The three forces shown in the figure below act on a particle (with F1 = 50.0 N, F2 = 60.0 N, θ1 = 60.0°, and θ2 = 30.0°). If the particle is in translational equilibrium, find F3, the magnitude of force 3 and the angle θ3.
To find the magnitude of force 3 (F3) and the angle θ3, we need to use the concept of equilibrium. In translational equilibrium, the net force acting on an object is zero. Therefore, the vector sum of all the forces acting on the particle should be zero.
Let's break down the forces into their x and y components.
Force 1 (F1) can be broken down into F1x and F1y. F1x is the horizontal component of force 1, and F1y is the vertical component of force 1. To find the values of F1x and F1y, we can use trigonometry:
F1x = F1 * cos(θ1)
F1y = F1 * sin(θ1)
Similarly, Force 2 (F2) can be broken down into F2x and F2y:
F2x = F2 * cos(θ2)
F2y = F2 * sin(θ2)
Finally, let's analyze Force 3 (F3). Since we only know its magnitude (F3), we can represent it as the sum of two components, F3x and F3y:
F3x = F3 * cos(θ3)
F3y = F3 * sin(θ3)
Now, for the particle to be in translational equilibrium, the sum of all the forces in the x-direction and the sum of all the forces in the y-direction should each be equal to zero:
∑Fx = F1x + F2x + F3x = 0
∑Fy = F1y + F2y + F3y = 0
Substituting the values we know:
F1x + F2x + F3x = 0
F1y + F2y + F3y = 0
Now let's substitute the values we know and solve for F3 and θ3.
F1x = F1 * cos(θ1) = 50.0 N * cos(60.0°)
F1y = F1 * sin(θ1) = 50.0 N * sin(60.0°)
F2x = F2 * cos(θ2) = 60.0 N * cos(30.0°)
F2y = F2 * sin(θ2) = 60.0 N * sin(30.0°)
Let's calculate these values:
F1x = 50.0 N * cos(60.0°) = 25.0 N
F1y = 50.0 N * sin(60.0°) = 43.3 N
F2x = 60.0 N * cos(30.0°) = 51.96 N
F2y = 60.0 N * sin(30.0°) = 30.0 N
Now we have the following:
25.0 N + 51.96 N + F3x = 0
43.3 N + 30.0 N + F3y = 0
Let's solve for F3x and F3y:
F3x = -25.0 N - 51.96 N = -76.96 N (opposite sign because it is in the opposite direction)
F3y = -43.3 N - 30.0 N = -73.3 N (opposite sign because it is in the opposite direction)
Since we know that F3x = F3 * cos(θ3) and F3y = F3 * sin(θ3), let's solve for F3 and θ3 using these equations.
F3 * cos(θ3) = -76.96 N
F3 * sin(θ3) = -73.3 N
Dividing the two equations, we get:
tan(θ3) = (-73.3 N) / (-76.96 N)
θ3 = tan^(-1)[(-73.3 N) / (-76.96 N)]
Calculating this value:
θ3 ≈ tan^(-1)(0.951) ≈ 44.3°
Now, solving for F3 using either of the equations:
F3 * cos(44.3°) = -76.96 N
F3 = (-76.96 N) / cos(44.3°)
F3 ≈ -112.0 N (magnitude will be positive)
Therefore, the magnitude of force 3 (F3) is approximately 112.0 N, and the angle θ3 is approximately 44.3°.