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Resultant Force Forces with magnitudes of
125 newtons and 300 newtons act on a hook (see
figure). The angle between the two forces is 45°. Find
the direction and magnitude of the resultant of these
forces. (Hint: Write the vector representing each force
in component form, then add the vectors.)


Sagot :

The resultant of the two forces is 398.32 N and the direction is 12.8⁰.

The given parameters:

  • first force, F₁ = 125 N at 45⁰
  • second force, F₂ = 300 N at 0⁰

The vertical component of the forces is calculated as;

[tex]F_y = 125 \times sin(45) \ + \ 300 \times sin(0)\\\\F_y = 88.388 \ N[/tex]

The horizontal component of the forces is calculated as follows;

[tex]F_x = 125 \times cos(45) \ + \ 300 \times cos(0)\\\\F_x = 388.388 \ N[/tex]

The resultant of the two forces is calculated as follows;

[tex]F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{88.388^2 + 388.388^2} \\\\F = 398.32 \ N[/tex]

The direction of the forces is calculated as follows;

[tex]\theta = tan^{-1} (\frac{F_y}{F_x} )\\\\\theta = tan^{-1}(\frac{88.388}{388.388} )\\\\\theta = 12.8\ ^0[/tex]

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