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A particle at the origin is experiencing three forces. The first one has a magnitude of 57 lb and is directed towards (2,-4,5), while the second one has a magnitude of 32 lb and is directed towards (-1,-2,-8). If the net force is 41 lb in the direction of (-5,-1,4), what is the x-component of the third force acting on the particle (in lb)?

Sagot :

onyebuchinnaji

The x-component of the third force acting on the particle is 41.2 lb.

The given parameters;

  • (2, -4, 5) = 57 lb
  • (-1, -2, -8) = 32 lb
  • (-5, -1, 4) = 41 lb

The vector components of the forces is calculated as;

[tex]F_1 = 57, (2,-4, 5) = 57\times \frac{2i -4j+ 5k}{\sqrt{2^2 + 4^2 + 5^2} } = 8.5(2i -4j+ 5k)\\\\F_1 = (17 i -34j + 42.5k)lb\\\\\\ F_2 = 32, (-1, -2, -8) = 32\times \frac{-i -2j-8k}{\sqrt{1^2 + 2^2 + 8^2} } = 8.3(-i-2j-8k)\\\\ F_2 = (-8.3i - 16.6j - 66.4k)lb\\\\\\ F_{net}= 41, (-5, -1, 4) = 41\times \frac{-5i -j+ 4k}{\sqrt{5^2 + 1^2 + 4^2} } = 6.5(-5i-j+ 4k)\\\\ F_{net} = (-32.5i -6.5j+ 24k)lb[/tex]

The x-component of the third force is calculated from the resultant of the forces;

[tex]F_1+F_2 + F_3 = F_{net}\\\\F_3 = F_{net} - (F_1 + F_2)\\\\F_3 = (-32.5i -6.5j + 24k) - (17i -34j + 42.5k) - (-8.3i -16.6j -66.4k)\\\\F_3 = (-32.5 -17 + 8.3)i +(-6.5 + 34+ 16.6)j + (24 -42.5+66.4)k\\\\F_3=(-41.2 i) +(44.1)j + (47.9)k\\\\F_3 = (-41.2i + 44.1 j + 47.9k)lb[/tex]

Thus, the x-component of the third force acting on the particle is 41.2 lb.

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