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Question

Three forces F1, F2 and F3 act on a body. F1 acts vertically downwards with magnitude 3 N , F2 acts horizontally leftwards with magnitude 1 N and F3 acts at an angle θ with horizontal as shown in the figure. Find the magnitude of F3 and θ for the body to remain in equilibrium.


A
2 N, 30
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B
2 N, 60
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C
1 N, 30
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D
1 N, 60
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Solution

The correct option is B 2 N, 60
Given, F1=3 N and F2=1 N
Let the magnitude of F3 be p N.
Let us resolve this in horizontal and vertical direction.
We can write after the resolution
F3=(pcosθ)+(psinθ)
Now the net force along x must be 0.
pcosθ1=0 ----------(1)
Similarly, the net force along y must be 0
psinθ3=0 ----------(2)

From eq (1) and (2) we get,
tanθ=3 θ=60
Putting this value in equation (1) we get
p=2 N.

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