Two forces with magnitudes of 150 and 75 pounds act on an object at angles of 30° and 150°, respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

Question

Two forces with magnitudes of 150 and 75 pounds act on an object at angles of 30° and 150°, respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

Having lots of trouble with this question. I started with
150 (cos(30), sin(30)) + 75 (cos(150), sin(150))
F = 150 (√3/2, 1/2) + 75 (−√3/2, 1/2)
F = (75√3+37.5, 75+75√3/2)
But I'm having trouble finding the magnitude after that

Answer 1

You have messed up on your last line:

F = 150 (√3/2, 1/2) + 75 (−√3/2, 1/2)
F = (75√3-37.5√3, 75+75/2)
F = (37.5√3,112.5)
= 75 (√3/2,3/2)
= 75√3 (1/2,√3/2)
= 130 at 60°

Answer 2

steve and maria fell in love the end

Answer 3

To find the magnitude of the resultant force, you can use the magnitude formula:

Magnitude of a vector F = √(F_x^2 + F_y^2)

First, let's calculate the x-components and y-components of the resultant force by multiplying the magnitudes with their respective cosine and sine values:

For the 150-pound force:
F_1x = 150 * cos(30°)
F_1y = 150 * sin(30°)

For the 75-pound force:
F_2x = 75 * cos(150°)
F_2y = 75 * sin(150°)

Calculating these values:

F_1x = 150 * cos(30°) = 150 * √3/2 ≈ 129.90
F_1y = 150 * sin(30°) = 150 * 1/2 = 75

F_2x = 75 * cos(150°) = 75 * (-√3/2) ≈ -64.95
F_2y = 75 * sin(150°) = 75 * 1/2 = 37.50

Now, let's find the resultant force by adding the x-components and y-components:

F_x = F_1x + F_2x
F_y = F_1y + F_2y

F_x = 129.90 + (-64.95) ≈ 64.95
F_y = 75 + 37.50 = 112.50

The resultant force can be represented as a vector (F_x, F_y). To find the magnitude of this vector, use the magnitude formula:

Magnitude of the resultant force F = √(F_x^2 + F_y^2)

Magnitude of F = √(64.95^2 + 112.50^2) ≈ 130.11

Therefore, the magnitude of the resultant force is approximately 130.11 pounds.

To find the direction of the resultant force, you can use the arctan function:

θ = arctan(F_y / F_x) = arctan(112.50 / 64.95) ≈ 59.11°

The direction of the resultant force is approximately 59.11°.

Answer 4

To find the magnitude of the resultant force, you can use the Pythagorean theorem. The magnitude of the resultant force, denoted as R, can be calculated using the equation:

|R| = √(Rx² + Ry²)

Here, Rx represents the x-component of the resultant force and Ry represents the y-component of the resultant force.

First, let's find the x-components and y-components of the two forces:

For the first force:
Magnitude = 150 lbs
Angle = 30°

The x-component of the force (Rx1) can be calculated using the equation:
Rx1 = 150 * cos(30°) = 150 * √3/2 ≈ 129.90 lbs

The y-component of the force (Ry1) can be calculated using the equation:
Ry1 = 150 * sin(30°) = 150 * 1/2 = 75 lbs

For the second force:
Magnitude = 75 lbs
Angle = 150°

The x-component of the force (Rx2) can be calculated using the equation:
Rx2 = 75 * cos(150°) = 75 * (-√3/2) ≈ -64.95 lbs

The y-component of the force (Ry2) can be calculated using the equation:
Ry2 = 75 * sin(150°) = 75 * 1/2 = 37.50 lbs

Now, let's calculate the x-component and y-component of the resultant force:

Rx = Rx1 + Rx2
= 129.90 lbs + (-64.95 lbs) ≈ 64.95 lbs

Ry = Ry1 + Ry2
= 75 lbs + 37.50 lbs = 112.50 lbs

Now, we can find the magnitude of the resultant force:

|R| = √(Rx² + Ry²)
= √((64.95 lbs)² + (112.50 lbs)²)
≈ √(4225.05 + 12656.25)
≈ √16881.30
≈ 129.92 lbs (rounded to two decimal places)

Therefore, the magnitude of the resultant force is approximately 129.92 lbs.