Question

A force of 5 newton and a force of 6 newton are inclined to each other by angle 60डिग्री than find the resultant of them?​

Answer 1

Given :-

A force F of 5N is inclined to another force F' of 6N at 60°. And now we are required to find out the resultant force of this.

As we know that,

R = √F² + (F')² + 2FF' cos60°

R = √ 5² + 6² + 2.5.6.1/2

R = √25 + 36 + 30

R = √91

R = 9.5 N

Some other informations regarding Vectors :-

1. Magnitude of resultant vector R = A - B is given by :-

R = | A - B | = √A² + B² + 2AB cos(180° - Ø)

or

R = √ A² + B² - 2AB cosØ

2. If two vectors have equal magnitude.

| A | = | B | = a, then Ø is the angle between them,

| A-B | = √a² + a² - 2a² cosØ

| A - B | = 2a sin(Ø/2)

In this also when Ø = 60°, then

= 2a sin(Ø/2) = a

3. If A + B = A - B, then

B = null vector.

4. The vector substraction doesn't follow commutative law i.e. A - B ≠ B - A .

5. The vector substraction doesn't follow associative law i.e.

A-(B-C) ≠ (A-B) - C

Answer 2

Answer:

$\bigstar{\bold{Resultant\:Force\:=\:9.54\:N}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Force A = 5 N
• Force B = 6N
• Angle (θ) = 60°

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Resultant between the vectors (R)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Resultant between two vectors is given by the formula

  $R\:=\:\sqrt{A^{2} +B^{2} +2ABcos\theta}$

→ Substituting the given datas, we get

  $R\:=\:\sqrt{5^{2}+6^{2} +2\times 5\times 6\times cos60 }$

  $R\:=\:\sqrt{25+36+60\times \frac{1}{2} }$

 $R\:=\:\sqrt{25+36+30}$

 $R\:=\:\sqrt{91}$

 $R\:=\:9.54$

 $\boxed{\bold{Resultant\:Force\:=\:9.54\:N}}$

$\Large{\underline{\underline{\bf{Solution:}}}}$

• Resultant is given by the formula
• $R\:=\:\sqrt{A^{2} +B^{2}+2ABcos\theta }$