Question

10. The resultant of two forces has magnitude 20 N. One of the forces is of magnitude 20√3 N and makes an angleof 30° with the resultant. Then what is the magnitude ofthe other force?
need steps...​

Answer 1

Answer:-

Given:

Magnitude of resultant of two forces (R) = 20 N

Magnitude of first force (A) = 20√3

Angle between A & R = 30°

Let the magnitude of other force be B.

We know that,

R = √(A² + B² + 2ABcos ∅)

According to the question,

→ B = √(A² + R² + 2ARcos 30°)

• cos 30° = √3/2

→ B = √(20√3)² + (20)² + 2 * 20√3 * 20 * √3/2

→ B = √(1200 + 400 + 1200)

→ B = √2800

→ B = √7 * √400

→ B = 20√7 N

Therefore, the magnitude of other force will be 20√7 N.

Answer 2

Given :-

• The resultant of two forces has magnitude (R) = 20N.
• One of the force is of magnitude (A) = 20√3N.
• Angle between R & A = 30°

To Find :-

• The magnitude of the other force (B) = ?

Solution :-

As we know that,

${\red {\boxed{ \tt{Resulting \: Force (R) = \sqrt{ (A² + B² + 2ABCos\theta)}}}}}$

According to Question :

$\tt{B \:= \:√(A^{2}+ R^{2}+ 2ARCos30°)}$

[ Putting values ]

$\tt{B \:= \:√[(20√3)^{2}+ (20)^{2} + 2 × 20√3 × 20 × √3/2)]}$

$\tt{B \:=\: √(1200 + 400 + 1200)}$

$\tt{B = √2800}$

$\tt\blue{B = 20√7 N}$

Hence,

• The magnitude of the other force (B) is 20√7 N.