What is the magnitude of tension on each cord
18 degrees
4.75 kg
To find the magnitude of tension on each cord, we need to consider the forces acting on the system.
Let's assume there are two cords attached at an angle of 18 degrees with each other. The system also has a mass of 4.75 kg.
Now, draw a free body diagram of the system. Label the forces acting on the system. Since we are looking for the tension in each cord, we can label those forces as T1 and T2.
Note that the weight of the system acts downward, and we can label that force as mg, where m is the mass of the system (4.75 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The angle between the vertical direction and one of the cords can be calculated by subtracting 90 degrees (since the vertical direction is perpendicular to the horizontal plane) from the given angle of 18 degrees. This gives us an angle of 72 degrees.
Now, using trigonometry, we can find the components of the weight force mg along the vertical and horizontal directions. The vertical component of the weight force is given by mg * cos(72 degrees) and the horizontal component is given by mg * sin(72 degrees).
Since the system is in equilibrium, the sum of the vertical components of the tensions in the cords must balance the weight force. Therefore, we have:
T1 * cos(18 degrees) + T2 * cos(72 degrees) = mg * cos(72 degrees)
Similarly, the sum of the horizontal components of the tensions in the cords must balance the horizontal component of the weight force. Therefore, we have:
T1 * sin(18 degrees) = mg * sin(72 degrees)
Now, we can solve these two equations simultaneously to find the values of T1 and T2, which will give us the magnitudes of tension on each cord.