When solving systems of equations for first-year physics, I am told that to be able to find the solution for a system of n unknowns (if one exists), I need at least n equations.In this system, I am looking for a. The known quantities are θ 0 and θ 1 . T 0 cos ⁡ θ 0 = m g T 0 sin ⁡ θ 0 = m v 2 ℓ sin ⁡ θ 0 T 1 cos ⁡ θ 1 − m g = m a T 1 sin ⁡ θ 1 = m v 2 ℓ sin ⁡ θ 1 In this system of equations, I had 5 unknown quantities: T 0 , T 1 , m, ℓ, a. But I was able to successfully solve for a with four equations. Where is the logical error in my understanding?

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When solving systems of equations for first-year physics, I am told that to be able to find the solution for a system of n unknowns (if one exists), I need at least n equations.In this system, I am looking for   a. The known quantities are       θ    0   and       θ    1  .      T    0    cos  ⁡      θ    0    =  m  g      T    0    sin  ⁡      θ    0    =            m              v              2                    ℓ      sin      ⁡              θ        0                  T    1    cos  ⁡      θ    1    −  m  g  =  m  a      T    1    sin  ⁡      θ    1    =            m              v              2                    ℓ      sin      ⁡              θ        1            In this system of equations, I had 5 unknown quantities:       T    0  ,       T    1  ,   m,   ℓ,   a. But I was able to successfully solve for   a with four equations. Where is the logical error in my understanding?

robegarj · Accepted Answer

You can solve for   ℓ from the start by plugging in the first equation to the second equation, so I don&#039;t think you made a mistake. Ie. from the first equation get an expression for m, plug into second equation, eliminate T0 and then just compute   ℓ.

Gaaljh · Accepted Answer

Divide all the equations by   m. You get a system of four equations in the four unknowns       T    0    /  m,       T    1    /  m,   ℓ and   a.It&#039;s been a while since I did any even semi-serious mechanics, so take the following with a grain of salt. The physical reason why the above works is that the system scales with the mass m. If you double the mass, you will simply double the forces (tensions?)       T    0   and       T    1  . Any other scaling factor works the same way. The object will have the same acceleration irrespective of its mass, when it is undergoing the same motion. Only the forces involved will scale accordingly.

When solving systems of equations for first-year physics, I am told that to be able to find the solu

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