Question
Forces of magnitudes 
and 
act at a point parallel to the sides of an equilateral triangle taken in order. The resultant of these forces, is
- 3P
Hint:
In this question, we have given a equilateral triangle and the force of magnitude is parallel with the sides of triangle. Which is P- Q, P, P + Q. We have to find the resultant force. Find the component of force for each magnitude.
The correct answer is:
Here,
Þ 
= 
…..(i)
Þ 
= 
…..(ii)
Squaring and adding (i) and (ii), we get
.
Here we have to find the resultant force.
Now, we have 3 magnitudes of force,
a = P, b = P – Q, c = P + Q,
Now the component of force,
At A, θ= 60°
acos θ = P cos 60° =
asin θ = P sin 60° = 
At C, θ= 60°
c cos θ = P+Q cos 60° = 
c sin θ = P+Q sin 60° = 
No component for B because it is at x- axis.
So Resultant force is,
Fi = ( 
+ P – Q ) i^
Fi = ( - Q/2 + P – Q ) i^
Fi = ( 
) i^
And
Fj = (
) j^
Fj = ( 
) j^
Fj = ( 
) j^
Therefore,
|F| = 
|F| = 
|F| = 
|F| = 
|F| = 
Therefore, the resultant force is
The correct answer is .
Here we have to find the resultant force of the given magnitude. Firstly, find the component of the force which is Fsin θ and Fcos θ.
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