I have this physics problem and I need to solve for α\alpha, but I can’t seem to figure out how. I have been trying for a while and I keep running into dead ends. This is how far I get before I start having problems.

R2(m2g−m2R2α)−R1(m1R1α+m1g)=α(M1R212+M2R222)R_2(m_2g-m_2R_2\alpha)-R_1(m_1R_1\alpha+m_1g)=\alpha\left(\frac{M_1R_1^2}{2}+\frac{M_2R_2^2}{2}\right)

After this point, I usually try to divide by what’s being multiplied against α\alpha on the right side, but then I can’t figure how to isolate α\alpha on the left side! Any help will be very appreciated!

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1

Put all terms containibng α\alpha in the lhs and the remaining in the rhs. You could probably simplify notations to make life easier.

– Claude Leibovici

2 days ago

In this document, try looking up the term “distributive property” edu.gov.on.ca/eng/curriculum/elementary/math18curr.pdf

– John Joy

yesterday

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2 Answers

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Did you multiply out the left hand side of the equation, collect terms involving خ± and then move that term over to the right hand side of the equation and simplify?

I see some 3/2’s in your future.

Following what was mentioned in the comments, let’s first multiply everything out.

R2m2g−m2R22α−m1R21α+R1m1g=αM1R212+αM2R222.R_2m_2g-m_2R^2_2\alpha-m_1R^2_1\alpha+R_1m_1g= \frac{\alpha M_1R_1^2}{2}+\frac{\alpha M_2R_2^2}{2}.

Next, bring everything with an α\alpha in it to one side. Let’s go with the right side.

R2m2g+R1m1g=αM1R212+αM2R222+m2R22α+m1R21α.R_2m_2g+R_1m_1g= \frac{\alpha M_1R_1^2}{2}+\frac{\alpha M_2R_2^2}{2} +m_2R^2_2\alpha+m_1R^2_1\alpha.

Now, factor out an α\alpha from the right side.

R2m2g+R1m1g=α(M1R212+M2R222+m2R22+m1R21).R_2m_2g+R_1m_1g= \alpha\left(\frac{ M_1R_1^2}{2}+\frac{ M_2R_2^2}{2} +m_2R^2_2+m_1R^2_1\right).

Finally, divide both sides by the parenthesis to find α\alpha.