A 15.5 kg block is pulled by two forces. The first is 11.8 N at a 53.7° angle and the second is 22.9 N at a -15.8° angle. What is the magnitude of the acceleration?

Answer 1

I'm going to name F₁ = 11.8 N , F₂ = 22.9 N , θ₁ = 53.7° , θ₂ = -15.8°, m =15.5 kg

This problem truly only relies on two points. Firstly is the equation F = ma, or more appropriately, the sum of the forces ∑F = ma. Next is the magnitude equation which is Pythagorean's Theorem a² + b² = c², or more appropriately, a² = a_x² + a_y² where a_x is the horizontal acceleration and a_y is the vertical acceleration. Thus the forces F₁ and F₂ will need to be broken into components to find ∑F_x and ∑F_y :

F₁cos(θ₁) + F₂cos(θ₂) = 11.8 N * cos(53.7°) + 22.9 N * cos(-15.8°) = 29.021 N = ∑F_x

F₁sin(θ₁) + F₂sin(θ₂) = 11.8 N * sin(53.7°) + 22.9 N * sin(-15.8°) = 3.275 N = ∑F_y

Now we can find a_x and a_y with ∑F = ma by dividing the mass (m) :

a_x = ∑F_x / m = 29.021 N / 15.5 kg = 1.872 m/s²

a_y = ∑F_y / m = 3.275 N / 15.5 kg = 0.211 m/s²

And finally a² = a_x² + a_y² → a = SQRT(a_x² + a_y²) = SQRT( (1.872 m/s²)² + (0.211 m/s²)² ) = 1.884 m/s²

Alternatively, the same magnitude concept can be used to instead find a single consolidated force (F_R) then using that F_R = ma which saves one division-by-m step as shown :

F_R = SQRT(∑F_x² + ∑F_y²) = SQRT( (29.021 N)² + (3.275 N)² ) = 29.205 N

a = F_R / m = 29.205 N / 15.5 kg = 1.884 m/s²

However, as the systems become more complex, it actually becomes easier to solve by analyzing the motions caused kept in separate x and y directions as shown in the former way.