So the simplified formula for force is F=MA, or force equals mass times acceleration.
All you have to do is do some simple algebra to figure this out.
I assure you it isn't that hard.
If both boxes are observed with the same acceleration then for the purposes of this problem they are the same.
Switching the variables around for the force formula we get F/M=MA/M, which is F/M=A.
If the first box has 3 times the force applied to it,
then we can distinguish between these by denoting them 3F and F.
We also need to note the masses as they are different, M1 and M2.
If both of the box's accelerations are the same we can set 3(F)/(M1)=A=(F)/(M2).
Multiply the masses across the equation, 3(F)(M2)=(F)(M1).
Finally divide by force to get 3(M2)=(M1).
The ratio the problem was talking about isn't any named convention, it's just the ratio of the box's masses.
Which by solving the problem, we can see is 1:3.
I made the mistake of making the problem seem bigger than it actually is.
Really you can imagine it being a lot harder to push a heavy table than a lighter one.
It just so happens though that a table 3 times as heavy takes 3 times the force to push.
That's all this problem is stating.
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