A uniform rod 8m long weighing 5kg 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 thesame end. What is tension in each vertical string. Please help am confused.
P=36.25N,Q=50.75N
The tension of vertical string of P=59.17N
and that of vertical string of Q=15.83N
no no no p=32.92N and Q=9.16N rather
To find the tension in each vertical string, you need to consider the torque equilibrium of the rod.
Torque is the product of force and distance. For an object to be in rotational equilibrium, the sum of the torques acting on it must equal zero.
In this case, the torques acting on the rod are due to the weight of the rod itself and the weights attached to it.
Let's calculate the torque due to the weight of the rod. The weight of the rod can be considered as acting at its center of mass, which is at a distance of 4m from the end. The weight of the rod is 5kg, and the acceleration due to gravity is 9.8 m/s^2. Therefore, the torque due to the weight of the rod is:
Torque_rod = weight_rod * distance_rod
= 5kg * 9.8 m/s^2 * 4m
= 196 Nm
Now let's calculate the torques due to the weights attached to the rod. The torques due to the 1kg, 1.5kg, and 2kg weights are:
Torque_1kg = weight_1kg * distance_1kg
= 1kg * 9.8 m/s^2 * 1m
= 9.8 Nm
Torque_1.5kg = weight_1.5kg * distance_1.5kg
= 1.5kg * 9.8 m/s^2 * 5m
= 73.5 Nm
Torque_2kg = weight_2kg * distance_2kg
= 2kg * 9.8 m/s^2 * 7m
= 137.2 Nm
Now, based on the torque equilibrium, the sum of the torques acting clockwise must be equal to the sum of the torques acting counterclockwise. Since the rod is in rotational equilibrium, the torques acting on it must balance out.
The tension at point P will create a clockwise torque, and the tension at point Q will create a counterclockwise torque.
Therefore, we can set up the equation:
Tension_P * distance_P = Tension_Q * distance_Q (1)
Applying the torque equilibrium condition, we have:
Torque_1kg + Torque_1.5kg + Torque_2kg + Torque_rod = Tension_P * distance_P + Tension_Q * distance_Q (2)
Substituting the values we calculated, we have:
9.8 Nm + 73.5 Nm + 137.2 Nm + 196 Nm = Tension_P * 2m + Tension_Q * 6m
Simplifying the equation, we get:
416.5 Nm = 2m * Tension_P + 6m * Tension_Q
Since the two strings are vertical and parallel, the sum of the tensions in the strings is equal to the weight of the rod.
Therefore, we have:
Tension_P + Tension_Q = weight_rod
Tension_P + Tension_Q = 5kg * 9.8 m/s^2
Now we have a system of two equations:
416.5 Nm = 2m * Tension_P + 6m * Tension_Q
Tension_P + Tension_Q = 5kg * 9.8 m/s^2
Now you can solve these equations simultaneously to find the tensions in each vertical string.