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Answer:
Explanation:
Since the surface is frictionless therefore there will be no friction force on block but there will be weight of block which we can divide in to two components i.e. mgcosθ &mgsinθ which is perpendicular and parallel to the surface respectively.
In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.
Therefore Ramp will apply a Force of mgcosθ on block where m is the mass of block.
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
