Question

A ball on a plane surface has a force of 100 N at an angle of 35 degrees north of west. A second force of 150 N acts at an angle of 45 degrees south of east. A third force of 80 N acts on the ball pushing it due north. What is the net force on the ball?

Answer 1

To find the net force on the ball, we need to break down the given forces into their x and y components.

Let's start with the first force of 100 N at an angle of 35 degrees north of west. To find its x component, we can use cosine:

cos(35) = adjacent / hypotenuse

adjacent = cos(35) * 100 N

Similarly, to find the y component, we can use sine:

sin(35) = opposite / hypotenuse

opposite = sin(35) * 100 N

Now let's move on to the second force of 150 N at an angle of 45 degrees south of east. Since this force is south of east, we'll need to break it down into its negative x and y components.

To find the x component, we can use cosine:

cos(45) = adjacent / hypotenuse

adjacent = cos(45) * 150 N

And to find the y component, we can use sine:

sin(45) = opposite / hypotenuse

opposite = sin(45) * 150 N

Finally, we have the third force of 80 N pushing the ball due north. Since this force is already in the y direction, we don't need to break it down into components.

Now, let's sum up the x and y components separately:

x_component = adjacent1 + (-adjacent2)

y_component = opposite1 + opposite2 + 80 N

Calculate the x_component and y_component using the given values, and then find the magnitude of the net force using the Pythagorean theorem:

net_force = sqrt(x_component^2 + y_component^2)

Remember to consider the direction of the net force by finding the angle it makes with the positive x-axis:

angle = atan(y_component / x_component)

By following these steps, you can determine the net force on the ball.

Step-by-step explanation:

Answer 2

The net force on the ball is approximately \(39.24 \, \text{N}\) directed at an angle north of west.

To find the net force on the ball, we need to break each force into its horizontal (x) and vertical (y) components and then sum these components separately.

Let's break down each force:

1. **Force 1 (100 N at 35 degrees north of west):**

- Horizontal Component (\(F_{1x}\)): \(100 \, \text{N} \cdot \cos(35^\circ)\) (westward is negative)

- Vertical Component (\(F_{1y}\)): \(100 \, \text{N} \cdot \sin(35^\circ)\) (northward is positive)

2. **Force 2 (150 N at 45 degrees south of east):**

- Horizontal Component (\(F_{2x}\)): \(150 \, \text{N} \cdot \cos(45^\circ)\) (eastward is positive)

- Vertical Component (\(F_{2y}\)): \(150 \, \text{N} \cdot \sin(45^\circ)\) (southward is negative)

3. **Force 3 (80 N due north):**

- Horizontal Component (\(F_{3x}\)): 0 (no horizontal component)

- Vertical Component (\(F_{3y}\)): 80 N (northward is positive)

Now, let's calculate each component:

- \(F_{1x} = 100 \, \text{N} \cdot \cos(35^\circ) \approx 82.31 \, \text{N}\) (westward)

- \(F_{1y} = 100 \, \text{N} \cdot \sin(35^\circ) \approx 57.48 \, \text{N}\) (northward)

- \(F_{2x} = 150 \, \text{N} \cdot \cos(45^\circ) \approx 106.07 \, \text{N}\) (eastward)

- \(F_{2y} = 150 \, \text{N} \cdot \sin(45^\circ) \approx 106.07 \, \text{N}\) (southward)

- \(F_{3x} = 0 \, \text{N}\) (no horizontal component)

- \(F_{3y} = 80 \, \text{N}\) (northward)

Now, sum the horizontal and vertical components separately:

- Horizontal Component (\(F_x\)): \(F_{1x} + F_{2x} + F_{3x} = 82.31 \, \text{N} - 106.07 \, \text{N} + 0 \, \text{N} = -23.76 \, \text{N}\) (westward)

- Vertical Component (\(F_y\)): \(F_{1y} + F_{2y} + F_{3y} = 57.48 \, \text{N} - 106.07 \, \text{N} + 80 \, \text{N} = 31.41 \, \text{N}\) (northward)

The net force is given by the magnitude and direction of the resultant vector:

\[ \text{Net Force} = \sqrt{F_x^2 + F_y^2} \]

\[ \text{Net Force} = \sqrt{(-23.76 \, \text{N})^2 + (31.41 \, \text{N})^2} \]

\[ \text{Net Force} \approx 39.24 \, \text{N} \]

So, the net force on the ball is approximately \(39.24 \, \text{N}\) directed at an angle north of west.