I can’t solve for this variable?

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.


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!




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


2 Answers


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.