Two forces are applied to a car in an effort to move it, as shown in the following figure (cannot attach picture), where F1 = 416 N, 10 degrees to the left of north and F2 = 356 N, 30 degrees to the right of north.
What is the sum of these two forces acting on the car?
Magnitude=
Direction=
I found the following:
F1x=416*sin(10)+-226.3128
F2x=356*sin(30)=-351.7393
F1x+F2x=-578.0521
and
F1y=416*cos(10)=-349.0538
F2y=356*cos(30)=54.9135
F1y+F2y=-294.1403
I tried finding directing using:
Tan=-294.1403/-578.0521=0.50885
So direction should be =26.969 but was wrong.
I'm not sure where to go from here
x is generally taken positive to the right (east)
F1x= -416 sin 10 = -72.2
F2x = 356 sin 30 = 178
so Fx = 106
F1y = 416 cos 10 = 410
F2y = 356 cos 30 = 308 We must be operating on different number systems :(
Fy = 718
tan angle up from x axis (east)
= 718/106 =
so angle up from east = 81.6 deg
so angle clockwise from north = 8.4 degrees right from north
magnitude = sqrt (106^2+718^2)
= 726
To find the sum of these two forces acting on the car, you need to find the x and y components of each force and then add them separately.
First, let's find the x and y components of F1:
F1x = 416 * sin(10°)
F1y = 416 * cos(10°)
F1x ≈ 71.792
F1y ≈ 407.990
Now, let's find the x and y components of F2:
F2x = 356 * sin(30°)
F2y = 356 * cos(30°)
F2x ≈ 178.000
F2y ≈ 307.692
To find the sum of the x components, add F1x and F2x:
F1x + F2x ≈ 71.792 + 178.000 ≈ 249.792
To find the sum of the y components, add F1y and F2y:
F1y + F2y ≈ 407.990 + 307.692 ≈ 715.682
So, the magnitude of the sum of these two forces acting on the car is √(249.792^2 + 715.682^2) ≈ 763.773 N.
To find the direction, use the inverse tangent function:
tanθ = (F1y + F2y) / (F1x + F2x)
θ ≈ tan^(-1)((F1y + F2y) / (F1x + F2x))
θ ≈ tan^(-1)(715.682 / 249.792)
θ ≈ 71.309°
Since the forces are applied in opposite directions (one to the left of north and the other to the right of north), the direction should be 180° - 71.309° = 108.691°.
Therefore, the sum of these two forces acting on the car is approximately 763.773 N in the direction of 108.691°.
To find the sum of the two forces acting on the car, you need to break down each force into its horizontal and vertical components, and then add the corresponding components together.
Let's go through the calculations step by step:
1. Resolve Force F1:
Force F1 has a magnitude of 416 N and is 10 degrees to the left of north.
To find the horizontal component (F1x) of F1:
F1x = F1 * sin(10)
F1x = 416 * sin(10)
F1x ≈ -71.75 N (rounded to two decimal places)
To find the vertical component (F1y) of F1:
F1y = F1 * cos(10)
F1y = 416 * cos(10)
F1y ≈ 413.62 N (rounded to two decimal places)
2. Resolve Force F2:
Force F2 has a magnitude of 356 N and is 30 degrees to the right of north.
To find the horizontal component (F2x) of F2:
F2x = F2 * sin(30)
F2x = 356 * sin(30)
F2x = 178 N (rounded to two decimal places)
To find the vertical component (F2y) of F2:
F2y = F2 * cos(30)
F2y = 356 * cos(30)
F2y = 308.86 N (rounded to two decimal places)
3. Find the sum of the horizontal components:
Ftotalx = F1x + F2x
Ftotalx = -71.75 N + 178 N
Ftotalx ≈ 106.25 N (rounded to two decimal places)
4. Find the sum of the vertical components:
Ftotaly = F1y + F2y
Ftotaly = 413.62 N + 308.86 N
Ftotaly ≈ 722.48 N (rounded to two decimal places)
5. Find the magnitude of the total force:
Magnitude = sqrt(Ftotalx^2 + Ftotaly^2)
Magnitude = sqrt((106.25)^2 + (722.48)^2)
Magnitude ≈ 738.82 N (rounded to two decimal places)
6. Find the direction of the total force:
Direction = tan^(-1)(Ftotaly/Ftotalx)
Direction = tan^(-1)(722.48/106.25)
Direction ≈ 81.24 degrees (rounded to two decimal places)
So, the sum of the two forces acting on the car is approximately 738.82 N in a direction of 81.24 degrees.