Finding Minimum A for Equilibrium in a Two Block Problem

[goodOrBad]

Homework Statement

Given the weight of block A, friction coefficients and an angle determine the max.weight of B for which the system will stay in balanc

Relevant Equations

fg=µ1N

T2=T1e^(µ2β)

 

[haruspex]

You have $\beta=\alpha+90^o$. Check that.

 

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So it should be -90? I don really understand how I have to determine the angle

 

[haruspex]

So it should be -90? I don really understand how I have to determine the angle

Did you draw a diagram of the circle with radii to the ends of where the string arcs over it?

 

[goodOrBad]

no, but it does seem like the rope arcs over a 90 degree area

 

[haruspex]

View attachment 269851
no, but it does seem like the rope arcs over a 90 degree area

If the upper section of the string were horizontal it would be 90 degrees. But it isn't.
Mark the ends of string contact on that circle.

 

[goodOrBad]

Okay so
cos30T1=0.3*500
T1=173.205
B=173.205*e^(0.2*300* π/180)=493.574N

 

[haruspex]

cos30T1=0.3*500

That looks like a big step backwards.
What is wrong with the equations you had previously, $T_1\cos(30)=\mu_1N=\mu_1(500+T_1\sin(30))$?

e^(0.2*300* π/180)

300? Where did that come from? Or did you mean 30^o?
No, not 30^o either.
Did you draw the diagram I described?

 

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then beta should be 60 right?

 

[haruspex]

then beta should be 60 right?View attachment 269878

Yes.

 

[goodOrBad]

Great then this should be correct

 

[haruspex]

Great then this should be correct
View attachment 269879

Looks good.

 

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Thank you now I have one more question, what if almost everything was the same but... I was given weight of B and needed to find A, would I be able to use T2=T1e^(µ2β)

 

[haruspex]

View attachment 269881
Thank you now I have one more question, what if almost everything was the same but... I was given weight of B and needed to find A, would I be able to use T2=T1e^(µ2β)

To find the minimum A for equilibrium? Sure - why should the equations change?