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jasenka [17]
3 years ago
14

{question \hookleftarrow}}" alt=" \sf{ \huge{question \hookleftarrow}}" align="absmiddle" class="latex-formula">
Need help with this one ~

Thanks for Answering !​

Physics
1 answer:
erica [24]3 years ago
7 0

Answer:

Hey There!

Let's solve...

I think your diagram must be similar to whatever i drew on attachment...

Now.....

n_{1} =  ma_{1} \\  \\ mg -  f_{1} - t =  2ma_{1} \\  \\ mg -  \mu _{1} ma_{1} - t =  2ma_{1} \\  \\ t = mg  -   ma_{1}(  \mu_{1} + 2 )

Lets solve bigger block which is along vertical

n_{2} = mg +  f_{1} + t \\  \\  n_{2} = mg +   \mu_{1} ma_{1} + mg -  \\  ma_{1} (  \mu_{1} + 2 )  \\  \\  n_{2} = (m + m)g -  2ma_{1} \\  \\

Lets solve it horizontally

2t -  n_{1} -  f_{2} =  ma_{1} \\  \\ 2mg -  2ma_{1}(  \mu_{1} + 2)  -  ma_{1} \\  -   \mu_{2} n_{2} =  ma_{1}

2mg -   \mu_{2} n_{2} =  a_{1}(m + m +  2m\mu_{1} +  4_{m})

2mg -   \mu_{2}^{(m + m)}g +  \mu_{2}2ma_{1} \\  \\  =  a_{1}(m + 5m + 2 \mu_{1}) \\  \\ g(2m -  \mu_{2}m -   \mu_{2}m) =  \\  \\  a_{1}(m + 5m +  2 \mu_{1} -  2 \mu_{2})

Final answer is just make fraction of that

a_{1} =  \frac{g(2m -   \mu _{2}m -     \mu_{2}m )}{(m + 5m + 2m \mu_{1} - 2m \mu_{2})}

so it is your answer.....

<h3>I hope it is helpful to you....</h3><h3>Cheers!___________</h3>

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