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For the left figure below, replace the distributed loads by an equivalent resultant force and a couple moment acting at point A. (See the right figure below.) Let a = 3.85m , w1 = 5.70kN/m , and w2 = 4.10kN/m.
Calculate the resultant force, FR, and the couple moment, MR,A. Don't forget to include the appropriate signs with your numerical answers.

Express your answers numerically in kilonewtons and kilonewton-meters to three significant figures separated by a comma.


Sagot :

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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