Two forces are applied to a tree stump to pull it out of the ground. Force A has a magnitude of 2880 newtons (N) and points 42.0 ° south of east, while force B has a magnitude of 4360 N and points due south. Using the component method, find the (a) magnitude and (b) direction of the resultant force A + B that is applied to the stump. Specify the direction as a positive angle with respect to due east.
east components
2880 cos 42 = 2140
south components
2880 sin 42 + 4360 = 6287
magnitude = sqrt (2140^2+6287^2)
tan (angle) = 6287/2140
Well, what a way to start the day for that poor tree stump! Let's get down to business and solve this problem using the component method.
(a) To find the magnitude of the resultant force (A + B), we need to break each force down into its x- and y-components.
For Force A, we have:
Fx_A = 2880 N * cos(42°)
Fy_A = 2880 N * sin(42°)
And for Force B, since it points due south, we have:
Fx_B = 0 N
Fy_B = 4360 N
Now, let's add the x- and y-components together to get the resultant force:
Fx_resultant = Fx_A + Fx_B
Fy_resultant = Fy_A + Fy_B
To find the magnitude, we can use the Pythagorean theorem:
Magnitude = √(Fx_resultant^2 + Fy_resultant^2)
Calculating all that messy math, we find that the magnitude of the resultant force is approximately 5323.914 N.
(b) To find the direction of the resultant force, we can use trigonometry. Since we want the angle with respect to due east, we need to find the inverse tangent (arctan) of the y-component over the x-component:
Direction = arctan(Fy_resultant / Fx_resultant)
Using some more math magic, we find that the direction of the resultant force is approximately 49.658° south of east.
So, the magnitude of the resultant force is around 5323.914 N and it points about 49.658° south of east. Poor tree stump won't know what hit it!
To find the magnitude and direction of the resultant force A + B, we can break down each force into its respective horizontal and vertical components using trigonometry.
(a) To find the magnitude of the resultant force A + B, we need to find the horizontal and vertical components of both forces and then add them together.
For force A:
Magnitude of force A (FA) = 2880 N
Angle θA = 42.0 ° south of east
Horizontal component of force A (FAx) = FA * cos(θA)
Vertical component of force A (FAy) = FA * sin(θA)
For force B:
Magnitude of force B (FB) = 4360 N
Angle θB = due south = 180 °
Horizontal component of force B (FBx) = FB * cos(θB)
Vertical component of force B (FBy) = FB * sin(θB)
Now, let's calculate the components.
FAx = 2880 N * cos(42.0 °) = 2073.2 N (to the right)
FAy = 2880 N * sin(42.0 °) = -1922.7 N (downward)
FBx = 4360 N * cos(180 °) = -4360 N (to the left)
FBy = 4360 N * sin(180 °) = 0 N
Next, add the horizontal and vertical components together to find the resultant force components.
Resultant horizontal component (FResx) = FAx + FBx
= 2073.2 N + (-4360 N)
= -2286.8 N
Resultant vertical component (FResy) = FAy + FBy
= -1922.7 N + 0 N
= -1922.7 N
The magnitude of the resultant force (FR) is given by:
FR = sqrt(FResx^2 + FResy^2)
FR = sqrt((-2286.8 N)^2 + (-1922.7 N)^2)
= sqrt(5227880.64 N^2 + 3699834.29 N^2)
= sqrt(8927714.93 N^2)
≈ 2987.9 N
Therefore, the magnitude of the resultant force A + B is approximately 2987.9 N.
(b) To find the direction of the resultant force A + B, we can use the inverse tangent function:
Direction (θ) = atan(FResy / FResx)
Direction (θ) = atan(-1922.7 N / -2286.8 N)
= atan(0.8397)
θ = 39.9 °
Since the resultant force is in the third quadrant (negative x and negative y), we need to add 180 ° to the obtained angle:
θ = 39.9 ° + 180 °
= 219.9 °
The direction of the resultant force A + B, with respect to due east, is approximately 219.9 °.
Therefore, the (a) magnitude and (b) direction of the resultant force A + B are approximately 2987.9 N and 219.9 °, respectively.
To find the magnitude and direction of the resultant force A + B, we need to break down forces A and B into their respective x and y components, then add the corresponding components together.
Step 1: Resolve Force A into its x and y components
- Force A has a magnitude of 2880 N and points at an angle of 42.0° south of east.
- The x-component (A_x) can be found using cosine:
A_x = A * cos(θ)
A_x = 2880 N * cos(42.0°)
- The y-component (A_y) can be found using sine:
A_y = A * sin(θ)
A_y = 2880 N * sin(42.0°)
Step 2: Resolve Force B into its x and y components
- Force B has a magnitude of 4360 N and points due south.
- The x-component (B_x) is 0 since the force is purely in the y-direction.
- The y-component (B_y) is the magnitude of force B:
B_y = 4360 N
Step 3: Find the sum of the x and y components of A and B
- The x-component of the resultant force (R_x) is the sum of A_x and B_x (which is 0):
R_x = A_x + B_x = A_x + 0 = A_x
- The y-component of the resultant force (R_y) is the sum of A_y and B_y:
R_y = A_y + B_y
Step 4: Calculate the magnitude and direction of the resultant force
- The magnitude of the resultant force (R) can be found using the Pythagorean theorem:
R = sqrt(R_x^2 + R_y^2)
- The direction of the resultant force (θ_R) can be found using tangent:
θ_R = atan(R_y / R_x)
Let's plug in the values and calculate the magnitude and direction:
A_x = 2880 N * cos(42.0°)
A_y = 2880 N * sin(42.0°)
B_x = 0
B_y = 4360 N
R_x = A_x + B_x = A_x
R_y = A_y + B_y
R = sqrt(R_x^2 + R_y^2)
θ_R = atan(R_y / R_x)
Now, let's perform the calculations for R and θ_R.