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substitute an equivalent resultant force and a couple of moment acting at point A. Allow w1 to be 5.70 kN/m, w2 to be 4.10 kN/m, and a to be 3.85 m. the pair moment is 77.08 kN m, and the resulting force is FR=4.8125 kN.
The force delivered by the distributed loads is given as,
F₁=[tex]\frac{1}{2}[/tex] w₁a
=[tex]\frac{1}{2}[/tex](5.70kN/m)(3.85m)
=10.9725 kN
F₂=w₂a
=(4.10 kN/m)(3.85m)
=15.785 kN
The resultant force is given as,
[tex]F_{R}[/tex]= -F₁+F₂
= -10.9725 kN+15.785 kN
=4.8125kN
The resultant moment exerted on the beam is calculated as,
M=-F₁(a/3)+F₂(a+(a/2))
M=-(10.9725 kN)(3.85m/3)+15.785 kN(3.85+(3.85/2))
= -14.081 kN m+91.158 kN m
=77.08 kN m
In mechanics, a force is any action that seeks to preserve, modify, or deform a body's motion. Isaac Newton's three laws of motion, which are outlined in his Principia Mathematica, are frequently used to illustrate the concept of force (1687). Newton's first law states that a body at rest or traveling at a constant speed in a straight line will stay in that state unless a force is applied to it. According to the second law, when an external force acts on a body, the body accelerates (changes velocity) in the force's direction.
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