Moment Calculations for a Machine

[ChicagoEagle8]

Hi, this may seem like a silly question to ask, but it has to do with +/- signing when taking the moment about a point. I have images attached to this post to make things a little clearer.

Homework Statement

The compound-lever pruning shears shown can be adjusted by placing pin A at various ratchet positions on blade ACE. Knowing that 1.5-kN vertical forces are required to complete the pruning of a small branch, determine the magnitude P of the forces that must be applied to the handles when the shears are adjusted as shown.

Homework Equations

(Moment about C): [tex]\sum[/tex]MC = 0

(Moment about D):[tex]\sum[/tex]MD = 0

The Attempt at a Solution

Using the FBD for the Blade ACE, I took the moment about C to find FAx and FAy, the horizontal and vertical components of FA, respectively:

[tex]\sum[/tex]MC = 0 = (32 mm)x(1.5 kN) + (-10 mm)x(FAx) + (-28 mm)x(FAy)
0 = 48 kN.mm - 10FAx - 28FAy

[tex]\rightarrow[/tex] The dimensions provided (11 mm vertically and 13 mm horizontally) allow me to solve for the length of member AB through the Pythagorean theorem.

c= [tex]\sqrt{a^2 + b^2}[/tex]
c=[tex]\sqrt{(13)^2 + (11)^2}[/tex]
c= 17.03 mm

Therefore, instead of using FA sin[tex]\theta[/tex] and FA cos[tex]\theta[/tex] for FAy and FAx, respectively, I can just use their geometric equivalents:

0 = 48 - 10([tex]\frac{13}{17.03}[/tex])FA - 28([tex]\frac{11}{17.03}[/tex])FA
0 = 48 - 7.63FA - 18.09FA
25.7FA = 48
FA = 1.86 kN
[tex]\rightarrow[/tex]Therefore, FAx = [tex]\frac{13}{17.03}[/tex](1.87 kN) = 1.43 kN
[tex]\rightarrow[/tex]And FAy = [tex]\frac{11}{17.03}[/tex](1.87 kN) = 1.21 kN

Now, based off the free body diagram for Handle ABD, in which member AB is a two-force member. With that in mind, I took the moment about D

[tex]\sum[/tex]MD = 0 = (-70 mm)x(P) + (5 mm)x(1.43 kN) + (-15 mm)x(-1.21 kN)
0 = -70P + 7.15 + 18.15
70P = 25.3
P = 0.361 kN

However, when I checked my answer on the instructor's site, I saw that P actually equalled 0.157 kN, and that the Moment about D was calculated like this:

[tex]\sum[/tex]MD = 0 = (-70 mm)x(P) - (5 mm)x(1.43 kN) + (15 mm)x(1.21 kN)
0 = -70P - 7.15 + 18.15
70P = 11
P = 0.157 kN

What I don't understand about this moment calculation is why we SUBTRACT the cross product of (5 mm)x(1.43 kN). Point B is located 5 mm above Point D, and up is positive in my frame of reference, and that the 1.43 kN force is moving toward the right, which is also positive in my my frame of reference. What reason is there to have a subtraction sign for that cross product, then?

Sorry that my post is really long, but I really appreciate the time that anyone takes to read, think about, and answer this question that I have.

 

[jbsteele]

Thank you.Your frame of reference is not the only frame of reference. Your instructor's frame of reference must be different from yours, so the sign for the cross product must also change. The direction of the force and the distance between B and D may be positive in your frame of reference, but it could be negative in the instructor's frame of reference. Therefore, the sign for the cross product must also change.

 

[thomate1]

Hi there,

First of all, there is no such thing as a silly question in science. Asking questions and seeking clarification is an important part of the scientific process.

In regards to your moment calculations, it is important to remember that the sign convention for moments is based on the right-hand rule. This means that moments are positive if they cause counterclockwise rotation and negative if they cause clockwise rotation.

In your first calculation, you took the moment about D and included the cross product of (5 mm)x(1.43 kN). This means that you are considering the moment caused by this force to be clockwise, and therefore it has a negative sign.

In the second calculation, you also took the moment about D, but this time you subtracted the cross product of (5 mm)x(1.43 kN). This means that you are considering the moment caused by this force to be counterclockwise, and therefore it has a positive sign.

So why did you get different answers for P? This is because the two calculations are considering different directions for the moment caused by the 1.43 kN force. In the first calculation, it is considered to be clockwise, while in the second calculation it is considered to be counterclockwise. This results in a different magnitude of moment and therefore a different value for P.

I hope this helps to clarify the +/- signing when taking moments. Keep asking questions and keep up the good work with your calculations!