Answer:
12.75N
Step-by-step explanation:
Let's consider the forces acting on the bar:
Weight of the bar (W = 30.0 N): It acts vertically downward, passing through the center of gravity of the bar.
Tension in cord 1 (T1): It acts horizontally and to the left, making an angle of 90° with the bar. Since cord 1 is massless, it does not contribute to the torque.
Tension in cord 2 (T2): It acts at an angle φ = 40.0° with the vertical.
To find the tension in cord 1 and cord 2, we need to set up torque equilibrium equations. The torque of the weight about the point of suspension must be balanced by the torques of the tensions in cords 1 and 2.
Taking the left end of the bar as the reference point (pivot), the torque equilibrium equation can be written as:
Torque due to weight = Torque due to T1 + Torque due to T2
The torque due to the weight is calculated as follows:
Torque due to weight = Weight of the bar * Perpendicular distance between the weight and the pivot point
The torque due to T1 is zero since it acts along the line of action passing through the pivot point.
The torque due to T2 can be calculated as follows:
Torque due to T2 = T2 * Perpendicular distance between the cord and the pivot point
Using the given values:
Weight of the bar (W) = 30.0 N
Length of the bar (L) = 3.0 m
Distance of the center of gravity from the left-hand end of the bar (d) = 2.2 m
Angle between cord 2 and the vertical (φ) = 40.0°
We can calculate the perpendicular distance between the weight and the pivot point as:
Perpendicular distance = L/2 - d
Using these values, we can solve for T2:
30.0 N * (L/2 - d) = T2 * L * sin(φ)
Let's substitute the given values and solve for T2:
30.0 N * (3.0/2 - 2.2) = T2 * 3.0 * sin(40.0°)
T2 ≈ 12.75 N
Therefore, the tension in cord 2 (T2) is approximately 12.75 N.