Find the tension T1 in the diagram above if the system is in equilibrium. [G = 10ms-2]

Find the tension T1 in the diagram above if the system is in equilibrium. [G = 10ms-2]

Answer Details

To find the tension T1, we need to analyze the forces acting on the system and use the fact that it is in equilibrium, which means the net force in all directions is zero. First, let's consider the vertical forces. The weight of the hanging mass is acting downwards with a force of 100N. The tension T2 is also acting downwards with a force of T2*cos(30°) because it is at an angle of 30° to the vertical. Therefore, the net vertical force is: Net vertical force = T2*cos(30°) - 100N Since the system is in equilibrium, the net vertical force must be zero. Thus, we can solve for T2: T2*cos(30°) = 100N T2 = 100N/cos(30°) = 115.5N Next, let's consider the horizontal forces. The tension T2 is acting to the left with a force of T2*sin(30°) because it is at an angle of 30° to the horizontal. The tension T1 is acting to the right. Therefore, the net horizontal force is: Net horizontal force = T1 - T2*sin(30°) Since the system is in equilibrium, the net horizontal force must be zero. Thus, we can solve for T1: T1 = T2*sin(30°) = 115.5N*sin(30°) = 57.8N Therefore, the tension T1 is 57.8N. Option (B)  100/√3N
is closest to this value, but it is not the correct answer.