Question

the maximum value of resultat vector is 5 N and the minimum value is 1 N find the resultant of the forces when they act at 60 °​

Answer 1

Answer:

$\large{\longrightarrow{\bf{\sqrt{19}\:N}}}$

Explanation:

•let the first force be F1

• second force be F2

Now,

as per the question;

→F1+F2= 5....(1)

→ F1-F2= 1...(2)

solving eqⁿ (1) and (2)

we get

→ F1= 3 and

→ F2= 2

Now, finding the resultant of forces

→√(F1²+F2²+2F1F2cos∅)

→√(9+4+2×6×cos60°)

→ √(9+4+2×6×1/2) ( cos 60°= 1/2)

→ √(13+6))

→ √19

→ √19 N

Answer 2

QuestioN :

The maximum value of resultant vector of force is 5 N and the minimum value is 1 N. Find the resultant of the forces when they act at 60 °.

GiveN :

• F₁ = 5 N
• F₂ = 1 N
• Angle between the forces (θ) = 60°

To FinD :

• Resultant force between them

SolutioN :

We are given,

⇒F₁ + F₂ = 5.....(1)

⇒F₁ - F₂ = 1.....(2)

Solving (1) and (2)

⇒2F₁ = 6

⇒F₁ = 3 N

Similarly,

⇒F₂ = 2N

Use Parallelogram law of Vector Addition :

$\boxed{\sf{|F_{net}| \: = \: \sqrt{F_1 \: + \: F_2 \: + \: 2 F_1 F_2 \cos \theta}}} \\ \\ \implies {\sf{|F_{net}| \: = \: \sqrt{3^2 \: + \: 2^2 \: + \: 2( 3 \: \times \: 2 \: \times \: \cos 60^{\circ})}}} \\ \\ \implies {\sf{F_{net} \: = \: \sqrt{9 \: + \: 4 \: + \: 12 \cos 60^{\circ}}}} \\ \\ \implies {\sf{F_{net} \: = \: \sqrt{13\: + \: 12 \cos 60^{\circ}}}} \\ \\ \implies {\sf{F_{net} \: = \: \sqrt{13 \: + \: 12 \: \times \: \dfrac{1}{2}}}} \\ \\ \implies {\sf{F_{net} \: = \: \sqrt{13 \: + \: 6}}} \\ \\ \implies {\sf{F_{net} \: = \: \sqrt{19}}} \\ \\ \underline {\sf{\therefore \: The \: resultant \: force \: is \: \sqrt{19} \: N}}$