Forces with magnitudes of 250 pounds and 130 pounds, act on an object at angles 45°and -61° respectiviley , with the positive axis. Find the direction and magnitude of the resultant of these forces.
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To find the magnitude and direction of the resultant of these forces, we can use vector addition.
First, let's resolve the forces into their x and y components.
For the force with a magnitude of 250 pounds and an angle of 45°, we can find the x and y components as follows:
Fx = 250 * cos(45°)
= 250 * √2 / 2
= 125√2 pounds (approximately 176.78 pounds, rounded to two decimal places)
Fy = 250 * sin(45°)
= 250 * √2 / 2
= 125√2 pounds (approximately 176.78 pounds, rounded to two decimal places)
Similarly, for the force with a magnitude of 130 pounds and an angle of -61°, we can find the x and y components:
Fx = 130 * cos(-61°)
= 130 * cos(61°)
= 130 * 0.5
= 65 pounds
Fy = 130 * sin(-61°)
= 130 * sin(61°)
= 130 * 0.866
= 112.46 pounds (rounded to two decimal places)
Now we can add the x and y components of the forces to find the resultant:
Rx = Fx1 + Fx2
= 125√2 + 65
≈ 249.01 pounds (rounded to two decimal places)
Ry = Fy1 + Fy2
= 125√2 + 112.46
≈ 288.24 pounds (rounded to two decimal places)
The magnitude of the resultant force (R) is given by the magnitude of the vector formed by the components:
R = √(Rx^2 + Ry^2)
= √(249.01^2 + 288.24^2)
≈ 376.48 pounds (rounded to two decimal places)
The direction of the resultant force, θ, is given by the angle formed by the resultant vector with the positive x-axis:
θ = tan^(-1)(Ry / Rx)
= tan^(-1)(288.24 / 249.01)
≈ 49.14° (rounded to two decimal places)
Therefore, the direction of the resultant force is approximately 49.14° with respect to the positive x-axis, and its magnitude is approximately 376.48 pounds.