Question

A mass of 15 kg is resting on a horizontal, frictionless surface. Force 1 of 206 N is applied to it at some angle above the horizontal, force 2 has a magnitude of 144 N and is applied vertically downward, force 3 has a magnitude of 5 N and is applied vertically upwards, and force 4 has a magnitude of 42 N and is applied in the -x direction to the object. When these forces are applied to the object, the object is moving at 20 m/s in the x direction in a time of 3 seconds. What is the normal force acting on the mass in Newtons

Answer 1

N = 136.77 N

Step-by-step explanation:

This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.

Let's use trigonometry to decompose the force F1

cos θ = F₁ₓ / F₁

sin θ = F_{1y} / F₁

F₁ₓ = F₁ cos θ

F_{1y} = F₁ sin θ

now let's apply Newton's second law to each axis

X axis

F₁ₓ - F4 = m a

Y axis

N + F3 + F_{1y} -F₂ -W = 0

the acceleration can be calculated with kinematics

v = v₀ + a t

since the object starts from rest, the initial velocity is zero v₀ = 0

a = v / t

a = 20/3

a = 6.667 m / s²

we substitute in the equation

F₁ₓ = F₄ + m a

F₁ₓ = 42 + 15 6,667

F₁ₓ = 142 N

F₁ cos θ = 142

cos θ = 142/206 = 0.6893

θ = cos⁻¹ 0.6893

θ = 46.42º

now let's work the y axis

N = W + F₂ - F₃ - F_{1y}

N = 15 9.8 + 144 -5 - 206 sin 46.42

N = 286 - 149.23

N = 136.77 N