A uniform rod 8m long weighing 5kg is supported horizontally by 2 vertical parallel strings at P and Q and at distances 2m and 6m from one end. Weights of 1kg, 1.5kg ,and 2kg are attatched at distances of 1m, 5m, and 7m respectively from the same end. Find the tension in each vertical string
Solve pls
To find the tension in each vertical string, we can analyze the forces acting on the rod and use the principles of equilibrium.
Let's consider the left side of the rod (from 0 to 2 meters) as Segment A and the right side (from 2 meters to 8 meters) as Segment B.
To maintain equilibrium, the sum of the clockwise torques (moment of forces) must be equal to the sum of the counterclockwise torques.
First, let's calculate the weight of the rod. The weight can be calculated using the formula:
Weight = Mass * Acceleration due to gravity
Weight = 5 kg * 9.8 m/s^2 (taking acceleration due to gravity as approximately 9.8 m/s^2)
Weight = 49 N
Next, let's calculate the torques due to the weights on Segment A:
For the 1 kg weight at 1 m:
Torque = Weight * Distance from the pivot point
Torque = 1 kg * 9.8 m/s^2 * 1 m = 9.8 Nm
For the 1.5 kg weight at 5 m:
Torque = Weight * Distance from the pivot point
Torque = 1.5 kg * 9.8 m/s^2 * 5 m = 73.5 Nm
Now, let's calculate the torques due to the weights on Segment B:
For the 2 kg weight at 7 m:
Torque = Weight * Distance from the pivot point
Torque = 2 kg * 9.8 m/s^2 * 7 m = 137.2 Nm
Now, we can calculate the total clockwise and counterclockwise torques:
Clockwise Torque = Torque of 1 kg weight + Torque of 1.5 kg weight = 9.8 Nm + 73.5 Nm = 83.3 Nm
Counterclockwise Torque = Torque of 2 kg weight = 137.2 Nm
Since the rod is in equilibrium, the sum of the clockwise torques must be equal to the sum of the counterclockwise torques. Therefore, we can set up the following equation:
83.3 Nm = 137.2 Nm + Tension at point Q * (8 m - 6 m)
Simplifying the equation, we get:
83.3 Nm - 137.2 Nm = 2 m * Tension at point Q
-53.9 Nm = 2 m * Tension at point Q
Tension at point Q = -53.9 Nm / 2 m
Tension at point Q = -26.95 N
The negative sign indicates that the tension at point Q is acting in the opposite direction (downwards) to the assumed positive direction. However, tension cannot be negative, so we take its magnitude as 26.95 N.
To find the tension at point P, we can use the fact that the sum of the vertical components of tension at P and Q must equal the weight of the rod.
Tension at point P + Tension at point Q = Weight of the rod
Tension at point P + 26.95 N = 49 N
Tension at point P = 49 N - 26.95 N
Tension at point P = 22.05 N
Therefore, the tension in the vertical string at point P is 22.05 N and at point Q is 26.95 N.