FORCE AND ACCELERATION: 3-4-5 Right-Triangle Mathematics

Q: Two men attempt to pull a box in the diagram below:

What is the resultant force in Newtons ( N ) exerted on the box?

A: This is a classic question involving a 3-4-5 right triangle. Trigonometry and the Pythagorean Theorem enable us to solve the problem using the numbers 3, 4, and 5. We begin by re-orienting the force vectors into the opposite and adjacent sides of a right triangle:

? N = √( 4² N x 3² N ) x 100 N

? N = √( 25 N ) x 100 N

? N = 5 N x 100

? N = 500 N

Thus, the resultant force of F_t = F₁ + F₂ will always be some multiple of five when 3-4-5 right triangles are involved. We use the ( tan θ ) function to determine the angle ( F_t ) makes with the west-to-east axis:

tan θ = opp/adj

tan θ = ( 300 N/400 N ) = ( ¾ )

θ = tan^-1 ( ¾ )

θ = 36.9⁰ north of east

Check:

( 500 N )( sin 36.9⁰ ) = 300 N

( 500 N )( cos 36.9⁰ ) = 400 N