Two forces F1 = (62.13i − 54.63j) N and F2 = (19.73i − 69.15j) N are exerted on a particle. The particle's mass is 11.30 kg.
(a) Find the particle's acceleration in component form. (Express your answer in vector form.)
(b) Find the magnitude and direction of the acceleration (degree counterclockwise from the x axis)
for each F, extract the components
add them up to get the resultant force
F = ma
Now, having a,
(b) get |a|
the direction is tanθ = y/x (from a)
To find the particle's acceleration, we can use Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. In this case, the net force is the sum of the two forces.
(a) To find the particle's acceleration in component form, we can add the x-components and the y-components of the forces separately:
For the x-component:
Fx = F1x + F2x = 62.13 + 19.73 = 81.86 N
For the y-component:
Fy = F1y + F2y = -54.63 - 69.15 = -123.78 N
Now, we can divide the net force components by the mass of the particle to get the acceleration:
ax = Fx / m = 81.86 N / 11.30 kg = 7.245 m/s^2
ay = Fy / m = -123.78 N / 11.30 kg = -10.937 m/s^2
Therefore, the particle's acceleration in component form is (7.245i - 10.937j) m/s^2.
(b) To find the magnitude of the acceleration, we can use the Pythagorean theorem:
|a| = sqrt(ax^2 + ay^2) = sqrt((7.245)^2 + (-10.937)^2) = 13.12 m/s^2
To find the direction of the acceleration, we can use trigonometry:
θ = arctan(ay / ax) = arctan((-10.937) / 7.245) = -56.32 degrees
However, the direction should be measured counterclockwise from the x-axis. Since the angle is negative, we can add 360 degrees to get its positive equivalent.
θ = -56.32 degrees + 360 degrees = 303.68 degrees
Therefore, the magnitude of the acceleration is 13.12 m/s^2 and the direction is 303.68 degrees counterclockwise from the x-axis.
To find the particle's acceleration, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
(a) First, let's find the net force acting on the particle by adding the two given forces together:
Net force, F_net = F1 + F2
= (62.13i - 54.63j) N + (19.73i - 69.15j) N
= (62.13i + 19.73i) N + (-54.63j - 69.15j) N
= 81.86i - 123.78j N
Now, we can find the particle's acceleration by dividing the net force by its mass:
Acceleration, a = F_net / m
= (81.86i - 123.78j) N / 11.30 kg
≈ 7.24i - 10.94j m/s²
So, the particle's acceleration in vector form is approximately a = (7.24i - 10.94j) m/s².
(b) To find the magnitude of acceleration (|a|), we can use the Pythagorean theorem:
|a| = √(aₓ² + aᵧ²)
= √((7.24 m/s²)² + (-10.94 m/s²)²)
≈ √(52.4976 + 119.6836) m/s²
≈ √172.1812 m/s²
≈ 13.10 m/s²
To find the direction of the acceleration in degrees counterclockwise from the x-axis, we can use trigonometry:
θ = arctan(aᵧ / aₓ)
= arctan((-10.94 m/s²) / (7.24 m/s²))
≈ arctan(-1.51)
≈ -55.62°
Since the angle is negative, we need to convert it to a positive angle by adding 360°:
θ ≈ -55.62° + 360°
≈ 304.38°
Therefore, the magnitude of the acceleration is approximately 13.10 m/s², and the direction is approximately 304.38° counterclockwise from the x-axis.