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Mechanics question help

I know this is physics but seeing as mechanics is maths at A-level, I thought some of you could help me ? It is part of a classical mechanics MOOC I am doing.

7ad18f2386dff2dcd0fec34851653e9b.png

Thanks
Reply 2
Original post by Patrick2810
I know this is physics but seeing as mechanics is maths at A-level, I thought some of you could help me ? It is part of a classical mechanics MOOC I am doing.

7ad18f2386dff2dcd0fec34851653e9b.png

Thanks


Let the force by the wind on each block be TT.
Let F1,F2F_1, F_2 be the friction between the blocks and the friction between the lower block and the floor respectively.


The upper block is moving to the right at constant speed thus F1F_1 is acting towards the left and F1=TF_1 = T. Hence

T=F1=μM1gT = F_1 = \mu M_1 g

M1=F1μg\Rightarrow M_1 = \frac{F_1}{\mu g}


The lower block is moving to the left at constant speed thus F1F_1 on this block is acting towards the right and F2F_2 is also acting towards the right. Hence:

F=F1+T+F2=2F1+F2F = F_1 + T + F_2 = 2F_1 +F_2

F=2F1+μ(M1+M2)g\Rightarrow F = 2F_1 + \mu (M_1 + M_2)g

F1=Fμ(M1+M2)g2\Rightarrow F_1 = \frac{F - \mu (M_1 + M_2)g}{2}


Substitute this into the above expression of M1M_1, we have:

M1=Fμ(M1+M2)g2μgM_1 = \frac{F - \mu (M_1 + M_2)g}{2 \mu g}

Rearrange to make M1M_1 the subject, we get:

M1=F3μgM23M_1 = \frac{F}{3 \mu g} - \frac{M_2}{3}


:smile:
Original post by NDVA
Let the force by the wind on each block be TT.
Let F1,F2F_1, F_2 be the friction between the blocks and the friction between the lower block and the floor respectively.


The upper block is moving to the right at constant speed thus F1F_1 is acting towards the left and F1=TF_1 = T. Hence

T=F1=μM1gT = F_1 = \mu M_1 g

M1=F1μg\Rightarrow M_1 = \frac{F_1}{\mu g}


The lower block is moving to the left at constant speed thus F1F_1 on this block is acting towards the right and F2F_2 is also acting towards the right. Hence:

F=F1+T+F2=2F1+F2F = F_1 + T + F_2 = 2F_1 +F_2

F=2F1+μ(M1+M2)g\Rightarrow F = 2F_1 + \mu (M_1 + M_2)g

F1=Fμ(M1+M2)g2\Rightarrow F_1 = \frac{F - \mu (M_1 + M_2)g}{2}


Substitute this into the above expression of M1M_1, we have:

M1=Fμ(M1+M2)g2μgM_1 = \frac{F - \mu (M_1 + M_2)g}{2 \mu g}

Rearrange to make M1M_1 the subject, we get:

M1=F3μgM23M_1 = \frac{F}{3 \mu g} - \frac{M_2}{3}


:smile:


Thanks!! I'll read and try to understand when I'm more awake. Out of interest, which out of m1,m2,m3,m4,m5 would this sort of Q come up ?

thanks again :smile:

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