Solve Force F Applied to Objects with Masses m₁ and m₂

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To find the acceleration of two merged objects under the same force F, first determine their individual masses using the equations F = m₁ × 2 and F = m₂ × 3, leading to m₁ = F / 2 and m₂ = F / 3. When combined, the total mass is m₁ + m₂, allowing the equation F = (m₁ + m₂) × a to be used. By substituting the values of m₁ and m₂, the equation simplifies to F = (F/2 + F/3) × a. Solving this yields the acceleration a, confirming the calculations and clarifying any misunderstandings.
Ockonal
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Hello, I have a little question:
Power F applies for object with mass m₁ and it gaves acceleration 2 m/s²;
The same power F applies for another object with mass m₂, a = 3 m/s².
Find acceleration of both object (merged) with the same force F.
I wrote this:
F = m×a;

=>

F = m₁ × 2;
F = m₂ × 3;
F = (m₁ + m₂) × a;

a - ?
What's right way to solve that?
 
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Hello Ockonal! :smile:

First find m1 and m2 :wink:
 
Yeah, I already did this:
F = m₁ × 2;
F = m₂ × 3;
F = (m₁ + m₂) × a;
=>
m₁ = F / 2;
m₂ = F / 3;

F ≠ 0
F = (F/2 + F/3) * a : F
0 = (1/2 + 1/3) * a;

By logical I thing there should be:
a = (a₁ + a₂) / a₁×a₂
But how to get it from last my calculations?
 
Yeah, I already did this:
F = m₁ × 2;
F = m₂ × 3;
F = (m₁ + m₂) × a;
=>
m₁ = F / 2;
m₂ = F / 3;

F ≠ 0
F = (F/2 + F/3) * a : F
0 = (1/2 + 1/3) * a;

By logical I thing there should be:
a = (a₁ + a₂) / a₁×a₂
But how to get it from last my calculations?
 
F = (F/2 + F/3) * a : F
0 = (1/2 + 1/3) * a;
You divide both sides by F, so you get F/F = 1 on the left side.

1 = (1/2 + 1/3) * a.

And that you should be able to solve.
 
Oh, thanks :) Now I understand. Silly mistake.
 
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