There are two forces acting on an object. One force has a magnitude of 39 N and is directed along the +x axis. The other force has a magnitude of 25 N and is directed along the +y axis. What is the direction of the acceleration of the object? Express your answer by giving the angle between a and the x axis.
The direction of the acceleration is the direction of the resultant force. That direction makes an angle of
arctan(25/39) = 32.7 degrees with the +x axis, measured counterclockwise from the +x axis.
Thank you!
To determine the direction of the acceleration of the object, we can use vector addition. The two forces can be represented as vectors:
- Force 1: 39 N along the +x-axis
- Force 2: 25 N along the +y-axis
We can use the Pythagorean theorem to find the resultant vector, which represents the net force acting on the object. The magnitude of the resultant vector is given by:
|F_res| = sqrt((F1x + F2x)^2 + (F1y + F2y)^2)
Where F1x and F1y are the x and y components of Force 1, and F2x and F2y are the x and y components of Force 2.
For Force 1, since it is directed along the +x-axis, its x-component is 39 N and its y-component is 0 N.
For Force 2, since it is directed along the +y-axis, its x-component is 0 N and its y-component is 25 N.
Now we can use the equation for the magnitude of the resultant vector:
|F_res| = sqrt((39 + 0)^2 + (0 + 25)^2)
= sqrt(39^2 + 25^2)
= sqrt(1521 + 625)
= sqrt(2146)
≈ 46.35 N
Next, we need to find the angle between the resultant vector (net force) and the +x-axis. This can be done by finding the angle theta, using the equation:
tan(theta) = (F1y + F2y) / (F1x + F2x)
tan(theta) = (25 N) / (39 N)
theta = tan^(-1)(25 N / 39 N)
≈ 33.69 degrees
Therefore, the direction of the acceleration is approximately 33.69 degrees above the +x-axis.