How to find moment at a certain point?

[SagarPatil]

I have tired it multiple times but I keep getting 228, 12 ,-8N

How do in what direction the force is moving? clockwise or counter clockwise?

What to do with the 20 N in the y comp?

In the book it says if it moves clockwise its negative and if it moves counter clock wise its negative.

 

[DEvens]

Sorry dude. I can't read your question. Can you possibly type in the actual text? Or do a better picture?

As some background that may help. Or not. Torque is a (pseudo) vector. That means it has a magnitude and direction. You usually have a rule that tells you if the vector points "this way" then it is "clockwise." Usually it is called something like "the right hand rule" where you put your thumb in the direction of the vector and your fingers then point the way the torque will be turning the thing.

Torque from a force applied at a distance from a turning point is usually defined as a cross product. So it takes into account the distance from the centre, the magnitude of the force, and the relative angles of the distance from the centre and the force.

Torque will add as a vector.

 

[SagarPatil]

39*12/13=36N in the Y comp
39*5/13=15N in the X comp

60*4/5=48 in the Y comp
60*3/5=36 in the X comp

1 square = 1 Meter

What I did

-3(1) = 3(36) = -108N

2(1) = 2(36) = 72N
(1) = 1(48) = 48N

-108+72+48 = 12N

The question says: Determine the moment about point A for the force system.

I did it using this method : http://www.mathalino.com/reviewer/engineering-mechanics/226-moment-force-about-different-points

 

[SagarPatil]

Anyone?

 

[NascentOxygen]

The moment of a force about a point = force x perpendicular distance to the line of action of the force

There are 3 forces here. I was expecting to see among your calculations 20 x 4, attributable to the 20N force, but I don't. I haven't figured out what you've done. Maybe you can explain?

 

[SagarPatil]

I will explain one 60 Lb one, reset are the same.

So first I found how much force was going in the x component and the y component.

60*4/5=48 in the Y comp
60*3/5=36 in the X comp

In the Y comp, it is going 1 meter up, so I did 1*48 = 48N
Then in the X component, it is going 2 meters to the right, so I did 36*2 = 72N

So the 60N force has 48 N in the y-comp and 72N in the x component.

I did the same thing for the other componentsFor the 20N component I did (-20*3) + (20*4) = 20NWhen I add all the components it gives me 24N which is wrong.

I hope you understand.

 

[NascentOxygen]

I will explain one 60 Lb one, reset are the same.

So first I found how much force was going in the x component and the y component.

60*4/5=48 in the Y comp
60*3/5=36 in the X comp

In the Y comp, it is going 1 meter up, so I did 1*48 = 48N
Then in the X component, it is going 2 meters to the right, so I did 36*2 = 72N

The last two lines above are wrong, you are multiplying by the wrong distances. After you resolve the force into horizontal and vertical components, you should then multiply by the perpendicular distance to the line of action of each component. This means you multiply the x-component by the vertical distance to point A, and the y-component by the horizontal distance.

The result of this product has units of Newton metres.