1. Jul 9, 2007

### pokmiuhy

ok...so i got some review questions for an upcoming test, and i'm lost on this one. i have no clue where to begin..

any help is good!!! thank you!!

2. Jul 9, 2007

### pokmiuhy

i just need an idea of where to start... i thought about taking moments about points, then i got really confused and started over.

and now i'm just staring at the paper, hoping, that by osmosis, it will complete itself. then i can learn it

3. Jul 9, 2007

4. Jul 9, 2007

### K.J.Healey

I'm not completely sure, but at a glance I'd say yes, sum the moments about certain points.

Lets take the top one, we have a 500KG mass at a R of 1100mm from the top left pivot point, right? Thats a moment of (500kg)(9.81m/s^2)(1100mm) = 5,395,500 kg*m*mm/s^2.

Then we look at that one point(B) that will have a piston holding the entire weight. Well that moment is at a distance of 350mm, so take yuour 5.4million and divide by 350mm to get a 15.5 KiloNewton Force to the left due to the weight of the arm about point A.

The second piston is more complicated, but i think its the same idea. Nothing is moving so you can find forces by figuring out the moments about other pieces.

5. Jul 9, 2007

### K.J.Healey

For the second E and I are attached together, so the arm must rotate about point E with a force acting on H. We must first figure out what this is before even looking at J.

Draw a FBD for forces on H about E.
Id say its going to be somethign like WSin(30)*distanceR, which R is something.

I feel like this piece is mislabeled, and you cant technically figure out the length of E-J bar without making assumptions due to the left scales being un-attached to the right. You can't tell where point E is in reference to the right scales.

6. Jul 9, 2007

### pokmiuhy

ok. so if you use the moment about A to calculate the force at I?

what if i sum the forces in the x and y for point I? would that work?

7. Jul 9, 2007

### pokmiuhy

well that's why i was confused about IJ. parts of the drawing you have to make assumptions on to get measurements.

i am guess that the vertical distance between H and E is 200, as H and C seem to be inline with each other.

so if i use that to figure out the forces on H and E, then from there I can get I can't i?

8. Jul 9, 2007

### K.J.Healey

Well you find the force on H about E ( which is a moment) then you use that to find the moment, or Force on I about E. Its almost like unit conversion.
Cause you have, where "M" will mean moment, and R means distance from.

M_HE = M_IE
(Force_H * R_HE) = (Force_I * R_IE)

Solve for Force_I. Thats what its asking.
Figure out the R's. Remember Force = Mass*Accel, so 500KJ*9.81m/s^2

The most important part is figuring out the Forces. Because its not just 10KN downward at any point. Its the moment we're worried about, which is a cross product. You need to find the tangential force component to the angled bar. so the "force" and the "R" are at 90 degrees to eachother. This i "think" would be a mgSin(Theta) for the second piston part. This is due to the given angle of 30 degrees. so mgsin(30). If that angle 30 went to zero, and the mass pulled on the rod up and down, the moment would be zero right? And if the angle went to 90 and the mass force was at a 90 degree angle to the part, then moment would be maximum right? Well, we MUST then be talking about a Sin, since Sin(0) = 0 and Sin(90) = 1(max).

Does that help at all?

9. Jul 9, 2007

### pokmiuhy

yes...thanks a lot!!!

if i have more questions after i finish working it, i'll ask.

thanks!