An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.2 N ; a second force has a magnitude of 5.0 N and points in the negative y direction
Find the magnitude of the third force acting on the object. F= N 2SF
Find the direction of the third force acting on the object.
θ = from the +x-direction
If it moves with constant V the forces add to zero.
to counter first
Fx = -6.2 N
to counter second
Fy = +5 N
|F| =sqrt(5^2 + 6.2^2 )
= 7.96 N
quadrant II
tan T = 5/-6.2
T = 180 - 38.9
To find the magnitude of the third force acting on the object, we need to analyze the forces acting on the object.
Given:
Magnitude of the first force (Fx) = 6.2 N
Magnitude of the second force (Fy) = 5.0 N
Since the object is moving with a constant velocity, the net force acting on the object is zero. Therefore, the sum of the forces in the x-direction and y-direction must cancel each other out.
In the x-direction:
The force in the positive x-direction is given as 6.2 N. Since the object is moving with constant velocity, there must be an equal and opposite force in the negative x-direction to cancel it out. So, the net force in the x-direction is 6.2 N - Fx = 0. Therefore, Fx = -6.2 N.
In the y-direction:
The force in the negative y-direction is given as 5.0 N. Since the object is moving with constant velocity, there must be an equal and opposite force in the positive y-direction to cancel it out. So, the net force in the y-direction is 5.0 N + Fy = 0. Therefore, Fy = -5.0 N.
Now, we can find the magnitude of the third force using the principle of vector addition.
Let Fz be the magnitude of the third force acting on the object. Since the net force in the x-direction and the net force in the y-direction are equal to zero, we can write:
Fx + Fz = 0 (1)
Fy + Fz = 0 (2)
Substituting the values of Fx and Fy, we get:
-6.2 N + Fz = 0 (1)
-5.0 N + Fz = 0 (2)
Solving these equations, we find:
Fz = 6.2 N (magnitude of the third force)
To find the direction of the third force, we need to find the angle (θ) it makes with the positive x-direction.
θ = arctan(Fy / Fx)
θ = arctan(-5.0 N / -6.2 N)
θ ≈ 39.8 degrees from the +x-direction
Therefore, the magnitude of the third force is 6.2 N, and the direction of the third force is approximately 39.8 degrees from the +x-direction.