Agebra Question

[HollowSuperet36]

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I feel really dumb.

This was given as a rational as to why a block on a frictionless table would accelerate at a rate less than g when attached to another block by a string hanging off the table.

I understand F=mg-T

But can't derive the expression for 'a' without canceling both out when dividing. This should be really easy. What am I missing?

MAg-mBa=mAa

a=mAg/(mB+mA)

Help much appreciated.

P.S.
I know it says Agebra. I said I felt dumb didn't I.

 

[411309]

is there air resistance involved?

 

[MedPR]

I'm assuming you need another equation.

 

[HollowSuperet36]

Discrete

Blocks A and B are connected across a frictionless pulley by a cord of negligible weight, W. If block B lies on a horizontal frictionless surface and block A hangs vertically, what can be said about the acceleration of block B?

A. It would accelerate at approximately 9.8 m/s2
B. It would accelerate at some constant rate other than 9.8 m/s2
C. Its acceleration would be a function of the initial velocity of block A.
D. It would not accelerate but rather move with constant velocity.

Rational is lengthy, prefer not to type all out. It consisted of the equations presented and convoluted messaging.

No other equations involved other than definition of T=mBg and substitution in F=ma-T

 

[411309]

Is it B because the vertical block experiences air resistance upon decline therefore having an acceleration of slightly less than 9.8? This gives the block on the table an acceleration of less then 9.8 as well.

 

[MedPR]

I mean, just looking at F=ma-T, where a=g, you can see that it is some value not equal to g..

 

[HollowSuperet36]

Yeah it is B. Common sense tells us this. (Though it's not due to air resistance)

But, my question was pertaining to the steps through which you derive the second equation from the first one.

MAg-mBa=mAa

a=mAg/(mB+mA)


Of the later, common sense left me in the dark.

 

[MedPR]

Yeah it is B. Common sense tells us this. (Though it's not due to air resistance)

But, my question was pertaining to the steps through which you derive the second equation from the first one.

MAg-mBa=mAa

a=mAg/(mB+mA)


Of the later, common sense left me in the dark.

Oh ok.

I'm going to rename the variables so it's easier to type.

Mass A = M
mass B = m

So you make your force diagram and see that Fx=T and Fy=Mg-T
You already know that T=ma

So you just add the equations and come out with:

Ma=Mg-ma
Ma+ma=Mg
a(M+m)=Mg
a=Mg/(M+m)

 

[HollowSuperet36]

Oh ok.

I'm going to rename the variables so it's easier to type.

Mass A = M
mass B = m

So you make your force diagram and see that Fx=T and Fy=Mg-T
You already know that T=ma

So you just add the equations and come out with:

Ma=Mg-ma
Ma+ma=Mg
a(M+m)=Mg
a=Mg/(M+m)

Thank you so much!

I feel dumb having not known that step was 'legal' yet Ive taken all the practice tests and not had to use it, and scored high.

One more tool for the box. I see the Kaplan topicals use this step alot in rational, didn't make sense to me. Though their math seems harder than any I've encountered on the AAMC exams or AAMC MCAT guide.

Thanks again MedPR!

 

[MedPR]

Thank you so much!

I feel dumb having not known that step was 'legal' yet Ive taken all the practice tests and not had to use it, and scored high.

One more tool for the box. I see the Kaplan topicals use this step alot in rational, didn't make sense to me. Though their math seems harder than any I've encountered on the AAMC exams or AAMC MCAT guide.

Thanks again MedPR!

Yea, a lot of the prep math is probably going to be harder than anything on the MCAT because they're trying to get you to understand the concept/logic so that you won't have to do the math on the MCAT. As we agreed earlier, you could intuitively tell that it would be some fraction of gravity, you just couldn't do the math. Doing the math might not be necessary on the real thing if you can identify the concept early on.

 

[EnginrTheFuture]

Just to add to this for people not interested in the math but still having trouble...

Using a limiting case would be very useful here. If you made the block sitting on the frictionless surface infinitely massive, the little block dangling isn't going to accelerate with a high magnitude because it's having to tug the entire universe with it. If you made it infinitely NOT massive, it's as if the dangling block is free falling and thus both will accelerate with the usual gravitational value.

 

[HollowSuperet36]

Very good analogy^^

Thanks