Question

Two forces, each equal to P. act at right angles. Their
effect may be neutralised by a third force acting along their
bisector in the opposite direction with a magnitude of​

Answer 1

Answer:

√2P

Explanation:

Two equal forces each equal P are acting at right angle. Their components along bisector will be- Pcos45, Pcos45.

then,

R (bisector)= Pcos45 + Pcos45

= P (1/√2) + P (1/√2)

= P (1/√2 + 1/√2)

= P (2/√2)

= P*√2

= √2P

A/Q,

Effective force or third force (f3) is along the bisector

but in opposite direction, So magnitude of required third force (f3) will be same as value of R (bisector) ..

thus,

effective force(third force) = R (bisector)

required third force, F3 = √2P

.....

Answer 2

Answer:

The correct answer to this question is $\sqrt{2}$P.

Explanation:

Given that there are two equal forces acting at an angle of P. They will consist of Pcos45 and Pcos45 along the bisector.

Then, R (bisector) = Pcos45 + Pcos45

= P (1/$\sqrt{2}$) + P (1/$\sqrt{2}$)

= P (1/$\sqrt{2}$ + 1/$\sqrt{2}$)

= P (2/$\sqrt{2}$)

= P×$\sqrt{2}$

= $\sqrt{2}$P

The bisector is where the effective force, or third force(f3), is located. But in the opposite direction, therefore the third force's (f3) magnitude will be equal to R's value (bisector).

So, effective force(third force) = R(bisector)

And, the required third force will be, f3 = $\sqrt{2}$P

A particle's effective force is calculated by multiplying its mass by its acceleration.

Thus, it will be demonstrated that the system of internal forces and the system of external forces operating on a system of particles are equivalent.