You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Second Law of Motion

Two forces ...

Question

Two forces $F_{1} = 1 N$ and $F_{2} = 2 N$ act along the lines $x = 0$ and $y = 0$ , respectively. Then find the resultant of forces.

Open in App

Solution

$F_{1} = 1 N F_{2} = 2 N$
angle between both the forces is $90^{0}$ .
$R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 90^{0}}$
$R = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$
$\to F = 2^i +^j$
$t a n θ = \frac{1}{2} \Rightarrow θ = {tan}^{- 1} (\frac{1}{2})$ direction of force resultant.

Suggest Corrections

Similar questions

Q. Three forces

_{1}

_{2}

and

_{3}

act on a body.

_{1}

acts vertically downwards with magnitude

\sqrt{3} N

_{2}

acts horizontally leftwards with magnitude

1 N

and

_{3}

acts at an angle

θ

with horizontal as shown in the figure. Find the magnitude of

_{3}

and

θ

for the body to remain in equilibrium.

Q. Two forces

_{1} = (2^i + 2^j) N

and

_{2} = (3^j + 4^k) N

are acting on a particle. The component of force

_{1}

along force

_{2}

Q. Forces

X, Y,

and

Z

have magnitudes

10 N, 5 (\sqrt{3} - 1) N

and

5 (\sqrt{3} + 1) N

, respectively. The forces

Y

and

Z

act in the same direction as shown in the figure. The resultant of

X

and

Y

and the resultant of

X

and

Z

have the same magnitude. Find

θ

, the angle between

X

and

Y

Q. Two forces

F_{1}

and

F_{2}

of magnitude

5 N

each inclined to each other at

60^{o}

, act on a body. Find the resultant force acting on the body.

A particle is observed from two frames $S_{1} a n d S_{2}$ . The frame $S_{2}$ moves with respect to $S_{1}$ with an acceleration a. Let $F_{1} a n d F_{2}$ be the pseudo forces on the particle when seen from $S_{1} a n d S_{2}$ respectively. Which of the following are not possible?

Join BYJU'S Learning Program

Two forces F1=1N and F2=2N act along the lines x=0 and y=0, respectively. Then find the resultant of forces.

F1=1NF2=2Nangle between both the forces is 900.R=√F21+F22+2F1F2cos900R=√12+22=√5→F=2^i+^jtanθ=12⇒θ=tan−1(12) direction of force resultant.

Two forces $F_{1} = 1 N$ and $F_{2} = 2 N$ act along the lines $x = 0$ and $y = 0$ , respectively. Then find the resultant of forces.

$F_{1} = 1 N F_{2} = 2 N$
angle between both the forces is $90^{0}$ .
$R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 90^{0}}$
$R = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$
$\to F = 2^i +^j$
$t a n θ = \frac{1}{2} \Rightarrow θ = {tan}^{- 1} (\frac{1}{2})$ direction of force resultant.