let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
1) 12 cm
2) 3 N
Explanation:
1)
The relationship between force and elongation in a spring is given by Hooke's law:
where
F is the force applied
k is the spring constant
x is the elongation
For the spring in this problem, at the beginning we have:
So the spring constant is
Later, the force is tripled, so the new force is
Therefore, the new elongation is
2)
In this second problem, we know that the elongation of the spring now is
From part a), we know that the spring constant is
Therefore, we can use the following equation to find the force:
And substituting k and x, we find:
So, the force to produce an elongation of 6 cm must be 3 N.