Forces of 11.3N north, 19.1N east, and 14.5N south are simultaneously applied to a 4.12kg mass as it rests on an air table. What is the magnitude of its acceleration?
What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive. Enter an angle between -180 degrees and +180 degrees.)
I got 4.70 m/s^2 for the acceleration which is correct. The problem is that I can't find the right angle. I got 9.51 degrees below east (ie. -9.51 degrees) but this is not correct.
To find the magnitude and direction of the acceleration, we need to break down the given forces into their horizontal and vertical components.
Let's start by calculating the horizontal and vertical components of each force:
Force 1 (11.3N north):
Horizontal component = 0
Vertical component = 11.3N
Force 2 (19.1N east):
Horizontal component = 19.1N
Vertical component = 0
Force 3 (14.5N south):
Horizontal component = 0
Vertical component = -14.5N (negative because it is pointing downwards)
Now, we can calculate the total horizontal and vertical components of the forces:
Horizontal net force = 19.1N + 0 + 0 = 19.1N
Vertical net force = 0 + 11.3N - 14.5N = -3.2N
To find the magnitude of the acceleration, we use Newton's second law: F = ma
Since F = net force, and mass (m) equals 4.12kg, we can rearrange the equation to solve for acceleration (a):
a = F/m
Plugging in the values, we have:
a = sqrt((19.1N)^2 + (-3.2N)^2)/4.12kg
a = sqrt(365.81N^2 + 10.24N^2)/4.12kg
a = sqrt(376.05N^2)/4.12kg
a = 19.38N/4.12kg
a ≈ 4.70 m/s^2
So, you are correct. The magnitude of the acceleration is approximately 4.70 m/s^2.
To find the direction of the acceleration, we can use trigonometry. Since the horizontal net force is positive (pointing east) and the vertical net force is negative (pointing downwards), the direction will be in the fourth quadrant.
We can calculate the angle using the arctan function:
tan(angle) = (Vertical net force / Horizontal net force)
angle = arctan((-3.2N) / (19.1N))
angle ≈ -9.74 degrees
The negative sign indicates that the direction is below the positive x-axis (east).
Therefore, the correct direction of the acceleration is approximately -9.74 degrees below east (counterclockwise from east). So, your calculation of -9.51 degrees was close, but the correct answer is approximately -9.74 degrees.
To find the magnitude of the acceleration, we can use Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration:
Net force = mass × acceleration
In this case, we have three forces acting on the object simultaneously: 11.3N north, 19.1N east, and 14.5N south.
To calculate the net force, we need to combine these forces. Since they are acting in different directions, we'll need to break them down into their horizontal and vertical components.
The 11.3N force acts in the north direction, so its vertical component (y-direction) will be positive, and its horizontal component (x-direction) will be zero.
The 19.1N force acts in the east direction, so its vertical component is zero, and its horizontal component is positive.
The 14.5N force acts in the south direction, so its vertical component is negative, and its horizontal component is zero.
Now, let's calculate the vertical and horizontal components:
Vertical components sum = 11.3N - 14.5N = -3.2N
Horizontal components sum = 19.1N + 0N = 19.1N
Next, we can calculate the net force in each direction:
Net vertical force = -3.2N
Net horizontal force = 19.1N
To find the magnitude of the acceleration, we can use the Pythagorean theorem:
Acceleration = √((Net horizontal force)^2 + (Net vertical force)^2)
Acceleration = √((19.1N)^2 + (-3.2N)^2)
Acceleration = √(365.21N^2 + 10.24N^2)
Acceleration = √375.45N^2
Acceleration ≈ 19.37 N
Therefore, the magnitude of the acceleration is approximately 19.37 m/s^2.
To find the direction of the acceleration in degrees, we can use trigonometry.
Direction angle = atan(Net vertical force / Net horizontal force)
Direction angle = atan(-3.2N / 19.1N)
Direction angle ≈ atan(-0.1675)
Using a calculator, we find that atan(-0.1675) ≈ -9.64 degrees.
Therefore, the direction of the acceleration is approximately -9.64 degrees below east.