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.
v=const => ΣF=0
F₁₂=sqrt(F₁²+F₂²)
F₃=F₁₂
F₃(x)= -F₂
F₃(y)= -F₁
tan α= F₃(y)/ F₃(x)=
=F₁/F₂
α is the angle that F₃ makes with +x direction
(south of east)
Components of a force P due North-West acting at an angle A from North will be PcosA and PsinA in North and West directions respectively and should be equal in magnitude to the force acting on the object. Thus PcosA=83.4 and PsinA=47.2. Therefore tanA=47.2/83.4 =0.566 and A=tan inv 0.566 =29.5 deg. Now to neutralize the acting forces, it must act in opposite direction that is SE and will be at an angle of 29.5 deg from South which means 90-29.5=60.5 deg from East. I feel this is the simplest way to find it.
To find the magnitude and direction of the third force, we need to analyze the net force acting on the object.
Given:
Force 1 (F1) = 83.4 N, directed due north
Force 2 (F2) = 47.2 N, directed due west
Since the object is moving with a constant velocity, we know that the net force acting on it is zero.
Step 1: Resolve the forces into their components.
- The force F1 directed north can be written as F1 = 83.4 N (0°)
- The force F2 directed west can be written as F2 = 47.2 N (-90°)
Step 2: Combine the forces along the x and y-axis separately.
- Along the x-axis:
There are no forces acting in the east-west direction, so the sum of forces should be zero, which means that the x-component of the third force is also zero.
- Along the y-axis:
The sum of forces should be zero since the object is moving with constant velocity.
So, we can write:
0 = F1 sin(90°) + F3 sinθ
0 = 83.4 N + F3 sinθ
Step 3: Solve for F3 and θ.
Since the x-component of the third force is zero and the y-component should be opposite and equal to the y-components of F1 and F2:
0 = F2 cos(-90°) + F3 cosθ
0 = 47.2 N + F3 cosθ
Simplifying these equations, we get:
83.4 N + F3 sinθ = 0
47.2 N + F3 cosθ = 0
Step 4: Rearrange and solve for F3.
Rearrange the first equation to get F3 sinθ = -83.4 N.
Rearrange the second equation to get F3 cosθ = -47.2 N.
Now, square both equations and add them:
(F3 sinθ)^2 + (F3 cosθ)^2 = (-83.4 N)^2 + (-47.2 N)^2
F3^2 (sin^2θ + cos^2θ) = 6948.56 N^2 + 2227.84 N^2
F3^2 = 9176.4 N^2
Taking the square root of both sides:
F3 = √9176.4 N
F3 ≈ 95.8 N (magnitude of the third force)
Step 5: Solve for θ.
Now, we can solve for θ using the equation F3 sinθ = -83.4 N:
sinθ = -83.4 N / F3
sinθ ≈ -0.8686
Taking the inverse sine of -0.8686, we find θ:
θ ≈ -60.3°
Step 6: Convert the angle to a positive angle south of east.
Since the angle is negative, we can add 180° to get the positive angle south of east:
θ = -60.3° + 180°
θ ≈ 119.7°
Thus, the magnitude of the third force is approximately 95.8 N, and the direction (angle south of east) is approximately 119.7°.
To find the magnitude and direction of the third force, we need to analyze the forces acting on the object. Since the object is moving with a constant velocity, it means that the net force acting on it is zero.
Let's break down the given forces:
Force 1: 83.4 N, directed due north
Force 2: 47.2 N, directed due west
To find the net force, we add these two forces together. Since they are acting in perpendicular directions, we can use the Pythagorean theorem to determine the magnitude of the net force:
Net force = sqrt[(83.4 N)^2 + (47.2 N)^2]
Calculating this, we find that the magnitude of the net force is approximately 96.19 N.
Now, to find the direction of the net force, we can use trigonometry. Recall that the angle θ can be found using the equation:
tan(θ) = opposite/adjacent
In this case, we want to find the angle south of east, so we need to consider the opposite and adjacent sides accordingly.
The opposite side can be determined by subtracting the y-component (north-south) of the net force (83.4 N) from the y-component (north-south) of the second force (-47.2 N):
Opposite side = -47.2 N - 83.4 N = -130.6 N
The adjacent side can be determined by adding the x-component (east-west) of the net force (0 N) with the x-component of the second force (0 N):
Adjacent side = 0 N + 0 N = 0 N
Applying these values to the equation, we get:
tan(θ) = -130.6 N / 0 N
Since the adjacent side is zero, the tangent function gives us an undefined value. Therefore, we can conclude that the angle θ is 90 degrees, or due east.
However, the problem asks for the angle south of east. Since due east is 90 degrees, we can subtract this angle from 180 degrees to get the angle south of east:
Angle south of east = 180 degrees - 90 degrees = 90 degrees
Therefore, the magnitude of the third force must be approximately 96.19 N, and the direction of the third force must be 90 degrees south of east.