# Tough moments question

1. Feb 17, 2014

### yugeci

Hello friends, I have this problem I've been stuck on for a while:

Here are a few relevant equations:

Moment = FxDy + FyDx
Fx = F cos θ
Fy = F sin θ
etc.

I can't even get past part (a). Resolving the 100 lb force is easy. It would be 100 sin 60 for the vertical component and - 100 cos 60 for the horizontal component. And the P component is a horizontal force, with no vertical component. The distance between P and C is 8". The distance between the horizontal component of the 100 lb force would be 8" too I guess (not sure), but what would be the vertical distance?

Part (b) doesn't look too hard once P is figured out.

I am not sure about part (c). Where will point A be if the moment is a maximum?

I don't understand part (d) either.

Help would greatly be appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 17, 2014

### BvU

Hello friend, a request here: make it a little easier on potential helpers and use the template. It also helps you to sort out things, check if you have the right equations (you do) and especially to be conscious of where you get stuck.

Your tackle the exercise in the right way. Your translation from mm to inch is unnecessary, so don't do it. Especially not if you have the wrong conversion factor ( should be 1 mm = 1/25.4 inch) . Same with N to lbf

The vertical distance is the distance between a line through the point where the force acts that is perpendicular to the axis of rotation, and the axis of rotation. Same way as with P, actually.

3. Feb 17, 2014

### CWatters

It says the moments sum to zero. The horizontal and vertical components don't necessarily sum to zero because you don't know the forces on the pivot point C.

My first reaction is to work out the torque caused by the 100N force about point C.

4. Feb 17, 2014

### BvU

Not sure what you are not sure about. You repeat the question; that is not an attempt at a solution.
By the time you can calculate the sum of moments around C you can also do so around a point D on the rim straight under C, right?
This helps you with c) and with d).
The fact that c) comes before d) suggests that you don't have to calculate this sum first to answer c). Some smart way to look at c) is indicated.

5. Feb 17, 2014

### BvU

Yes. One can sum the Fy times Dx (there is only one) and the Fx times Dy. The sum has to be zero.

6. Feb 17, 2014

### yugeci

OK, I'll use the template from next time. I managed to get solve part (a)

My = 100 sin 60 * 8
Mx = (- 100 cos 60 * 4) + (-8P)

Since they both anti-clockwise, My + Mx = 0

So I got P as 61.6N, and the resultant as 141.26N. Thank you guys for the help.

How will part (c) and (d) be solved?

Edit - just saw the 2 new posts, I'll try.

7. Feb 17, 2014

### yugeci

I get what you mean. But I don't get what approach to take for part (c). can think of..

My = 100 sin 60 * x
Mx = (-100 cos 60 * y) + (-P * y)

Where x and y are some values of horizontal distance and vertical distance that give a max moment M = Mx + My. Now what? We don't know what this max value is..

8. Feb 17, 2014

### BvU

Didn't find one yet, sorry. Thought about vector addition of P and 100 N and drawing a line perp to the sum. Comes close to answer. (addition introduces a small torque as well).

9. Feb 18, 2014

### yugeci

I saw the solution in the solution manual and they draw a perpendicular line from the resultant (attached the image). If so they used α to get the x and y values.. it actually makes sense to me since max torque happens when the distance is perpendicular. These questions are pretty insane and require so much thinking.

Edit - this is pretty much the same method you did BvU if I'm not mistaken.. where does the extra torque come from?

With that I think I will be able to do part (d) as well. Thanks for all of the help guys. :)

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10. Feb 18, 2014

### BvU

Well, this looks real good. Didn't think of moving F along its own line of action until F and P grab at the same point. Totally agree with the solutions manual.

I needed an extra torque because I moved P perpendicular to its axis to the x-axis, then to the point where the 100 N grabs and then added the two. This moving perpendicularly requires the introduction of an extra torque which spoiled an easy finding of the maximum moment point.

You indicate you understand the solution manual way makes sense, so you have learned something useful.
I have learned too: there had to be a shortcut easy way, and there is one. Just didn't find it. Next time (perhaps) both of us will do better.