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## 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

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