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
False’ because it is a force that makes a body follow a curved path
Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
Answer:
Magnitude of Vector = 79.3
Explanation:
When a vector is resolved into its rectangular components, it forms two vector components. These components are named as x-component and y-component, they are calculated by the following formulae:
x-component of vector = (Magnitude of Vector)(Cos θ)
y-component of vector = (Magnitude of Vector)(Sin θ)
where,
θ = angle of the vector with x-axis = 27°
Therefore, using the values in the equation of y-component, we get:
36 = (Magnitude of Vector)(Sin 27°)
Magnitude of Vector = 36/Sin 27°
<u>Magnitude of Vector = 79.3</u>