How can equations solve moments and equilibrium problems?

[OsiriS^]

Hello all. I'm not sure if this is the correct forum because I'm in the UK and I don't know what "Grade K12" is, but the sort of stuff in this forum seems to be the type of work we're doing.

We've just started some work on moments and equilibrium and I'm having difficulty solving the problems. I understand the theory work, such as the principle of moments, requirements for equilibrium and so on, I just need help in understanding why certain things solve the problems. Anyway, all is revealed below

I've had help from the lecturer and so I have the answers but I'm still a little confused...well a lot actually

Here are the two questions:

http://img.photobucket.com/albums/v216/rigdon/questions.jpg

And here are my answers:

http://img.photobucket.com/albums/v216/rigdon/question5.jpg
http://img.photobucket.com/albums/v216/rigdon/question6.jpg

So, for question 5, I understand that Moment = Force x Perpendicular Distance from pivot and that 100N x 3m will give the moment of the weight acting down. But I don't understand how writing the equation 100x3 = Rx8 would allow you to solve R when we are taking points about pivot 'A'. I get it mathematically i.e obviously 100x3 = Rx8 can easily find R, but I mean the moments/equilibrium part of it.

Now for question 6. How does 500x0.4 = T Sin12 x 0.47 allow you to find T? Again I don't mean mathematically, I mean how does it logically work out that that equation can find T...

I know my questions are weird and I would be eternally grateful if someone can make me understand these moments/equilibrium problems.

Thanks in advance

 

[Doc Al]

So, for question 5, I understand that Moment = Force x Perpendicular Distance from pivot and that 100N x 3m will give the moment of the weight acting down. But I don't understand how writing the equation 100x3 = Rx8 would allow you to solve R when we are taking points about pivot 'A'. I get it mathematically i.e obviously 100x3 = Rx8 can easily find R, but I mean the moments/equilibrium part of it.

I'm not sure what you are looking for, but here are a few comments. One of the conditions for equilibrium is that the moments taken about any point must sum to zero. Clockwise and counter-clockwise moments have opposite signs, of course, so another way of expressing the equilibrium condition is: clockwise moments must equal counter-clockwise moments (in magnitude).

One way to find the moment of a force about a point is this. (I'll call the moment by another name, torque.) [itex]\tau = F D_{perp}[/itex], or the force times the perpendicular distance that the force makes with the pivot point. Clockwise torques = 100x3; counter-clockwise torques = Rx8. So, for equilibrium: 100x3 = Rx8.

Now for question 6. How does 500x0.4 = T Sin12 x 0.47 allow you to find T? Again I don't mean mathematically, I mean how does it logically work out that that equation can find T...

The rule above for finding the torque or moment of a force is fine, but there is another mathematically equivalent rule: [itex]\tau = F D sin\theta[/itex], where D is the distance from the pivot point to where the force is applied, and [itex]\theta[/itex] is the angle between the force vector and the vector drawn from the pivot point to the point where the force is applied. (You should convince yourself that [itex]D sin\theta[/itex] is the same as the perpendicular distance that the force makes with the pivot point.)

Once again, the equilibrium condition is: clockwise moments must equal counter-clockwise moments (in magnitude).

 

[OsiriS^]

I'm not sure what you are looking for, but here are a few comments. One of the conditions for equilibrium is that the moments taken about any point must sum to zero. Clockwise and counter-clockwise moments have opposite signs, of course, so another way of expressing the equilibrium condition is: clockwise moments must equal counter-clockwise moments (in magnitude).

One way to find the moment of a force about a point is this. (I'll call the moment by another name, torque.) [itex]\tau = F D_{perp}[/itex], or the force times the perpendicular distance that the force makes with the pivot point. Clockwise torques = 100x3; counter-clockwise torques = Rx8. So, for equilibrium: 100x3 = Rx8.

With question 5 I don't understand how the force R creates a moment when the pivot is point A. The part I'm having difficulty with is identifying the clockwise and anti-clockwise moments on the pivot.

 

[Doc Al]

With question 5 I don't understand how the force R creates a moment when the pivot is point A. The part I'm having difficulty with is identifying the clockwise and anti-clockwise moments on the pivot.

Any force exerted at B will exert a torque about A, provided that its line of action does not pass directly through A. The force R, since it points to the left, surely exerts some moment tending to turn the ladder counter-clockwise about point A.

 

[OsiriS^]

Any force exerted at B will exert a torque about A, provided that its line of action does not pass directly through A. The force R, since it points to the left, surely exerts some moment tending to turn the ladder counter-clockwise about point A.

Ah right, now that makes sense. I'll be sure to apply it to the my future moments and equilibrium problems. I suppose if I look at the whole moments thing as a circle, and since the weight of the ladder was already creating a clockwise moment then there would have to be a force acting the other way in order to keep the ladder in equilibrium i.e R.

Thanks