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Question: Three forces \({F_1}\), \({F_2}\) and \({F_3}\) together keep a body is equilibrium if \({F_1} = 3N\...

Three forces F1{F_1}, F2{F_2} and F3{F_3} together keep a body is equilibrium if F1=3N{F_1} = 3N along the positive X-axis F2=4N{F_2} = 4N along the positive Y-axis, then the third force F3{F_3} is:
A) 5N5N making an angle θ=tan⁡−1(34)\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)with negative Y-axis.
B) 5N5N making an angle θ=tan⁡−1(43)\theta = {\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right)with negative Y-axis.
C) 7N7N making an angle θ=tan⁡−1(34)\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)with negative Y-axis.
D) 7N7N making an angle θ=tan⁡−1(43)\theta = {\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right)with negative Y-axis.

Explanation

Solution

hint: We can solve this question by using simple vector laws of addition if a body is in equilibrium its means that the net force on that body must be zero.

Complete step by step solution:
In question it is given force F1=3N{F_1} = 3N in positive x-axis direction we can write this force in vector form
F⃗1=3i^{\vec F_1} = 3\hat i
And in question F2{F_2} is given 4N4N in the direction of positive y-axis in vector form it can written as
F⃗2=4j^{\vec F_2} = 4\hat j
We want to find F⃗3{\vec F_3}
If the body is in equilibrium means the net force on the body must be zero
Or can say that the vector sum of all three forces is zero net force F⃗net=0i^+0j^+0k^{\vec F_{net}} = 0\hat i + 0\hat j + 0\hat k
⇒F⃗1+F⃗2+F⃗3=F⃗net\Rightarrow {\vec F_1} + {\vec F_2} + {\vec F_3} = {\vec F_{net}}
Put values of forces
⇒3i^+4j^+F⃗3=0i^+0j^+0k^\Rightarrow 3\hat i + 4\hat j + {\vec F_3} = 0\hat i + 0\hat j + 0\hat k
Solving this
⇒F⃗3=−3i^−4j^\Rightarrow {\vec F_3} = - 3\hat i - 4\hat j .......... (1)
Equation (1) gives the vector form of F3{F_3}
Magnitude of any force given as ∣F⃗∣=(Fx)2+(Fy)2+(Fz)2\left| {\vec F} \right| = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2} + {{\left( {{F_z}} \right)}^2}}
Magnitude of F3{F_3} can be calculate by equation (1)
⇒∣F⃗3∣=(−3)2+(−4)2\Rightarrow \left| {{{\vec F}_3}} \right| = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 4} \right)}^2}}
⇒F3=9+16\Rightarrow {F_3} = \sqrt {9 + 16}
⇒F3=25\Rightarrow {F_3} = \sqrt {25}
So F3{F_3}
F3=5N{F_3} = 5N
To find the direction we draw F⃗3=−3i^−4j^{\vec F_3} = - 3\hat i - 4\hat j at origin as given

Let assume the force F3{F_3} make angle θ\theta with negative y-axis then
⇒tan⁡θ=PB\Rightarrow \tan \theta = \dfrac{P}{B}
From diagram it is clear that in triangle OBC perpendicular is 33 and base is 44
⇒tan⁡θ=(34)\Rightarrow \tan \theta = \left( {\dfrac{3}{4}} \right)
So the value of θ\theta
⇒θ=tan⁡−1(34)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)
Hence the force F3{F_3} is 7N7N at angle θ=tan⁡−1(34)\theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) with the negative y-axis.

Hence option (A) is correct.

Note: We can solve this question by another method first we calculate vector sum of F1{F_1} and F2{F_2} as given below
⇒F⃗12=3i^+4j^\Rightarrow {\vec F_{12}} = 3\hat i + 4\hat j
We know body is in equilibrium so then net force is zero means the third force must be equal and opposite to the F⃗12{\vec F_{12}} so that this cancel the resultant of F1{F_1} and F2{F_2} means
⇒F⃗3=−(3i^+4j^)\Rightarrow {\vec F_3} = - \left( {3\hat i + 4\hat j} \right)
By this we can calculate magnitude and direction as calculated in the above solution.