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1018
 one year ago
physics problem (question's in the comments)
1018
 one year ago
physics problem (question's in the comments)

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please see my answer to a similar question http://openstudy.com/users/michele_laino#/updates/55e295d1e4b03567e110576d

1018
 one year ago
Best ResponseYou've already chosen the best response.0oh hi ms michele! thank you for your help again. haha, ok i will :)

1018
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0please "ms" stands for?

1018
 one year ago
Best ResponseYou've already chosen the best response.0oh, that's just short for miss. haha. sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let's look at block m2 first forces acting on it are dw:1441042822826:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.0But then why is the block 1 given for?Maybe I'm missing out something :p

1018
 one year ago
Best ResponseYou've already chosen the best response.0i was just about to ask if that is the working equation already. haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you have an answer so we can at least verify by plugging in values?

1018
 one year ago
Best ResponseYou've already chosen the best response.0no 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

1018
 one year ago
Best ResponseYou've already chosen the best response.0but do you want to see the new? although it's literally the same problem only with different values

1018
 one year ago
Best ResponseYou've already chosen the best response.0or i could just type the new values. but really the same question.

1018
 one year ago
Best ResponseYou've already chosen the best response.0i 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 year ago
Best ResponseYou've already chosen the best response.0One 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah that's what i thought the first time, but i just assumed that it was not moving since there's a string involved.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you try the expression I got for Tension?see if it works

1018
 one year ago
Best ResponseYou've already chosen the best response.0ill try this equation. thanks! i'll let you know if its correct. haha

1018
 one year ago
Best ResponseYou've already chosen the best response.0the dot is multiplication right? just to be clear, with the mew(?) and g

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeep, I had to do it otherwise it was writing it as mug lol

1018
 one year ago
Best ResponseYou've already chosen the best response.0on calcu its negative, is that ok?

1018
 one year ago
Best ResponseYou've already chosen the best response.0oh hold on i think i made a mistake ill try again

1018
 one year ago
Best ResponseYou've already chosen the best response.0hey! it's correct! thank you so much!
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