How Do You Solve Equilibrium Problems Using Moments?

  • #1
TheePhysicsStudent
19
16
Homework Statement
Hi, I have been doing some questions from the textbook and I got the 2 right answers, however for different forces
Relevant Equations
M = FD
Question:
1704455126260.png
My answer:
1704455314692.png

The books answer:
1704455358742.png
 
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  • #2
(1.2)(0.4)=(SF2)(0.7)
 
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  • #3
Wow, thanks sir! I realised I was highly overcomplicating things (as per usual)
 
  • #4
TheePhysicsStudent said:
Wow, thanks sir! I realised I was highly overcomplicating things (as per usual)
No you weren't. You still need another equation to find ##SF_1##. Your solution was actually very close. Your mistake was in the way you wrote your ratios. The ratio of the support distances is not proportional to those two lengths. It's inversely proportional. Think about two kids playing on a seesaw. The heavier kid needs to be closer to the pivot, not further away, to get the seesaw to balance.

Equivalently, you can balance the moments about the center of the metre rule.
 
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  • #5
TheePhysicsStudent said:
Wow, thanks sir! I realised I was highly overcomplicating things (as per usual)
Here are some thoughts so that you do not overcomplicate things next time you see something like this.

First, when you have an equilibrium situation, the sum of moments about any point will be zero. That's because the system doesn't know (and doesn't care) what point you choose as reference for the moments. It will not acquire angular acceleration about that point simply because you chose it.

Second, you have two unknowns, namely the two support forces ##SF_1## and ##SF_2.## This means that you need two equations to find them, i.e. you need to solve a system of two equations and two unknowns. The first equation is the sum of forces equal to zero and the second the sum of moments equal to zero. The procedure for tackling this is to solve one equation, say the force equation, for one unknown in terms of the other, e.g. ##SF_2=1.2~(\text{N})-SF_1##, substitute that in the moments equation and solve for ##SF_1##.

However, you can take a shortcut and choose as reference point for the moments the point at which one of the forces is applied. This gives you an equation with only one unknown moment which you can solve for the unknown force. That's exactly what @Chestermiller did in post #2 by choosing as reference the point where ##SF_1## is applied. Of course, the remaining force can be found by substituting in the sum of forces is zero equation.

To summarize, in static equilibrium problems the moment balance is simplified by choosing a reference point for moments where one (or more) forces act.
 
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  • #6
Mister T said:
No you weren't. You still need another equation to find ##SF_1##. Your solution was actually very close. Your mistake was in the way you wrote your ratios. The ratio of the support distances is not proportional to those two lengths. It's inversely proportional. Think about two kids playing on a seesaw. The heavier kid needs to be closer to the pivot, not further away, to get the seesaw to balance.

Equivalently, you can balance the moments about the center of the metre rule.
Thanks Mister T, the see saw analogy really helped make it more clear for me where I went wrong, as it does ake more sense now
 
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  • #7
kuruman said:
Here are some thoughts so that you do not overcomplicate things next time you see something like this.

First, when you have an equilibrium situation, the sum of moments about any point will be zero. That's because the system doesn't know (and doesn't care) what point you choose as reference for the moments. It will not acquire angular acceleration about that point simply because you chose it.

Second, you have two unknowns, namely the two support forces ##SF_1## and ##SF_2.## This means that you need two equations to find them, i.e. you need to solve a system of two equations and two unknowns. The first equation is the sum of forces equal to zero and the second the sum of moments equal to zero. The procedure for tackling this is to solve one equation, say the force equation, for one unknown in terms of the other, e.g. ##SF_2=1.2~(\text{N})-SF_1##, substitute that in the moments equation and solve for ##SF_1##.

However, you can take a shortcut and choose as reference point for the moments the point at which one of the forces is applied. This gives you an equation with only one unknown moment which you can solve for the unknown force. That's exactly what @Chestermiller did in post #2 by choosing as reference the point where ##SF_1## is applied. Of course, the remaining force can be found by substituting in the sum of forces is zero equation.

To summarize, in static equilibrium problems the moment balance is simplified by choosing a reference point for moments where one (or more) forces act.
Thank Kuruman, for the shortcut method which explained it more than the other user (though i did sort of grasp it), I am gonna practise more questions with all of this in Mind, Thanks once again
 
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What is a moment?

A moment is a measure of the tendency of a force to rotate an object about an axis. It is calculated as the force multiplied by the perpendicular distance from the axis of rotation to the line of action of the force.

How do you calculate moments?

To calculate a moment, you multiply the force applied to an object by the perpendicular distance from the axis of rotation to the line of action of the force. The formula for calculating a moment is Moment = Force x Distance.

What is the unit of measurement for moments?

The unit of measurement for moments is the Newton-meter (Nm) in the International System of Units (SI). In some cases, moments may also be measured in foot-pounds (ft-lb) or pound-feet (lb-ft).

What are some real-life examples of moments?

Some real-life examples of moments include opening a door, turning a wrench, or using a seesaw. In each of these cases, a force is applied at a distance from an axis of rotation, resulting in a moment that causes the object to rotate.

How can moments be used in engineering and physics?

Moments are crucial in engineering and physics for analyzing the stability of structures, designing machines, and understanding the behavior of objects under the influence of forces. Engineers and physicists use moments to calculate torques, moments of inertia, and other important parameters in various applications.

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