1. A man pulls a crate with a rope. The crate slides along the floor in the horizontal direction

(x direction). The man exerts a force of 50 N on the rope, and the rope is at an angle q. Describe how the force components (Fx, Fy) change as the angle increases from 0° to 90° and use your graph to explain your answer. Give a detailed explanation of the forces at
q = 0°, 45°, and 90°. Show a sample calculation at one angle for both components.

Well, well, looks like we're diving into some physics! Don't worry, I'll help you out with a chuckle along the way. Now, let's talk about our crate-pulling friend and his beloved rope.

When the angle q is 0°, the rope is being pulled directly in the x-direction. That means all of our force is exerted in the x-component (Fx = 50 N), and there's no force in the y-component (Fy = 0 N). It's like a party for the horizontal forces, with the vertical forces crying in the corner.

Now, when the angle q is 90°, it's a role reversal. Our poor horizontal force (Fx) is non-existent, while the vertical force (Fy) gets to shine. So Fx = 0 N and Fy = 50 N. It's like the vertical forces finally get their moment in the spotlight, while the horizontal forces are taking a nap.

At an angle of 45°, Fx and Fy are equal. That's right, they're both about 35.36 N (rounded). This is because the force applied is being split equally between the x and y directions. Think of it like two restless kids sharing a single lollipop, sadly never reaching the full enjoyment of a whole lollipop each.

Now, let's do a sample calculation to seal the deal. Say we pick an angle of 30°. Using some sassy trigonometry, we find the value of Fx and Fy using the formula:

Fx = Force * cos(angle)
Fy = Force * sin(angle)

Substituting in our values, we get:

Fx = 50 N * cos(30°) ≈ 43.3 N
Fy = 50 N * sin(30°) ≈ 25 N

So, at an angle of 30°, our x-component (Fx) is around 43.3 N, while our y-component (Fy) is about 25 N. They're not equal, but they still get along pretty well.

Remember, as the angle increases from 0° to 90°, the x-component (horizontal force) decreases, while the y-component (vertical force) increases. It's all about finding the right balance between the two.

At q = 0°, the angle between the rope and the horizontal direction is 0°, which means the rope is perfectly horizontal. In this case, the force components Fx and Fy can be calculated as follows:

Fx = F * cos(q) = 50 N * cos(0°) = 50 N
Fy = F * sin(q) = 50 N * sin(0°) = 0 N

So, at q = 0°, only the Fx component exists, which is equal to the applied force of 50 N.

At q = 45°, the angle between the rope and the horizontal direction is 45°. In this case, the force components Fx and Fy can be calculated as follows:

Fx = F * cos(q) = 50 N * cos(45°) = 50 N * √2 / 2 ≈ 35.36 N
Fy = F * sin(q) = 50 N * sin(45°) = 50 N * √2 / 2 ≈ 35.36 N

So, at q = 45°, both Fx and Fy components exist, each approximately equal to 35.36 N.

At q = 90°, the angle between the rope and the horizontal direction is 90°, meaning the rope is vertically oriented. In this case, the force components Fx and Fy can be calculated as follows:

Fx = F * cos(q) = 50 N * cos(90°) = 50 N * 0 = 0 N
Fy = F * sin(q) = 50 N * sin(90°) = 50 N * 1 = 50 N

So, at q = 90°, only the Fy component exists, which is equal to the applied force of 50 N.

To illustrate how these force components change as the angle increases, we can plot a graph with q on the x-axis and the force components (Fx and Fy) on the y-axis.

- When q = 0°, Fx is at its maximum value of 50 N, and Fy is at its minimum value of 0 N.
- As q increases from 0° to 45°, both Fx and Fy increase, reaching approximately 35.36 N at q = 45°.
- As q increases further from 45° to 90°, Fx gradually decreases to 0 N, while Fy increases, reaching its maximum value of 50 N at q = 90°.

Overall, the graph of the force components would show Fx decreasing from its maximum to 0 N as q increases from 0° to 90°, while Fy increases from 0 N to its maximum value of 50 N during the same range.

Sample calculation:

Let's calculate the force components at q = 45°:

Fx = F * cos(q) = 50 N * cos(45°) ≈ 50 N * √2 / 2 ≈ 35.36 N
Fy = F * sin(q) = 50 N * sin(45°) ≈ 50 N * √2 / 2 ≈ 35.36 N

Therefore, at q = 45°, both Fx and Fy are approximately equal to 35.36 N.

To describe how the force components (Fx, Fy) change as the angle increases from 0° to 90°, we need to analyze the forces at different angles and break them down into their x and y components.

At q = 0° (or horizontally), the force exerted by the man is purely in the x-direction (Fx) because there is no vertical component (Fy). Therefore, at this angle, Fx = 50 N and Fy = 0 N.

As we increase the angle from 0° to 45°, the force will start to have both x and y components. To find the force components, we can use trigonometry. Let's assume the magnitude of the force (F) remains constant at 50 N.

At q = 45°, we can use the trigonometric identity sin(45°) = cos(45°) = 1/sqrt(2) = 0.7071 to calculate the components:

Fx = F * cos(q) = 50 N * cos(45°) = 50 N * 0.7071 = 35.355 N (approximately)
Fy = F * sin(q) = 50 N * sin(45°) = 50 N * 0.7071 = 35.355 N (approximately)

At q = 90° (or vertically), the force is purely in the y-direction (Fy) because there is no horizontal component (Fx). Therefore, Fx = 0 N and Fy = 50 N.

Now, let's plot a graph to visualize how Fx and Fy change as the angle increases.

On the x-axis, we'll represent the angle (q) from 0° to 90°, and on the y-axis, we'll represent the force components Fx and Fy.

At q = 0°, the graph will show Fx = 50 N and Fy = 0 N.
At q = 45°, the graph will show Fx = 35.355 N and Fy = 35.355 N.
At q = 90°, the graph will show Fx = 0 N and Fy = 50 N.

By increasing the angle from 0° to 90°, you'll notice that Fx decreases from 50 N to 0 N, while Fy increases from 0 N to 50 N. This indicates that as the angle increases, the horizontal force component decreases while the vertical force component increases. At 45°, both components are equal.

To summarize:
- At q = 0°, Fx = 50 N and Fy = 0 N.
- At q = 45°, Fx = Fy = 35.355 N.
- At q = 90°, Fx = 0 N and Fy = 50 N.

Note: The sample calculation for Fx and Fy at q = 45° used the trigonometric identities sin(45°) = cos(45°) = 1/sqrt(2) = 0.7071. These identities can be looked up in a trigonometric table or calculated using a scientific calculator.

F(x) =Fcosα

F(y)=F sinα