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
Hello There
Answers: T<span>he elastic potential energy can be increased by: </span>
<span>1) Getting a spring with a higher spring constant</span>
<span>2) Increasing the length at which the spring is compressed.
Reasons: Getting a stronger spring makes it stronger which equals more energy. While increasing the compression on the spring, increases the stored energy which makes it more powerful when its released
I hope this helps
-Chris</span>
Answer:
2442.5 Nm
Explanation:
Tension, T = 8.57 x 10^2 N
length of rope, l = 8.17 m
y = 0.524 m
h = 2.99 m
According to diagram
Sin θ = (2.99 - 0.524) / 8.17
Sin θ = 0.3018
θ = 17.6°
So, torque about the base of the tree is
Torque = T x Cos θ x 2.99
Torque = 8.57 x 100 x Cos 17.6° x 2.99
Torque = 2442.5 Nm
thus, the torque is 2442.5 Nm.
