- anonymous

do you know a way in which I can use my formula?

- anonymous

my final is 2.9m/s^2

- anonymous

ohhh but do you know how to answer the problem with my formula? ...

- anonymous

tension cancels out for this system which is why i didn't use it in any of my equations

- anonymous

your formula seems a little off but i will double check

- anonymous

ohh.. okay.

- anonymous

break down the system in 2 parts. hanging mass will be labeled 1.
\[m _{1}g-T _{1}=m _{1}a\] ^^^derived from F=ma
solve for T
\[T _{1}=m _{1}(g-a)\]
for second part...
\[T _{2}-m _{2}g=m _{2}a\]
solve for T...
\[T _{2}=m _{2}(g+a)\]

- anonymous

ohh..

- anonymous

so what you have to calculate a would work if there was no friction but since there is, the best way to do it is the way i did it

- anonymous

ohh, cos that formula I showed you was for no friction & I thought maybe to switch a few places would be okay.

- anonymous

Well then, thank you so much elica for putting up with me!! Thank you!!

- anonymous

np

- anonymous

btw, why is it: F=2g-2N? Why is it 2g?

- anonymous

after you calculate 2g (2kg*9.8m/s^2) it will equal to 19.6N. the unit N=kg*m/s^2

- anonymous

just remember you can't add or subtract unless they have the same units

- anonymous

ohhh, so it's like m2g?

- anonymous

yes

- anonymous

ohh, okay. Thank you!

- anonymous

np

- anonymous

so F = m2g - frictional resistance? (what's the variable for it?)

- anonymous

in this case, frictional force is denoted as
\[F _{k}=\mu mg\]
k for kinetic friction
and the u-looking sign is the friction coefficient which is usually given in a problem. in this problem, the Fk was already solved (2N)

- anonymous

ohhhh, I'm familiar with that! Oh, thank you!

- anonymous

yup

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