A uniform rod 8m long weighing 8kg is supported horizontally by two vertical parallel strings at P and Q and at distances of 2m and 6m from one end. Weights of 1kg,1.5kg and 2kg are attached at distances of 1m,5m and 7m respectively from the same end. Find the tension in each vertical strings.
total force down = 1 g + 8 g + 1.5 g + 2 g = 12.5 g
total force up = p+q
so p + q = 12.5 g
moments about left end
clockwise = 1*1 g + 8*4 g + 1.5*5 g + 2*7 g = 54.5 g
counterclockwise = 2 p + 6 q
so 2 p + 6 q = 54.5 g
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p + q = 12.5 g
p + 3 q = 27.25 g
-----------------------subtract
-2 q = -14.75 g
q = 7.375 g
p = 5.125 g
use whatever you use for g to get Newtons instead of kilograms 9.8 m/s^2 or 9.81 m/s^2 or 10 m/s^2
To find the tension in each vertical string, we need to consider the torque equilibrium of the rod.
Torque is the measure of a force's tendency to rotate an object around a particular axis. For an object to be in rotational equilibrium, the sum of the torques acting on it must be zero.
Here's how we can solve this problem step by step:
Step 1: Calculate the total weight of the rod.
The total weight is given by weight = mass × acceleration due to gravity.
The mass of the rod is given as 8 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
Therefore, the total weight of the rod is weight = 8 kg × 9.8 m/s^2 = 78.4 N.
Step 2: Calculate the torque due to the rod's weight.
The torque due to the weight of the rod acts at the center of the rod, and its value is weight × distance.
In this case, the rod is 8 m long, so the weight's distance from the point of rotation is 4 m.
Therefore, the torque due to the rod's weight is torque = 78.4 N × 4 m = 313.6 Nm.
Step 3: Calculate the torques due to the weights hanging from the rod.
The torque due to a weight hanging from the rod is weight × distance.
For the 1 kg weight hanging at 1 m from the same end, the torque is 1 kg × 9.8 m/s^2 × 1 m = 9.8 Nm.
For the 1.5 kg weight hanging at 5 m from the same end, the torque is 1.5 kg × 9.8 m/s^2 × 5 m = 73.5 Nm.
For the 2 kg weight hanging at 7 m from the same end, the torque is 2 kg × 9.8 m/s^2 × 7 m = 137.2 Nm.
Step 4: Calculate the tension in each vertical string.
Let's assume the tensions in the strings at P and Q are TP and TQ, respectively.
To maintain rotational equilibrium, the sum of the torques acting on the rod must be zero.
Therefore, TP × distance_TP - TQ × distance_TQ = torque_due_to_weights.
Using the given distances, we have:
TP × 2 m - TQ × 6 m = (9.8 Nm + 73.5 Nm + 137.2 Nm) - 313.6 Nm.
Simplifying this equation, we get:
2TP - 6TQ = -93.1 Nm.
Now let's use another equation to solve for TP and TQ.
Since the rod is in horizontal equilibrium, the sum of the vertical forces must be zero.
Therefore, TP + TQ = weight_of_the_rod.
Substituting the values, we have:
TP + TQ = 78.4 N.
Now, we have two equations with two variables:
2TP - 6TQ = -93.1 Nm,
TP + TQ = 78.4 N.
Solving these equations simultaneously will give us the tension in each vertical string.