Static Equilibrium Problem Dealing with Torque

[PeachBanana]

Homework Statement

The figure shows a model of a crane that may be mounted on a truck. A rigid uniform horizontal bar of mass = 90.00 kg and length = 5.200 m is supported by two vertical massless strings. String A is attached at a distance = 1.600 m from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass = 3000 kg is supported by the crane at a distance = 5.000 kg from the left end of the bar.

Throughout this problem, positive torque is counterclockwise and use 9.807 m/s^2 for the magnitude of the acceleration due to gravity.


Find the tension in string A.

Homework Equations

This problem let's you use hints and it suggested I use the following equation with having the Tension in String "B" as the origin.

The Attempt at a Solution

Tension A(1.6000m) - (90.00 kg)(9.807 m/s^2) 1/2(5.200 m) - (3000 kg)(9.807 m/s^2)(5.000 m) = 0

I don't understand the "1/2" part in front of the length mass one. Is that because of where it is placed or its moment of inertia?

 

[PhanthomJay]

If the weight of the 90 kg beam is uniformly distributed over its length, where does its resultant load act?

 

[PeachBanana]

I'm sorry, I'm not sure what you mean by resultant load.

 

[PhanthomJay]

I'm sorry, I'm not sure what you mean by resultant load.

If a uniform beam that is 2 m long has a mass of 0.5 kg/m, its total mass is 1 kg and it's resultant or total weight is 9.8 N. It can be represented as a single force of 9.8 N acting at its center of gravity.

 

[asteriatic]

The "1/2" in front of the length and mass in the equation is due to the fact that the torque equation for a point mass rotating about a fixed axis is given by τ = rFsinθ, where r is the distance from the axis of rotation to the point mass, F is the force acting on the point mass, and θ is the angle between the force and the lever arm (r). In this problem, since the bar is a rigid uniform object, we can consider it as a collection of point masses, each with a different distance from the axis of rotation (the left end of the bar). Therefore, we need to take into account the distribution of mass along the length of the bar by using the moment of inertia, which is given by I = 1/2mr^2. This is why the "1/2" is included in the equation. It represents the distribution of mass along the length of the bar.