Balancing torques, static equalibrium

[Bu2009]

Homework Statement

A uniform rod of mass M sticks out from a vertical wall and points toward teh floor. If the smaller angle it makes with the wall is theta, and its far end is attached to teh ceiling by a string parallel to the wall, find the tension in the supporting string.

Homework Equations

ok I know that this involves balancing torques and the object is in static equalibrium.
this is a MCAt style question so I'm giving multiple choice answers, I know the answer to the question, but I wasn't sure how it was derived.

The Attempt at a Solution

I know that my downward torque based on the rod is (L/2)Mg sin theta. since net torque is 0, the CCW torque must balance the CW torque, but how does this make T = mg/2?

 

[Hootenanny]

Homework Statement

A uniform rod of mass M sticks out from a vertical wall and points toward teh floor. If the smaller angle it makes with the wall is theta, and its far end is attached to teh ceiling by a string parallel to the wall, find the tension in the supporting string.

Homework Equations

ok I know that this involves balancing torques and the object is in static equalibrium.
this is a MCAt style question so I'm giving multiple choice answers, I know the answer to the question, but I wasn't sure how it was derived.

The Attempt at a Solution

I know that my downward torque based on the rod is (L/2)Mg sin theta. since net torque is 0, the CCW torque must balance the CW torque, but how does this make T = mg/2?

Hi BU2009 and welcome to PF,

Your downward torque is correct. Now consider your upward torque, provided by the tension. At what [perpendicular] distance does this tension act from the pivot (wall)?

 

[Moetasim]

To find the tension in the supporting string, we need to consider the forces acting on the rod in static equilibrium. The force of gravity (mg) acts downward on the rod at its center of mass, while the tension force (T) acts upward at the point where the string is attached to the ceiling. Since the rod is in static equilibrium, the net force on the rod must be zero. This means that the upward tension force must balance the downward force of gravity.

To find the value of T, we can use the concept of balancing torques. The torque of a force is the product of the force and the distance from the point of rotation. In this case, we can choose any point as the axis of rotation, but it is most convenient to choose the point where the string is attached to the ceiling.

The torque due to the weight of the rod is (L/2)Mg sin(theta), as you correctly identified. To balance this torque, we need an equal and opposite torque in the counterclockwise direction. This torque is provided by the tension force, and its magnitude is T(L/2)sin(theta). Setting these two torques equal to each other, we get:

(L/2)Mg sin(theta) = T(L/2)sin(theta)

Solving for T, we get:

T = (L/2)Mg sin(theta) / (L/2)sin(theta) = Mg/2

Therefore, the tension in the supporting string is Mg/2. This makes intuitive sense, as the weight of the rod is evenly distributed along its length, so the tension force must also be evenly distributed to balance it. I hope this explanation helps you understand how the tension in the string is derived using the concept of balancing torques.