- #1

vysero

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Block S (the sliding block) with mass M = 3.3 kg, is free to move along a horizontal friction-less surface and connected, by a cord that wraps over a friction-less pulley, to a second block H (the hanging block), with mass m = 2.1 kg. The cord and pulley have negligible masses. The hanging block H falls as the sliding block S accelerates to the right. Find A) the acceleration of block S, B) the acceleration of block H, and C) the tension in the cord.

Okay so T = Ma, got it. I will call it equation #1.

Now T-mg = -ma, got it as well and I will call it equation #2.

My book goes on: "Now note that Eqs. #1 and #2 are simultaneous equations with the same two unknowns, T and a. Subtracting these equations eliminates T. Then solving for a yields:

a = (m/M+m)(g)

Substituting this result into Eq. #1 yields:

T=((Mm)/(M+m))(g)

Putting in the numbers gives, for these two quantities: a = 9.8m/s^2 and T = 13 N."

Here is is my questions, why do I subtract the two equations? My book says nothing about simultaneous equations and therefore gives no explanations for why I should subtract them when I find them. My second question is could someone explain how exactly they are substituting the equations which has been solved for a into equation #1 and why I would want to do that anyway? Also this I am assuming the acceleration of 9.8m/s^2 is for both block S and black H right? Which makes sense because the table is friction-less and so is the pulley.

Its kind of ironic my book wastes a full page explaining how to set up the stupid diagrams and then when it comes to making the actual calculations there's no explanation what so ever. This makes me think that there is some simple logic behind it that I have yet to grasp.