Forces with magnitudes of 250 pounds and 130pounds act on an object at angles 45degrees and -60degrees respectively with the positive axis. Find the direction and magnitude of the resultant of these forces.
250 pounds at 45° ---> (250cos45, 250sin45)
130 pounds at -60° ---> (130cos(-60), 130sin(-60) )
Resultant = (250cos45, 250sin45)+(130cos(-60), 130sin(-60) )
= (125√2, 125√2) + (65 , -65√3)
= ( 125√2 + 65 , 125√2 - 65√3)
= appr (241.7767 , 64.1934)
maginitude = √(..^2 + ..^2) = 250.153
angle:
tanØ = 64.1934/241.7767
Ø = 14.87°
check my arithmetic
Well, if we have a force of 250 pounds at an angle of 45 degrees and another force of 130 pounds at an angle of -60 degrees, we can definitely say that these forces are going in different directions. It's such a "forceful" situation!
To find the magnitude and direction of the resultant force, we can break down each force into its horizontal and vertical components. The horizontal component (X) can be found by multiplying the force by the cosine of the angle, and the vertical component (Y) can be found by multiplying the force by the sine of the angle.
For the first force, the horizontal component (X1) will be 250 pounds * cos(45), and the vertical component (Y1) will be 250 pounds * sin(45). Similarly, for the second force, the horizontal component (X2) will be 130 pounds * cos(-60), and the vertical component (Y2) will be 130 pounds * sin(-60).
Now, let's add up the horizontal components and the vertical components separately. The resultant horizontal component (Xr) will be X1 + X2, and the resultant vertical component (Yr) will be Y1 + Y2.
Finally, we can use the Pythagorean theorem to find the magnitude (R) of the resultant force, which is given by the square root of (Xr^2 + Yr^2). And to find the direction (θ) of the resultant force, you can use the inverse tangent function, arctan(Yr/Xr). But make sure to consider the appropriate quadrants when finding the angle!
So, grab your calculator and let's compute those values. And remember, math can be a real "force" of nature!
To find the resultant of these two forces, we can use vector addition. Let's break down the forces into their x and y components.
Force 1: Magnitude = 250 pounds, Angle = 45 degrees
Force 2: Magnitude = 130 pounds, Angle = -60 degrees
First, we find the x-component of each force:
Force 1 x-component = 250 pounds * cos(45 degrees) = 250 pounds * √2 / 2 ≈ 176.78 pounds (rounded to 2 decimal places)
Force 2 x-component = 130 pounds * cos(-60 degrees) = 130 pounds * 0.5 ≈ 65 pounds (rounded to nearest whole number)
Now, let's find the y-component of each force:
Force 1 y-component = 250 pounds * sin(45 degrees) = 250 pounds * √2 / 2 ≈ 176.78 pounds (rounded to 2 decimal places)
Force 2 y-component = 130 pounds * sin(-60 degrees) = -130 pounds * 0.866 ≈ -112.82 pounds (rounded to 2 decimal places)
Next, we add the x and y-components separately to find the resultant x-component and resultant y-component:
Resultant x-component = 176.78 pounds + 65 pounds ≈ 241.78 pounds (rounded to 2 decimal places)
Resultant y-component = 176.78 pounds - 112.82 pounds ≈ 63.96 pounds (rounded to 2 decimal places)
To find the magnitude of the resultant, we use the Pythagorean theorem:
Magnitude of the resultant = sqrt((Resultant x-component)^2 + (Resultant y-component)^2)
= sqrt((241.78 pounds)^2 + (63.96 pounds)^2)
= sqrt(58491.12 + 4092.64)
= sqrt(62583.76)
≈ 250.17 pounds (rounded to 2 decimal places)
Finally, we can find the direction of the resultant using the arctan function:
Direction of the resultant = arctan(Resultant y-component / Resultant x-component)
= arctan(63.96 pounds / 241.78 pounds)
≈ 15.89 degrees (rounded to 2 decimal places)
Therefore, the direction of the resultant is approximately 15.89 degrees with the positive axis, and the magnitude of the resultant is approximately 250.17 pounds.
To find the direction and magnitude of the resultant of these forces, we can use the concept of vector addition.
Step 1: Resolve the forces into their horizontal and vertical components.
The horizontal component of a force F is given by Fx = F * cos(θ), where θ is the angle with the positive x-axis.
The vertical component of a force F is given by Fy = F * sin(θ), where θ is the angle with the positive x-axis.
For the force with a magnitude of 250 pounds and an angle of 45 degrees:
F1x = 250 * cos(45) = 250 * (√2/2) ≈ 250 * 0.7071 ≈ 176.8 pounds
F1y = 250 * sin(45) = 250 * (√2/2) ≈ 250 * 0.7071 ≈ 176.8 pounds
For the force with a magnitude of 130 pounds and an angle of -60 degrees:
F2x = 130 * cos(-60) = 130 * (1/2) = 65 pounds
F2y = 130 * sin(-60) = 130 * (-√3/2) ≈ 130 * (-0.866) ≈ -112.6 pounds
Step 2: Add the horizontal and vertical components separately to get the resultant components.
R_x = F1x + F2x = 176.8 + 65 ≈ 241.8 pounds
R_y = F1y + F2y = 176.8 + (-112.6) ≈ 64.2 pounds
Step 3: Use the resultant components to find the magnitude and direction of the resultant force.
The magnitude of the resultant R is given by R = sqrt(R_x^2 + R_y^2).
The direction θ of the resultant R is given by tan(θ) = R_y / R_x.
Calculating:
R = sqrt(241.8^2 + 64.2^2) ≈ sqrt(58506.24 + 4121.64) ≈ sqrt(62627.88) ≈ 250.25 pounds (approx.)
θ = tan^(-1)(64.2 / 241.8) ≈ 15.91 degrees (approx.)
Therefore, the magnitude of the resultant force is approximately 250.25 pounds, and its direction is approximately 15.91 degrees with the positive x-axis.