Resultant needed to neutralize two equal perpendicular forces In engineering mechanics, two concurrent forces each of magnitude T act at right angles to one another. If a single third force is applied along the internal bisector of the right angle, in the opposite direction to the resultant of the two forces, what magnitude must that third force have to exactly neutralize the effect?

Introduction / Context:
This problem checks the composition of two perpendicular concurrent forces and the idea of replacing them by a single equal-and-opposite balancing force applied along the angle bisector. It is a standard vector addition question in statics/dynamics.

Given Data / Assumptions:

• Two forces of equal magnitude T act at right angles and are concurrent.
• The balancing (neutralizing) force acts along the internal bisector of the angle formed by the two forces, opposite to their resultant.
• Planar vector addition applies; the body is treated as a particle (no couples).

Concept / Approach:

The vector resultant R of two perpendicular forces F1 and F2 of equal magnitude T is obtained from the Pythagorean relation. The required single balancing force must be equal in magnitude to R and opposite in direction, acting along the same line (the bisector for equal perpendicular forces).

Step-by-Step Solution:

Verification / Alternative check:

A head-to-tail triangle of forces with equal perpendicular sides length T forms an isosceles right triangle, whose hypotenuse is T√2, confirming the result geometrically.

Why Other Options Are Wrong:

2T and 3T overestimate the required balancing magnitude; T/2 underestimates it; 'None of these' is incorrect because a precise value exists.

Common Pitfalls:

Forgetting the right-angle relation and adding magnitudes arithmetically (T + T) instead of using the Pythagorean theorem; not aligning the balancing force along the bisector opposite the resultant.

Final Answer:

T√2