Forces of 10.6 N north, 20.8 N east, and 14.3 N south are simultaneously applied to a 3.76 kg mass as it rests on an air table. What is the magnitude of its acceleration?
N sum: 10.6-14.3 N or South 3.7N
E sum: 20.8N
angle, S of E, angle is arc tan 3.7/20.8
To find the magnitude of the acceleration, we can 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.
In this case, we have three forces acting on the object: 10.6 N north, 20.8 N east, and 14.3 N south. These forces are given in different directions, so we need to decompose them into their x and y components.
Let's select the north direction as the positive y-axis and the east direction as the positive x-axis. The south direction will be considered negative in the y-axis.
The net force acting on the object in the x-direction (east-west) can be found by summing the x-components of the individual forces:
Net force in the x-direction = force east - force west = 20.8 N east
The net force acting on the object in the y-direction (north-south) can be found by summing the y-components of the individual forces:
Net force in the y-direction = force north - force south = 10.6 N north - 14.3 N south
Since the south direction is considered negative in the y-axis, we subtract the force south from force north.
Now, we can calculate the net force vector using the x and y components:
Net force = √(Net force in x-direction)^2 + (Net force in y-direction)^2
Net force = √(20.8 N)^2 + (10.6 N - 14.3 N)^2
Net force = √(20.8 N)^2 + (-3.7 N)^2
Net force = √(432.64 N^2 + 13.69 N^2)
Net force = √446.33 N^2
Net force = 21.13 N
Now, we can use Newton's second law to find the acceleration:
Net force = mass * acceleration
21.13 N = 3.76 kg * acceleration
acceleration = 21.13 N / 3.76 kg
acceleration ≈ 5.62 m/s^2
Therefore, the magnitude of the acceleration of the 3.76 kg mass is approximately 5.62 m/s^2.