A piece of cord 2.5m is connected between two points 2m apart horizontally. A massless pulley is placed on the cord and a force 20N at angle - 80 degrees is applied to it.
1. Show that when the pulley reaches a state of rest, the angle that the positions of the cor make with the horizontal differ by 20 degrees.
2. Verify that the relevant equations are satisfied when the smaller angle is 28 degrees and hence determine the tension in the cord.
Thank you
To solve this problem, we can consider the equilibrium of forces acting on the pulley when it reaches a state of rest.
1. Let's analyze the forces acting on the pulley:
- The force applied to the cord (20N) forms an angle of -80 degrees (clockwise) with the horizontal.
- The tension in the cord applies an equal but opposite force on the pulley.
- The weight of the pulley is balanced by an equal and opposite normal force exerted by the supporting surface.
Now, let's determine the angles that the positions of the cord make with the horizontal:
- Let A be the angle that one segment of the cord makes with the horizontal, and B be the angle that the other segment of the cord makes with the horizontal.
Since both segments of the cord are connected to the pulley, the tension forces in the cord must be equal. Therefore, the vertical components of the tensions in the two cord segments must be equal.
2. From the above information, we can create the following equation:
T * cos(A) = T * cos(B) (equation 1)
Next, let's determine the horizontal components of the tension forces in the two cord segments.
- The horizontal component of the tension on segment A balances the force applied to the cord. Therefore, it can be expressed as:
T * sin(A) = 20N * cos(-80 degrees) (equation 2)
3. Now, let's verify the relevant equations when the smaller angle is 28 degrees and determine the tension in the cord.
Let's substitute 28 degrees for angle A in equations 1 and 2, and solve for angle B and the tension (T):
From equation 2:
T * sin(28 degrees) = 20N * cos(-80 degrees)
T * sin(28 degrees) = 20N * cos(80 degrees) (since cos(-θ) = cos(θ))
T * sin(28 degrees) = 20N * 0.1736 (using the known value of cos(80 degrees) = 0.1736)
T * 0.4695 = 3.472 (evaluating sin(28 degrees), approximately 0.4695)
T ≈ 7.39N
Therefore, when the smaller angle is 28 degrees, the tension in the cord is approximately 7.39N.
To solve these problems, we need to analyze the forces acting on the massless pulley and the tension in the cord. Here's how you can approach these questions step by step:
1. Finding the angle difference:
First, let's draw a diagram to visualize the situation. The two points connected by the cord are 2m apart horizontally. The force of 20N is acting on the cord at an angle of -80 degrees (measured counterclockwise from the positive x-axis). We'll assume the cord is being pulled towards the right.
To determine the angle difference, we can use trigonometry. Let's set up a right-angled triangle. The horizontal distance between the two points is the base, which is 2m. The vertical distance of the cord is the height, which is 2.5m. Let's call the angle between the cord and the horizontal x degrees.
Using trigonometry, we have:
tan(x) = height/base
tan(x) = 2.5/2
tan(x) = 1.25
Now, let's rearrange the equation to solve for x:
x = arctan(1.25)
x ≈ 51.34 degrees
Next, we need to find the angle at which the force is acting (counterclockwise from the positive x-axis). The angle given is -80 degrees, so we can convert it to a positive angle:
angle = 360 - 80
angle = 280 degrees
The difference between the two angles is:
angle difference = angle - x
angle difference = 280 - 51.34
angle difference ≈ 228.66 degrees
Therefore, when the pulley reaches a state of rest, the angle that the positions of the cord make with the horizontal will differ by approximately 228.66 degrees.
2. Verifying equations and determining tension:
Now, let's verify the relevant equations using a smaller angle (28 degrees) and find the tension in the cord.
We can start by drawing a new diagram, setting the angle between the cord and the horizontal to 28 degrees. We'll keep the horizontal distance between the two points as 2m and the cord's vertical distance as 2.5m.
Using trigonometry, we know that the tension in the cord can be found using the equation:
Tension = Force / cos(angle)
Plugging in the values in this case:
Tension = 20 N / cos(28)
Tension ≈ 23.11 N
So, when the angle between the cord and the horizontal is 28 degrees, the tension in the cord is approximately 23.11 Newtons.
These steps should help you solve the given questions. Let me know if you have any further queries or need additional explanations!