Find the force F^{4} , so as to give the resultant of the force as shown in the figure given below.

Given : Forces and their resultant

To find : Force F^{4}

#### Solution

Assume that force F_{4} acts at an angle θ

Taking forces having direction towards right as positive and forces having direction upwards as positive.

**Resolving forces along X direction :**

`-F_1sin30 – F_2cos30 + F_3cos45 + F_4cosθ = Rcos50`

`-500sin30 – 300cos30 + 400cos45 + F_4cosθ = 800cos50`

F_{4}cosθ = 741.195 …………(1) **Resolving forces along Y direction : **

`-F_1cos30 + F_2sin30 + F_3sin45 + F_4sinθ = -Rsin50 ` `-500cos30 + 300sin30 + 400sin45 +F_4sin θ = -800sin50 ` F_{4}sin θ = -612.6656 ……….(2)

**Squaring and adding (1) and (2)**

`(F_4sinθ)^2 + (F_4cos θ)^2 = (-612.6656)^2 + (741.195)^2`

`F_4^2(sin^2 θ + cos^2 θ) = 924729.1173`

`F_4 = 961.6284 N`

Dividing (2) by (1)

`(F_4sinθ)/(F_4cosθ) = (−612.6656)/(741.195) `

tan θ = -0.8266

θ = 39.5769^{o} (in fourth quadrant)

F_{4} = 961.6284 N ( at an angle 39.5769^{o} in fourth quadrant )

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