Rotational static equilibrium

  • #1

Homework Statement


I have a question about rotational static equilibrium. When we consider the static equilibrium of a system, I realized that most people only consider the net torque acting on the system. Why dont we need to consider the net force acting on the system?

For instance:
A 1.2m-wide sign hangs from a 5kg, 2-m-long pole. The mass of cable is negligible, What is the mass of the sign if the maximum tension in the cable without breaking is 300N?

View attachment 30392

Homework Equations





The Attempt at a Solution


I had two thoughts,
1) The net vertical force acting on the system is zero.(Horizontal forces are ignored)
T_cable = 300
mass_sign = unknown
mass_pole = 5kg
Vertical forces add up to be zero:
angle between cable and pole = arctan(2.5/2)
300sin(arctan(2.5/2)) = (mass_sign)g + 5g <-------- This equation allows me to calculate the mass of the sign, but it does not involve torque. So, are there any mistakes in my thought??

second thought)
The pivot point is at the wall

The net torque acting on the system is zero.
angle between cable and pole = arctan(2.5/2)
T_cable = 300
mass_sign = unknown
mass_pole = 5kg
torque exerted by the cable = 300 x 2 x sin(pi - arctan(2.5/2))
torque exerted by the sign = 1.4 meter x mg
torque exerted by the pole = 1meter x 5g

Net torque is zero:
300 x 2 x sin(pi - arctan(2.5/2)) = 1.4 meter x mg + 1meter x 5g <------- The calculated number is too large, is there anything wrong with this thought?


Sorry, i think the attachment failed to work. If it does, this is the link :
http://www.2shared.com/photo/vxJwBdL2/digram.html
 
Last edited:

Answers and Replies

  • #2
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ok im no expert but i got the sign to be 30.4 kg. My class is in the same subject right now. so if you know the answer and it is somewhat similar to mines I can tell you how I did it. I don't want to tell you the wrong thing until the answer is confirmed.
 
  • #3
kuruman
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The Attempt at a Solution


I had two thoughts,
1) The net vertical force acting on the system is zero.(Horizontal forces are ignored)
You can't ignore the horizontal forces.
T_cable = 300
mass_sign = unknown
mass_pole = 5kg
Vertical forces add up to be zero:
angle between cable and pole = arctan(2.5/2)
300sin(arctan(2.5/2)) = (mass_sign)g + 5g <-------- This equation allows me to calculate the mass of the sign, but it does not involve torque. So, are there any mistakes in my thought??
Yes, there is a vertical force at the point of support on the wall. You seem to have ignored that too.

second thought)
The pivot point is at the wall

The net torque acting on the system is zero.
angle between cable and pole = arctan(2.5/2)
T_cable = 300
mass_sign = unknown
mass_pole = 5kg
torque exerted by the cable = 300 x 2 x sin(pi - arctan(2.5/2))
torque exerted by the sign = 1.4 meter x mg
torque exerted by the pole = 1meter x 5g

Net torque is zero:
300 x 2 x sin(pi - arctan(2.5/2)) = 1.4 meter x mg + 1meter x 5g <------- The calculated number is too large, is there anything wrong with this thought?
This is the correct way to go about it if you are not interested in the reaction forces at the wall. Perhaps you made a mistake in the arithmetical calculation. Can you show the details of your work?
 

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