the 3 forces shown act on a object is in equilibrium. F1= 50N (E 60 degrees N), F2= 60N (E 30 degrees S ) Calculate the magnitude of the force F3 and the angle???? How to solve??
Fnetx = 0 (equilibrium) = T1x + T2x - T3x
Fnety = 0 (equilibrium) = T1y - T2y - T3y
T1
x component = 60cos * 50 = 25
y component = 60sin * 50 = 43.3
T2
x component = 30cos * 60 = 51.96
y component = 30sin * 60 = -30
T3
x component = 0 = T1x + T2x + T3x = 25 + 51.96 - T3x
T3x = 76.96 N
y component = 0 = T1y - T2y - T3y = 43.3 - 30 - T3y
T3y = 13.3 N
using pythagorean theorem: T3= √T3x^2+T3y^2
T3 = 78.1 N
Find angle using Tanθ = T3y/T3x
θ = 9.8°
Therefore T3 = 78.1 N [W 9.8°S]
To determine the magnitude and angle of force F3, we can use vector addition to find the net force acting on the object. Since the object is in equilibrium, the net force should be zero.
Step 1: Convert the given forces into their components.
Using the given information, we can find the components of F1 and F2.
F1 = 50N (E 60° N)
To convert this force into its components, we can use trigonometry. The horizontal component (F1x) can be calculated as:
F1x = 50N * cos(60°)
Similarly, the vertical component (F1y) can be calculated as:
F1y = 50N * sin(60°)
F2 = 60N (E 30° S)
To convert this force into its components, we can use trigonometry. The horizontal component (F2x) can be calculated as:
F2x = 60N * cos(150°)
Similarly, the vertical component (F2y) can be calculated as:
F2y = 60N * sin(150°)
Step 2: Find the net force components.
To find the net force components, we need to sum up the individual force components.
Net horizontal force (Fnetx) = F1x + F2x
Net vertical force (Fnety) = F1y + F2y
Step 3: Find the magnitude and direction of the net force.
Using Pythagoras' theorem, we can find the magnitude of the net force:
Magnitude of the net force (Fnet) = sqrt(Fnetx^2 + Fnety^2)
To find the angle of the net force (θ), we can use trigonometry:
θ = atan(Fnety / Fnetx)
Step 4: Calculate the magnitude and angle of F3.
Since the net force should be zero for the object to be in equilibrium, F3 will have the same magnitude as the net force but in the opposite direction.
Magnitude of F3 = -Fnet
Angle of F3 = θ + 180°
By following these steps, we can solve for the magnitude and angle of force F3.
To solve for the magnitude and angle of force F3, we need to use the concept of vector addition to find the resultant force.
Step 1: Draw a diagram
Start by drawing a diagram that represents the forces F1 and F2. Use a scale to accurately measure the angle and magnitude of each force. Make sure to correctly indicate the direction of each force on the diagram.
Step 2: Resolve the forces into x and y-components
Next, resolve the forces F1 and F2 into their x and y-components. This will allow us to add the forces more easily. To do this, you can use the trigonometric functions (sine and cosine).
For F1:
Fx1 = F1 * cos(60°)
Fy1 = F1 * sin(60°)
For F2:
Fx2 = F2 * cos(210°)
Fy2 = F2 * sin(210°)
Note that we used the reference angles for F2 since it is given in terms of a southern direction.
Step 3: Add the x and y-components
Now, add the x-components and y-components of the forces separately.
Fx3 = Fx1 + Fx2
Fy3 = Fy1 + Fy2
Step 4: Find the magnitude and angle of F3
To find the magnitude of F3:
|F3| = √(Fx3^2 + Fy3^2)
To find the angle of F3:
θ = atan(Fy3 / Fx3)
Step 5: Perform the calculations
Plug in the values calculated in step 2 into step 3 and step 4. Calculate the magnitude and angle of F3.
Magnitude of F3: |F3| = √(Fx3^2 + Fy3^2)
Angle of F3: θ = atan(Fy3 / Fx3)
By following these steps and performing the calculations, you can find the magnitude and angle of force F3.
F1+F2 = 50N[60o] + 60N[330o]
X = 50*Cos60 + 60*Cos330 = 73.3 N.
Y = 50*sin60 + 60*sin330 = 8.30 N.
(F1 + F2) +F3 = 0
73.3 + 8.3i + F3 = 0
F3 = -73.3 - 8.3i
Tan Ar = Y/X = -8.3/-73.3 = 0.11326
Ar = 6.46o N of E = Reference angle(Q1).
A = 6.46 + 180 = 186.46o(Q4) = 6.46o S of W
F3 = X/Cos A = -73.3/Cos186.46 = 73.8 N.
[186.46o] = 73.8N.[6.46o] S of W