“The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acti

Question

“The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body?”

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Nem 4 years 2021-09-05T09:32:14+00:00 1 Answers 12 views 0

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  1. thuthao
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    2021-09-05T09:33:21+00:00
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    Answer:

    The answer is F= 2.881 pounds

    Explanation:

    We will start this question by representing the force vectors in Cartesian coordinate system:

    • Force 1 acting on the right F_{1} = (2, 0)
    • Force 2 acting upward F_{2} = (0,5)
    • Force 3 acting at 45° F_{3} = (3cos(45),3sin(45))

    Let the single force we need be represented by F=(x,y)

    Since, the body is in equilibrium the summation of all the forces acting on it should be zero:

    ∑F_{x} = 0 (Sum of all forces in x direction is equal to zero)

    x + 2 - 3cos(45) =0 (x component of force 3 is negative because it is acting  in opposite direction)

    x = 0.1213

    ∑F_{y}=0 (Sum of all forces in y direction is equal to zero)

    y+5-3sin(45)=0 (y component of force 3 is negative because it is acting  in opposite direction)

    y=-2.8786

    So,

    F=(0.1213,-2.8786)

    Hence, to get the scalar quantity of a single force:

    F=\sqrt[2]{(0.1213)^{2}+(-2.8786)^{2} }

    F= 2.881 pounds

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