Force & Acceleration of a Ridgid Body

  • Thread starter joemama69
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  • #1
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Homework Statement


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The door has a weight of 200lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12ft to the right in 5 sec. starting from rest. Also, find the vertical reactions at the rollers A and B.


Homework Equations


F=ma


The Attempt at a Solution



200lb --> m = 6.21619 slug

F(x) = 6.21619a
F(y) = N(ab) - 6.21619(32.2) = 0, N(ab) = 200.161318 N

Moment @ G = F(2ft) - N(a)(6ft) + N(b)(6ft) = 0

N(a) + N(b) = 200.161318

Not sure where to go from here
 

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Answers and Replies

  • #2
PhanthomJay
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You forgot that the acceleration can be solved by the basic kinematic equation of motion from the givens: the box moves 12 feet in 5 seconds. Solve for the aceleration, then the rest looks OK (except please get rid of those decimals after the 200!).
 
  • #3
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Ok not sure which one to use. I'm confused about my variables.


heres what i tried

v = v(o) + at where v = 12/5, v(o) = 0, and t = 5, a = .48

s = s(o) + v(o)t + .5at where s = 12ft, s(o) = 0, v(o) = 0, t = 5, a = 4.8

v^2 = v(o)^2 + 2a(s - s(o)) where v = 12ft/5sec, v(o) = 0, s = 12ft, s(o) = 0, a = .24

Whats up with that

Also why not use the decimals. it is the force of gravity, ma
 
  • #4
PhanthomJay
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Ok not sure which one to use. I'm confused about my variables.


heres what i tried

v = v(o) + at where v = 12/5,
No, 12/5 is the average velocity, not the instantaneous velocity
s = s(o) + v(o)t + .5at^2 where s = 12ft, s(o) = 0, v(o) = 0, t = 5, a = 4.8
this is the right equation to use, but you forgot to square the 't' as noted in red , so you got the wrong answer
v^2 = v(o)^2 + 2a(s - s(o)) where v = 12ft/5sec, v(o) = 0, s = 12ft, s(o) = 0, a = .24
nahh, v is not 12/5
Whats up with that

Also why not use the decimals. it is the force of gravity, ma
The force of gravity, which is the objects weight, is given as 200 pounds. You complicated it .
 
  • #5
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ok

so my a = .96 ft/s^2

so then F = ma where m = 6.21619 slug and a = .96 ft/s^2
F = 5.97 slug ft/s^2

& the Vertical reaction on the rollers is 100lb each.

It does not add up though when u apply it to the moment at G
 
  • #6
PhanthomJay
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ok

so my a = .96 ft/s^2
OK
so then F = ma where m = 6.21619 slug and a = .96 ft/s^2
F = 5.97 slug ft/s^2
OK, but a slug ft/s^2 is called a 'pound' for short. F=5.97 pounds (call it 6 pounds)
& the Vertical reaction on the rollers is 100lb each.
No, that's not right. They are not equal. All you know right now is that the sum of the forces of each roller on the block must equal 200 pounds. Don't assume they are equal.
It does not add up though when u apply it to the moment at G
They should after you do the moment thing. I suggest, however, that you sum moments about one of the rollers instead of about G. That way, life will be a lot easier, because you'll be able to solve for one of the roller forces immediately, and the other will then come out of the previous 'sum of F_y = 0' equation.
 
  • #7
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Ok so i will get the moment about rolller a

5.97(9) + 12*F(b) - 200(6) = 0

F(b) = 95.5225
F(a) = 104.4775
 
  • #8
PhanthomJay
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Ok so i will get the moment about rolller a

5.97(9) + 12*F(b) - 200(6) = 0

F(b) = 95.5225
F(a) = 104.4775
Look good! But don't forget your units, and you should round off your values. I would say that F_a = 104 pounds (lb. or lbs.) and F_b = 96 lbs. And F = 6 lbs.
 
  • #9
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thanks for your help working through this sucker with me
 

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