Question

two 75.0-N forces act on an object at right angles to each other. what is the magnitude of the resultant of these two forces? what is the magnitude of their equilibrant

Answer 1

The magnitude of the resultant force when two 75.0 N forces act at right angles is 106.1 N. The equilibrant force, which is the force that balances the two original forces, also has a magnitude of 106.1 N.

Step-by-step explanation:

When two forces of 75.0 N act on an object at right angles to each other, we can find the magnitude of the resultant force by using the Pythagorean theorem. To do this, we calculate the square root of the sum of the squares of the individual forces:

Resultant force (R) = √(F1² + F2²)

Here, F1 and F2 are the two perpendicular forces, both with a magnitude of 75.0 N:

R = √(75.0² + 75.0²) = √(5625 + 5625) = √11250

R = 106.1 N

The equilibrant force is the force that would balance these two forces, essentially having the same magnitude as the resultant but acting in the opposite direction. So, the magnitude of the equilibrant is also 106.1 N.

Answer 2

Sine they are at right angles to each other, therefore “theta”= 90°. Using our cosine rule to look for our resultant force.

Step-by-step explanation:

A^2=B^2+C^2 - 2BC COSthetha

A=resultant force; B=3N; C=4N; COS90=0.

A^2=3^2+4^2 - 2×3×4×0

A^2=9+16–0

A^2=25

A=√25

A=5N.