Question

two forces 3n and 2n are at an angle theta such that the resultant is R the first force is now increase two 6n and the R become 2R the value of thetha is

Answer 1

Answer:

The value of $\theta=120^{\circ}$

Explanation:

Given that,

Angle= $\theta$

First force $F= 3 N$

Second force $F_{2}=2 N$

The resultant force is

$R= \sqrt{(F_{1})^2+(F_{2})^2+2F_{1}F_{2}cos\theta}$

$R= \sqrt{(3)^2+(2^2)+2\times3\times2\times cos\theta}$...(I)

Now, if the first force increases to 6 n and R become 2 R.

Then,

$2R= \sqrt{(6)^2+(2^2)+2\times6\times2\times cos\theta}$....(II)

Put the value of R in equation in equation (II)

$2(\sqrt{(13+12\times cos\theta})= \sqrt{(40+24\times cos\theta}$

On squaring both sides

$4(13+12\ cos\theta)=40+24\ cos\theta$

$cos\theta=\dfrac{-1}{2}$

$\theta=cos^{-1}\dfrac{-1}{2}$

Hence, The value of $\theta=120^{\circ}$  

Answer 2

Answer:

120 degree

Explanation:

The value of \theta=120^{\circ}θ=120∘

Explanation:

Given that,

Angle= \thetaθ

First force F= 3 NF=3N

Second force F_{2}=2 NF2=2N

The resultant force is

R= \sqrt{(F_{1})^2+(F_{2})^2+2F_{1}F_{2}cos\theta}R=(F1)2+(F2)2+2F1F2cosθ

R= \sqrt{(3)^2+(2^2)+2\times3\times2\times cos\theta}R=(3)2+(22)+2×3×2×cosθ ...(I)

Now, if the first force increases to 6 n and R become 2 R.

Then,

2R= \sqrt{(6)^2+(2^2)+2\times6\times2\times cos\theta}2R=(6)2+(22)+2×6×2×cosθ ....(II)

Put the value of R in equation in equation (II)

2(\sqrt{(13+12\times cos\theta})= \sqrt{(40+24\times cos\theta}2((13+12×cosθ)=(40+24×cosθ

On squaring both sides

4(13+12\ cos\theta)=40+24\ cos\theta4(13+12 cosθ)=40+24 cosθ

cos\theta=\dfrac{-1}{2}cosθ=2−1

\theta=cos^{-1}\dfrac{-1}{2}θ=cos−12−1

Hence, The value of \theta=120^{\circ}θ=120∘