Question on moments along an axis HW problem

[Noesis]

Well...I am able to work the problem just fine on the x' and y' axes using the standard vector analysis of r X F, I just have no idea how to transfer it to the x and y axes.

I provided a picture of the actual problem, alongside the actual answer in the solutions manual.

I don't understand how they get the -sin30 for translating the moment about the x' axis to the y direction...and why that is even done.

Shouldn't we just take the y' moment and cos30 it to get the moment about the y axis?

Those are my thoughts...if anyone can clarify I'll wash their car.

 

[Valkarie]

The answer in the solutions manual is correct, but it's not necessarily the only way to solve the problem. To answer your question, one way to solve the problem is by using the cross product of the two vectors. We can use the cross product to first calculate the moment about the x' and y' axes, then use the equation Mr = Mx'cosθ + My'sinθ to calculate the moment about the x and y axes. In this case, since θ = 30°, we have Mr = Mx'cos30 + My'sin30. This equation tells us that the moment about the x axis is equal to the moment about the x' axis times cos30, plus the moment about the y' axis times sin30. So, when we calculate the moment about the y axis, it will be equal to the moment about the x' axis times -sin30, plus the moment about the y' axis times cos30.

 

[piercas]

I can understand your confusion and frustration with this problem. Moments along an axis can be tricky to calculate, but with the right approach, it can be solved easily. Let me try to break it down for you.

First, it's important to understand that moments are not just about the magnitude of the force, but also the direction in which it acts. In this case, we are dealing with moments along the x and y axes, which are perpendicular to each other. This means that the force acting on the x' axis will have a different effect on the y axis compared to the x axis.

To transfer the moment from the x' axis to the y axis, we need to use the cross product formula, r x F. This formula takes into account the magnitude and direction of the force as well as the distance from the point of rotation. In this case, the distance from the point of rotation to the force is given by the y' component, which is why we use -sin30 in the calculation.

As for your question about using cos30 to get the moment about the y axis, keep in mind that this only works if the force is acting directly along the y axis. In this problem, the force is acting at an angle, so we need to take into account the component of the force that is acting perpendicular to the y axis, which is where the -sin30 comes into play.

I hope this helps clarify the process for you. Remember, moments along an axis can be challenging, but with practice and a solid understanding of the principles involved, you will be able to solve them with ease. And don't worry about washing my car, just keep practicing and you'll get the hang of it!