# Determine the moment of each of the three forces 8

Determine the moment of each of the three forces about point A. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Let us first draw the components of each force that is perpendicular to our reference point A. Components which pass through our reference will not be drawn as they do not create a moment about the said point. The dashed lines show each component of the force which creates a moment about point A.

Let us look at force $F_1$. In this force, only the y-component creates a moment about point A. The x-component of force $F_1$ passes through line of action of point A. Thus, the moment created by force $F_1$ is:

$\circlearrowright M_A=(2)(250\cos30^0)=433$ N$\cdot$m $\circlearrowright$

We will now look at force $F_2$. As with force $F_1$, only the y-component of force $F_2$ creates a moment about point A.

$\circlearrowright M_A=(5)(300\sin60^0)=1299$ N$\cdot$m $\circlearrowright$

Finally, let us look at force $F_3$. Notice how the x and y-components of force $F_3$ creates moments about point A. The x-component has a perpendicular distance of 4 m, and the y-component has a perpendicular distance of 5 m. It’s also important to note that the x-component will create a negative moment, or rather an opposite moment than the other forces because it will spin the bent arm counter clockwise about point A.

$\circlearrowright M_A=(500)(\dfrac{4}{5})(5)-(500)(\dfrac{3}{5})(4)=800$ N$\cdot$m $\circlearrowright$

$F_1$:$\circlearrowright M_A=433$ N$\cdot$m $\circlearrowright$

$F_2$:$\circlearrowright M_A=1299$ N$\cdot$$\circlearrowright$

$F_3$:$\circlearrowright M_A=800$ N$\cdot$$\circlearrowright$

## 8 thoughts on “Determine the moment of each of the three forces”

• Luis

Why is it that for F1, you used cos to find Fy and not sin? Also for the corner of F2, why does that become sin to find the Fy?﻿

• questionsolutions Post author

Hi,

If we look at force F1, we notice that only the y-component creates a moment about A. Remember that moment is found by multiplying the perpendicular distance times the force. In this case, the perpendicular force is the y-component of force F1 and the distance is 2 m. We don’t consider the x-component because it goes directly through our line of action at point A. The easiest way to calculate the y-component of force F1 is to use cosine, because it’s opposite over hypotenuse. You can also use sin, but then you would need to use 60° (90°-30° = 60°). So, you end up getting: MA=(2)(250 sin(60°))= 433 N•m (which is the same answer as before).

The same applies for force F2. We are using the given angles in the simplest manner possible, you can use sine for both, or cosine for both, the only difference is, you have to find the angle that applies to each case. For force F2, we are given an angle of 60° however, to use cosine, we need to subtract (90°-60°=30°) and use the 30° to find the answer. So if we use cosine, we get: MA=(5)(300 cos(30°))= 1299 N•m.

Hope that helps and thanks for the question 🙂 If this doesn’t clear it up, let us know, we will do our best to help you out.

• Luis

So the perpendicular distance determines if we use cosine or sine?

• questionsolutions Post author

So when we need to figure out the moment, we always consider the perpendicular distance, but that doesn’t determine whether we use cosine or sine. That is chosen by you. Usually, you will notice that sine or cosine is easier to use depending on the angle given. Again, let’s take a look at force F1 in the original diagram given to us. We know to find the moment at A, we need to figure out the y-component of force F1. To figure it out, we look at the angle given, and we see that it’s 30°. If we use sine 30°, we would get the x-component, because sine is opposite (x-component) over hypotenuse. So, to use the 30°, we have to use cosine to get the adjacent (y-component). Cosine is adjacent over hypotenuse. Seeing this, it’s much easier to use cosine, instead of sine. Hopefully that makes sense. It all depends on what is easier to use when doing the question, depending on how the angle is given.

To answer your question, it’s more of how the angle is given to us, that determines whether we use sine or cosine. Note that both can be used, you just need to figure out the corresponding angle value to use whichever you prefer.

https://goo.gl/f3CsWg

• Luis

Thank you for the quick reply!

• questionsolutions Post author

You are very welcome! 🙂

I think you might be having a bit of trouble trying to visualize the components of each force. Hopefully, the video on trigonometry will help clear up the confusion, and it will get a lot easier the more questions you do. Also, it might help to look at the first few chapters on how to break up forces into components and using sine, cosine, and tangent, to figure them out. Best of luck with your studies!

If you still have trouble getting it, leave us another comment, and we will try a different example, maybe one focusing on breaking forces up into components and using sine/cosine to figure it out.

• Abdifitaah Mohamed

Thanks for Help ! You saved My timeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

• questionsolutions Post author

You’re very welcome! 🙂