1018
  • 1018
physics problem (question's in the comments)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
1018
  • 1018
1 Attachment
Michele_Laino
  • Michele_Laino
1018
  • 1018
oh hi ms michele! thank you for your help again. haha, ok i will :)

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1018
  • 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
  • Michele_Laino
please "ms" stands for?
1018
  • 1018
oh, that's just short for miss. haha. sorry
anonymous
  • anonymous
Let's look at block m2 first forces acting on it are |dw:1441042822826:dw|
anonymous
  • 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
  • anonymous
But then why is the block 1 given for?Maybe I'm missing out something :p
1018
  • 1018
i was just about to ask if that is the working equation already. haha
anonymous
  • anonymous
Do you have an answer so we can at least verify by plugging in values?
1018
  • 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
  • anonymous
oh
1018
  • 1018
but do you want to see the new? although it's literally the same problem only with different values
1018
  • 1018
or i could just type the new values. but really the same question.
1018
  • 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
  • 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
  • 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
  • anonymous
Can you try the expression I got for Tension?see if it works
1018
  • 1018
ill try this equation. thanks! i'll let you know if its correct. haha
1018
  • 1018
the dot is multiplication right? just to be clear, with the mew(?) and g
anonymous
  • anonymous
yeep, I had to do it otherwise it was writing it as mug lol
anonymous
  • anonymous
\[mug\]
1018
  • 1018
on calcu its negative, is that ok?
1018
  • 1018
oh hold on i think i made a mistake ill try again
1018
  • 1018
hey! it's correct! thank you so much!

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