A light string has its ends tied to two walls separated by a distance equal to five-eighths the length of the string. A 53 kg mass is suspended from the center of the string, applying a tension in the string.
What is the tension in the two strings of length L/2 tied to the wall? The acceleration of gravity is 9.8 m/s^2.
Answer in units of N.
Let L be the total string length.
Use geometry/trig to determine the angle that the string makes with the horizontal, on both sides. That angle is
A = cos^-1 (5L/16)/(L/2)
= cos^-1(5/8) = 51.32 degrees
Then apply a static force balance. Both strings have the same tension force T because of symmetry.
2 T sin 51.32 = M g
Solve for T
To find the tension in the two strings of length L/2 tied to the wall, we can break down the problem into two parts:
1. Finding the tension in the horizontal part of the string:
Let's assume the length of the string is L. The distance between the two walls separated by a distance equal to five-eighths the length of the string is (5/8)L. Since the string is tied to the walls, the tension in the horizontal part of the string is equal to the force required to balance the weight of the mass.
The weight of the mass can be calculated using the formula:
Weight = mass * acceleration due to gravity
Weight = 53 kg * 9.8 m/s^2
Weight = 519.4 N
Since the weight is acting vertically downwards, the tension in the horizontal part of the string is also acting horizontally. Therefore, the tension in the horizontal part of the string is equal to the weight of the mass, which is 519.4 N.
2. Finding the tension in the vertical part of the string:
Since the vertical part of the string is supporting the weight of the mass, we can calculate the tension in the vertical part of the string using the concept of equilibrium.
The vertical component of the tension force is balancing the weight of the mass. This vertical component can be found using the trigonometric relationship between the angles and sides of a right-angled triangle.
In the current scenario, the angle between the vertical part of the string and the horizontal part of the string is 90 degrees, as the mass is suspended at the center of the string.
Using the concept of trigonometry, we can find the length of the vertical part of the string (L_v) and its tension (T_v) using the formula:
sin(theta) = opposite/hypotenuse
sin(90 degrees) = L_v / (L/2)
1 = L_v / (L/2)
L_v = L/2
Since the tension in the vertical part of the string is balancing the weight of the mass, we can write:
T_v = Weight = 519.4 N
Therefore, the tension in the two strings of length L/2 tied to the wall is equal to the tension in the vertical part of the string, which is 519.4 N.