Question

The greatest and the least resultant of two forces acting at a point are 25N and
5N, respectively. If each force is increased by 5N, find the resultant of two
new forces acting at right angles to each-other.

Answer 1

GiveN :

• Greatest Force $\sf{(F_1)}$ = 25 N
• Smallest Force $\sf{(F_2)}$ = 5 N

To FinD :

• Resultant of the forces of they are at right angles to each other.

Formula Used :

Here, we will use Parallelogram law or Triangle law of vector Addition :

$\bigstar \: \boxed{\sf{|F_{net}| \: = \: \sqrt{F_1^2 \: + \: F_2 ^2 \: + \: F_1 F_2 \cos \phi}}}$

Where,

• $\sf{F_1}$ and $\sf{F_2}$ are the forces.
• $\sf{F_{net}}$ Is Resultant force.
• $\phi$ is angle between the Forces.

Solution :

Greatest Force : F1 + F2 = 25....(1)

Smallest Force : F1 - F2 = 5.....(2)

Solving, (1) and (2)

⇒F1 = 15 N

⇒F2 = 10 N

A.T.Q,

Magnitude of forces are increased by 5 N. So, new forces will be :

• $\sf{F_1}$ = 20 N
• $\sf{F_2}$ = 15 N

and they are perpendicular to each other, means angle between them is 90°. So,

• $\phi$ = 90°

Use Parallelogram law of vector addition :

$\implies \sf{F_{net} \: = \: \sqrt{F_1^2 \: + \: F_2 ^2 \: + \: F_1 F_2 \cos \phi}} \\ \\ \\ \implies \sf{F_{net} \: = \: \sqrt{20^2 \: + \: 15^2 \: + \: 20 \: \times \: 15 \: \times \: \cos 90^{\circ}}} \\ \\ \\ \implies \sf{F_{net} \: = \: \sqrt{400 \: + \: 225 \: + \: 300 \cos 90^{\circ}}} \\ \\ \\ \implies \sf{F_{net} \: = \: \sqrt{625 \: + \: 300 \: \times \:0 }} \\ \\ \\ \implies \sf{F_{net} \: = \: \sqrt{625 \: + \: 0}} \\ \\ \\ \implies \sf{F_{net} \: = \: \sqrt{625}} \\ \\ \\ \implies \sf{F_{net} \: = \: 25}$

∴ Resultant of the forces is 25N. (approx.)