Two horizontal forces of 100 N [N] and 80 N [S 30degrees W] act on an object with a mass of 10 kg. What is the acceleration of the object? Note: this problem requires a full
vector solution
(0i + 100j) + (-40i - 69.28j) = -40i + 30.72j
F = ma, so
a = F/10 = -4.0i + 3.072j
what is i?
nevermind
thank you
To find the acceleration of the object, we first need to determine the net force acting on it.
Let's break down the given forces into their x and y-components:
Force 1: 100 N [N]
Since this force is purely vertical, it only has a y-component. The y-component can be calculated as:
F1y = 100 N * sin(90°) = 100 N * 1 = 100 N [upward]
Force 2: 80 N [S 30° W]
To determine the x and y-components of this force, we need to break it down into its horizontal and vertical components. The angle of 30° is measured clockwise from the south direction.
y-component:
F2y = 80 N * sin(30°)
= 80 N * 0.5
= 40 N [downward]
x-component:
F2x = 80 N * cos(30°)
≈ 80 N * 0.87
≈ 69.6 N [west]
Now, let's calculate the net force along the x and y-axes:
Net force in the x-direction:
Fnetx = F2x = -69.6 N (negative sign indicates the force is acting in the opposite direction of the positive x-axis)
Net force in the y-direction:
Fnety = F1y + F2y
= 100 N + (-40 N)
= 60 N
Now, we can find the magnitude and direction of the net force:
Magnitude of the net force:
|Fnet| = sqrt(Fnetx^2 + Fnety^2)
= sqrt((-69.6 N)^2 + (60 N)^2)
≈ sqrt(4832.16 N^2 + 3600 N^2)
≈ sqrt(8432.16 N^2)
≈ 91.8 N
Direction of the net force:
tanθ = Fnetx / Fnety
θ = atan(Fnetx / Fnety) [use the inverse tangent function]
θ ≈ atan(-69.6 N / 60 N)
θ ≈ atan(-1.16)
Since the net force is in the fourth quadrant, the angle θ is negative:
θ ≈ -47.5°
Now that we know the magnitude (91.8 N) and direction (-47.5°) of the net force, we can determine the acceleration of the object using Newton's second law:
Fnet = m * a
Substituting the values:
91.8 N = 10 kg * a
Solving for the acceleration (a):
a = 91.8 N / 10 kg
a ≈ 9.18 m/s²
Therefore, the acceleration of the object is approximately 9.18 m/s² in the direction opposite to the positive x-axis.