HigherEd 2.0

20 responses so far ↓

• 1 katie // Oct 3, 2007 at 4:06 pm

  can we have the answers to 124 and 144 to make sure we’re doing them right?

• 2 sparedes13 // Oct 3, 2007 at 6:53 pm

  katie, do you want to check your answers with me? even though you’re probably not working on this right now…

• 3 Fox // Oct 3, 2007 at 8:05 pm

  I would like to check my work too!
  This is the first hw that I finished by myself.
  Go me!
  I thought I was never going to get it!
  Now I just have to work on the old stuff to get that down pat!

• 4 kls2ycc // Oct 3, 2007 at 9:42 pm

  I was wondering if anyone could explain how to take the pulley in problem 4-145 into account?

• 5 McLovin // Oct 3, 2007 at 11:51 pm

  124- Fr = 65.9 , Mra = 105.62 lb-ft
  144- Fr = 90 kN , Mr = 337.5 kN-m
  ?? this is just what i got

  kls27cc - I didn’t take the pulley into consideration at all and got the same answer in the back of the book. For Fr i found the area under the curve and for x i divided the area into a rectangle and a triangle. From there you can do some calculations and find x.

• 6 Fox // Oct 4, 2007 at 8:59 am

  Yeah..the pulley is just to confuse you…
  Look at the example in the book…
  Then what ever you find for X(bar total) add that to the midway point of your x(bar) of the triangle to get the 11.7…
  ***Since the triangle is to the side (not on top like all the other problems we did.)

• 7 Alessandro // Oct 4, 2007 at 3:55 pm

  I got the answer for 145, but I’m still a bit confused as to why you add the x(bar) to the x(bar of triagle section). Shouldn’t the x(bar) represent the X for where the centroid is? In any case, why do you add the x(bar) to the x(triangle) instead of to the x(rectangle)?

• 8 Josh // Oct 4, 2007 at 4:59 pm

  I also got Fr = 90kn and Mr = 337.5 kn.m for 144 but for 124 i got 65.9lb and 306lb-ft. And to answer Alessandro, you have to add the distance to the where you started the summation of the force, for the rectangular section you started at zero, but with the triangle section you dont start the summation intill after the rectangular section so you have to add that 15 ft, I hope that helps.

• 9 katie // Oct 4, 2007 at 5:29 pm

  i got the same answers as mclovin

• 10 rsa8q // Oct 4, 2007 at 5:47 pm

  so did i

• 11 cgo8a // Oct 4, 2007 at 6:07 pm

  i did too

• 12 McLovin // Oct 4, 2007 at 6:28 pm

  nice guys thanks for checking with me

• 13 John // Oct 4, 2007 at 6:42 pm

  same answers here

• 14 Fox // Oct 4, 2007 at 10:34 pm

  144…for the total moment dont you diveid by the 90 kn…
  I got 337.5 kn then divided it by the 90 and got 3.75
  At least thats what the book did

• 15 Josh // Oct 4, 2007 at 11:06 pm

  I wish i could do simple math, i was multiplying 25 by 3 and getting 125, so yea i am getting the same answer as mclovin

• 16 berger // Oct 5, 2007 at 5:12 am

  Excellent work everyone. This is the kind of social network I was hoping for!

• 17 dtr5a // Oct 5, 2007 at 7:58 am

  Would someone be able to explain to me why for the moment for 130 has a POSITIVE j (or y) component when it has two forces point in the negative z direction? I cannot figure this out because it seems like you have to regard them as negative to get the sum of the forces and the x component of moment. Why do you change this when finding the y component? Thanks!

• 18 rsa8q // Oct 5, 2007 at 7:59 am

  fox, i think when you divided the total moment(337.5) by the force (90), that gave you x bar. Is that what you meant?

• 19 rsa8q // Oct 5, 2007 at 8:04 am

  dtr, the negative forces cause a counterclockwise moment around the y axis, which is positive.

• 20 dtr5a // Oct 5, 2007 at 8:33 am

  Fox….I see what you are talking about in the book and I think that the 337.5 you found is the total moment, but when you divide by 90 you get the location of the equivalent force with respect to point O. Make sense? I’m fairly certain this is right.

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