Question

Three forces of magnitude 40kn, 15kn and 20kn are acting at a point 'o' as shown in the figure the angles made by 40kn 15kn and 20kn forces with x-axis are 60degrees 120deg and 240deg respectively determine the magnitude and direction of the resultant force

Answer 1

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Answer 2

Answer: The magnitude of the resultant force will be 30.413 kN and its direction will be 92.16° from the direction of the positive x-axis.

Explanation:

Step 1: Prerequisite knowledge

Formula for resultant of forces= $\sqrt{(A^{2}+B^{2}+2(A)(B)cosa }$ where, a is the angle between two forces.

Formula for resultant angle = $\frac{Asina}{A+Bcosa}$

Step 2: Resultant of 40 kN and 20 kN

From the given diagram, we can conclude that,

The forces of 40 kN and 20 kN are acting in opposite direction so, without any calculations we can say the resultant of both will be 20 kN and will act in the direction of force of 40 kN.

Step 3: Calculating resultant of remaining two forces

Now, we have 2 forces one of 20 kN acting at 60° from positive x-axis and 15 kN acting at 120° from positive x-axis.

R=$\sqrt{(20^{2}+15^{2}+2(20)(15)cos60 }$

 =$\sqrt{400+225+300}$

 =$\sqrt{925}$

 =$30.41 kN$

Step 4: Calculating direction of resultant

Resultant of 20 kN and 15 kN can be calculated as:

tanФ=  $\frac{20sin60}{20+15cos60}$

       = $\frac{20\sqrt{3} }{55}$

       = $\frac{34.64}{55}$

      = $0.629$

Ф = 32.16°

From positive x- axis the angle would be 60° +Ф= 60° +32.16° = 92.16°