Three forces act on a moving object. One force has a magnitude of 83.4 N and is directed due north. Another has a magnitude of 47.2 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express your answer as a positive angle south of east.
if the object moves with constant velocity, there must be no acceleration, hence no net force.
If the two known forces are u and v, we need w = -(u+v) so that u+v+w = 0.
So, convert to rectangular components, add them up, and take the negative sum for the 3rd force.
F1+F2+F3 = M*a = M*0 = 0.
83.4i - 47.2 + F3 = 0.
F3 = 47.2 - 83.4i = 95.83 N.[60.5o] S. of E.
To find the magnitude and direction of the third force, we need to consider the concept of vector addition.
The object is moving with a constant velocity, which means the net force acting on it is zero. In other words, the sum of all the forces acting on the object must be zero.
Let's break down the given forces:
1. The force with a magnitude of 83.4 N acting due north can be represented as (0 N, 83.4 N) in Cartesian coordinates. Since it is directed due north, it only has a component in the y-direction.
2. The force with a magnitude of 47.2 N acting due west can be represented as (-47.2 N, 0 N) in Cartesian coordinates. Since it is directed due west, it only has a component in the x-direction.
Now, let's assume that the magnitude of the third force is F and the angle it makes with the positive x-axis (east) is θ.
We can represent the third force as (F * cos(θ), F * sin(θ)) in Cartesian coordinates, where cos(θ) represents the x-component and sin(θ) represents the y-component.
For the object to continue moving with a constant velocity (zero net force), the vector sum of the three forces should be equal to zero.
Adding up the x-components of the forces:
F * cos(θ) - 47.2 N = 0
Adding up the y-components of the forces:
F * sin(θ) + 83.4 N = 0
Now we can solve these two equations simultaneously to find the magnitude (F) and the direction (θ) of the third force.
From the first equation:
F * cos(θ) = 47.2 N
Divide both sides by cos(θ):
F = 47.2 N / cos(θ)
Substituting this value of F into the second equation:
(47.2 N / cos(θ)) * sin(θ) + 83.4 N = 0
Multiply through by cos(θ):
47.2 N * sin(θ) + 83.4 N * cos(θ) = 0
Rearranging this equation gives us:
tan(θ) = -83.4 N / 47.2 N
Taking the inverse tangent of both sides gives us:
θ = tan^(-1)(-83.4 N / 47.2 N)
Using a calculator, we find θ ≈ -59.8°.
To express the angle as a positive angle south of east, we can add 180° to it:
θ = -59.8° + 180°
Therefore, the direction of the third force is approximately 120.2° south of east.
To find the magnitude of the third force, substitute the value of θ into the first equation:
F * cos(120.2°) = 47.2 N
Divide both sides by cos(120.2°):
F = 47.2 N / cos(120.2°)
Using a calculator, we find F ≈ 82.1 N.
Therefore, (a) the magnitude of the third force is approximately 82.1 N, and (b) the direction is approximately 120.2° south of east.