Got the answer.TheunforgivenII wrote:
I have one more problem to ask for help with. The answer in the back of the book says F = 115N but I can't get that number.
In order to hold the wheelbarrow in the position shown, force F must produce a counterclockwise moment of 200 N·m about the axle at A. Determine the required magnitude of force F.
http://img171.imageshack.us/img171/6395/problem434.jpg
okay that's where I messed up, I didn't use -200 because I was just thinking about counterclockwise only not also clockwise........mcminty wrote:
http://farm4.static.flickr.com/3514/389 … 3f547d.jpg
http://farm3.static.flickr.com/2451/389 … 7ab7f3.jpg
You just need to break the force down into its horizontal and vertical components (it's why they gave you the angle). It tells you to produce a 200N.m moment - so that means there must be a 200N.m clockwise moment as well (which is generated by the wheelbarrow's weight). That is where the -200 comes from the in the equilibrium equation.
Last edited by TheunforgivenII (2009-09-06 16:59:29)
No worriesTheunforgivenII wrote:
okay that's where I messed up, I didn't use -200 because I was thinking about counterclockwise not clockwise........mcminty wrote:
http://farm4.static.flickr.com/3514/389 … 3f547d.jpg
http://farm3.static.flickr.com/2451/389 … 7ab7f3.jpg
You just need to break the force down into its horizontal and vertical components (it's why they gave you the angle). It tells you to produce a 200N.m moment - so that means there must be a 200N.m clockwise moment as well (which is generated by the wheelbarrow's weight). That is where the -200 comes from the in the equilibrium equation.
Thanks mcminty!
okay I really need help with this one now. I came up with 188.22 N·m but a friend of mine said that the answer is 200 N·m so I tried to work it out again but can't get 200 N·m. so i need help once again.TheunforgivenII wrote:
I have yet another question on a problem
http://img16.imageshack.us/img16/2417/problem460.jpg
rob = {(0.5-0)i + (2-0)k}m => {0.5i + 2k}m
roa = {(0-0)i + (1cos15°-0)j + (1sin15°-0)k}m => {.966j + .259k}m
so my question is did I pick the right points or rob and roa?
Chapter 4WldctARCHe wrote:
What chapter of Hibbeler are you in?
Last edited by TheunforgivenII (2009-09-10 21:31:30)
Determine the magnitude of the moment produced by the force of F = 200 N about the hinged axis (the x axis) of the door.WldctARCHe wrote:
So you are trying to solve for the moment of Force about the x-axis?
Okay I decided to try this problem again on my own but still couldn't get the right answer. The answer in the back of my book says the answer is (Mc)R = 260 lb·ft clockwise. Though I got 640 lb·ft clockwise.......this is what I tried to do: (-200(4/5)(2)) - (150cos30°)(4) - (200(4/5)(2)) + (150cos30°)(4). What am I doing wrong on this problem?TheunforgivenII wrote:
Sorry to ask yet another problem that I don't know how to do......I don't understand this new homework assignment that has been assigned to me which is about moment of a couple. So I need some help again...
If F = 200 lb, determine the resultant couple moment.
http://img143.imageshack.us/img143/3941/problem475.jpg
you're not helping and I really need help on this problem. I've Been working on this problem for hours without any progress.........CammRobb wrote:
42
Last edited by TheunforgivenII (2009-09-12 20:08:25)
thanks again WldctARCHe.....sorry to bug you again to help me with this stuffWldctARCHe wrote:
-200(4/5)*2 + 150sin(30)*4 - [200(3/5)*2] + 150cos(30)*4
Should be the equation you use. Your 1st term was correct, 2nd term you had was actually calculating the horizontal component of the force which won't actually produce a moment but the vertical force will (i.e. sin vs. cos) and will be in the positive direction. 3rd term you calculated vertical force which again won't produce a moment but the horizontal will (i.e (3/5) vs (4/5) and the 4th term you had correct. This solves out to 259.62 lb ft
