Question

Two equal forces are acting at a point with an
angle of 60° between them. If the resultant
force is equal to 40
$\sqrt{3}$
N, the magnitude of
each force is
[Kerala PMT 2007)
(A) 40 N
(B) 20 N
(C) 80 N
(D) 30 N​

Answer 1

Answer:-

Given:

Two equal forces are acting at a point with an angle of 60°.

Resultant force (R) = 40√3 N

Let the magnitude of first force = magnitude of second force = x.

We know that,

R = √(A² + B² + 2AB * Cos ∅)

Where,

• A is the magnitude of 1st force
• B is the magnitude of 2nd force
• ∅ is the angle between them.

Hence,

40√3 = √(x² + x² + 2*x*x * Cos 60°)

• Cos 60° = 1/2

→ √3 * 40 = √(2x² + 2x² * 1/2)

→ √3 * 40 = √(2x² + x²)

→ √3 * 40 = √(3x²)

→ √3 * 40 = √3 * (x)

→ (√3 * 40 )/√3 = x

→ 40 N = x

Therefore, the magnitude of each force will be 40 N (Option - A).

Additional Information:-

Vectors:

• The physical quantities which are expressed in both magnitude as well as direction are called Vectors.

• Vectors are generally written with a symbol and arrow head on it.

• For example, Velocity is a vector quantity. It is denoted by V ^ (→).

• They cannot be added , subtracted and multiplied by ordinary laws of algebra.

• There are different types of vectors based their magnitude and direction.

Answer 2

$\huge\bold{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• Two equal forces are acting at a point with an angle of 60° between them .

• The resultant force is $\underline\bold{40\sqrt{3}\:N}$

TO FIND :-

• The magnitude of each force .

CALCULATION :-

$<font color=blue>$Let,

✍️ The force acting on a point be “F” .

✍️ And the another force acting that point be “F” .

$\rm[\because\:{\purple{A.T.Q\:two\:forces\:are\:equal\:.]}}$

• $\rm\red{F_1\:=\:F}$

• $\rm\red{F_2\:=\:F}$ $<font color=baby>$

✍️ The resultant force = 40√3 .$<font color=green>$

$\pink\bigstar\:\rm{\purple{\boxed{\red{\implies\:Resultant\:Force\:=\:\sqrt{F_1^2\:+\:F_2^2\:+\:2\:.\:F_1\:.\:F_2\:.\:\cos{\theta}\:}}}}}$

• θ = 60° .

$\rm{\implies\:F_{resultant}\:=\:\sqrt{F^2\:+\:F^2\:+\:2.F.F\:\cos{60^{\degree}}\:}}$

$\rm{\implies\:F_{resultant}\:=\:\sqrt{2F^2\:+\:2F^2\:\times{\dfrac{1}{2}}\:}}$

$\rm{\implies\:F_{resultant}\:=\:\sqrt{2F^2\:+\:F^2\:}}$

$\rm{\implies\:F_{resultant}\:=\:\sqrt{3F^2}\:}$

$\rm{\implies\:F_{resultant}\:=\:\sqrt{3}\:F\:}$

$\rm{\implies\:40\sqrt{3}\:=\:\sqrt{3}\:F\:}$

$\rm{\red{\implies\:F\:=\:40\:N\:}}$----(A)

$\pink\therefore$ $<font color=purple>$ The resultant force is equal to “40 N” .