physics problem (question's in the comments)

- 1018

physics problem (question's in the comments)

- chestercat

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- 1018

##### 1 Attachment

- Michele_Laino

please see my answer to a similar question
http://openstudy.com/users/michele_laino#/updates/55e295d1e4b03567e110576d

- 1018

oh hi ms michele! thank you for your help again. haha, ok i will :)

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## More answers

- 1018

@Michele_Laino hi are you still online? im having trouble with the problem. if you're not too busy maybe you can help me? thanks

- Michele_Laino

please "ms" stands for?

- 1018

oh, that's just short for miss. haha. sorry

- anonymous

Let's look at block m2 first
forces acting on it are
|dw:1441042822826:dw|

- anonymous

for equilibrium resultant force on block m2=0
this implies
for vertical components
\[N=m_{2}g+F.\sin(34)\]
We know that friction is
\[f=\mu.N\]
\[\therefore f=\mu(m_{2}g+F.\sin(34))\]
For horizontal components we have
\[T+f=F.\cos(34)\]
From this you can find T
\[T=F.\cos(34)-f=F.\cos(34)-\mu(m_{2}g+F.\sin(34))\]

- anonymous

But then why is the block 1 given for?Maybe I'm missing out something :p

- 1018

i was just about to ask if that is the working equation already. haha

- anonymous

Do you have an answer so we can at least verify by plugging in values?

- 1018

no i dont sorry. actually the problem changed already, i mean same problem but with different values because i tried to answer it and it was wrong haha so im just trying to see the analysis of it

- anonymous

oh

- 1018

but do you want to see the new? although it's literally the same problem only with different values

- 1018

or i could just type the new values. but really the same question.

- 1018

i have an idea but i dont know. haha. there 2 tensions present in the string, so maybe it's tension (right) - tension (left) lol

- anonymous

One thing that I think is not given is whether the applied force is keeping the system in equilibrium or causing acceleration, which is a key deciding factor
Let's assume it causes acceleration a, cuz that seems to be the case here,
Therefore for block 1 we have
Force acting towards the right on it is the tension T and towards left have friction equal to normal reaction
\[T-\mu.m_{1}g=m_{1}a\]
Thus to find T we need to eliminate a
\[T=m_{1}(\mu.g+a)\]
Now consider the system of box 1 and 2 combined
We have
\[a=\frac{F_{res}}{m_{1}+m_{2}}\]
Where F is the resultant force on the system of 2 blocks
I think F is
\[F_{res}=T-\mu.m_{1}g+F.\cos(34)-\mu(m_{2}g+F.\sin(34))-T\]
The first 2 terms correspond to block 1 and other terms to block 2
cancelling out the Tension we get
\[F_{res}=F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))\]
\[\implies a=\frac{F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))}{m_{1}+m_{2}}\]
plugging this into equation for Tension we have
\[T=m_{1}(\mu.g+\frac{F.\cos(34)-\mu(m_{1}g+m_{2}g+F.\sin(34))}{m_{1}+m_{2}})\]
I'm not sure about it, but maybe worth a shot

- 1018

yeah that's what i thought the first time, but i just assumed that it was not moving since there's a string involved.

- anonymous

Can you try the expression I got for Tension?see if it works

- 1018

ill try this equation. thanks! i'll let you know if its correct. haha

- 1018

the dot is multiplication right? just to be clear, with the mew(?) and g

- anonymous

yeep, I had to do it otherwise it was writing it as mug lol

- anonymous

\[mug\]

- 1018

on calcu its negative, is that ok?

- 1018

oh hold on i think i made a mistake ill try again

- 1018

hey! it's correct! thank you so much!

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