Calculate the angle between a 2N force and a 3N force so that their resultant is 4N.

Open in App

Solution

Let the two vectors be A and B, and their resultant be R.

Let the angle between A and B be $θ$

$A = 2 N$

$B = 3 N$

$R = 4 N$

Now $R^{2} = A^{2} + B^{2} + 2 A B cos θ$

So, $4^{2} = 2^{2} + 3^{2} + 2 \times 3 \times 4 c o s θ$

So, $16 = 4 + 9 + 24 c o s θ$

So, $16 = 13 + 24 c o s θ$

So, $3 = 24 c o s θ$

So, $\frac{3}{24} = c o s θ$

So, $c o s θ = 1 / 8$

So, $θ = c o s^{-} 1 (1 / 8)$

Suggest Corrections

Similar questions

Q. Two forces of 3N and 4N are acting on a particle. The forces are acting at an angle of

30^{\circ}

between them. The magnitude of the resultant force will be

Calculate the angle between two vectors 2F and $\sqrt{2} F$ so that the resultant force is $F \sqrt{10} .$

Q. With two forces acting at a point, the maximum effect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are

Two forces 3N and 2 N are at an angle q such that the resultant is R. The first force is now increased to 6N and the resultant become 2R. The value of q is

Q. Two forces

3 N

and

2 N

at an angle

θ

such that resultant

R

. The first force is now increased to

6 N

and the resultant become

2 R

. The value of

θ

Join BYJU'S Learning Program

Calculate the angle between a 2N force and a 3N force so that their resultant is 4N.

Let the two vectors be A and B, and their resultant be R. Let the angle between A and B be θ A=2N B=3N R=4N Now R2=A2+B2+2ABcosθ So,42=22+32+2×3×4cosθ So, 16=4+9+24cosθ So, 16=13+24cosθ So, 3=24cosθ So, 324=cosθ So, cosθ=1/8 So, θ=cos−1(1/8)