Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From he question we are told that
The first mass is 
The second mass is 
From the question we can see that at equilibrium the moment about the point where the string holding the bar (where 
are hanged ) is attached is zero
Therefore we can say that
Making x the subject of the formula
Looking at the diagram we can see that the tension T on the string holding the bar where 
are hanged is as a result of the masses (
)
Also at equilibrium the moment about the point where the string holding the bar (where (
) and 
are hanged ) is attached is zero
So basically
Making subject
that statement is true
a Third class lever applied when the effort place between the load and the fulcrum.
For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = 
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
