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
The lateral displacement is I don’t know tbh I think 16.8
Answer:
0.54m
Explanation:
Step one:
given data
length of seesaw= 3m
mass of man m1= 85kg
weight = mg
W1= 85*10= 850N
mass of daughter m2= 35kg
W2= 35*10= 350N
distance from the center= (1.5-0.2)= 1.3m
Step two:
we know that the sum of clockwise moment equals the anticlockwise moment
let the distance the must sit to balance the system be x
taking moment about the center of the system
350*1.3=850*x
455=850x
divide both sides by 850
x=455/850
x=0.54
Hence the man must sit 0.54m from the right to balance the system
I don't know what's in your house! lol. try naming things like: ball, lamp, cup, bowl.
