A uniform aluminum beam 9.00 m long, weighing 300 N, rests symmetrically on two supports 5.00 m apart. A boy weighing 600 N starts at point A and walks toward the right. (a) In the same
diagram construct two graphs showing the upward forces FA and FB exerted on the beam at points A and B, as
functions of the coordinate x of the boy. Let 1 cm= 100 N vertically, and 1.00 cm= 100 m horizontally. (b) From
your diagram, how far beyond point B can boy walk before the beam tips? (c) How far from the right end of the
beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip?
diagram construct two graphs showing the upward forces FA and FB exerted on the beam at points A and B, as
functions of the coordinate x of the boy. Let 1 cm= 100 N vertically, and 1.00 cm= 100 m horizontally. (b) From
your diagram, how far beyond point B can boy walk before the beam tips? (c) How far from the right end of the
beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip?
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Assumption # 1: The 9m beam is center along the 5m-wide supports, i.e. support A is 2m from the left edge and support B is 2m from the right edge.
B) The easiest method to solve this is to understand that on the verge of tipping, the forces and torque at support A is zero, i.e. FA = 0, TA = 0
So we can use pt. B as the pivot point and determine x:
viewing counter-clockwise torque as negative:
T = 0 = -Wbeam * xbeam + Fboy * xboy
T = 0 = -300(7-4.5) + 600 * xboy
600 xboy = 750
xboy = 1.25 m (to the right of support B) OR
xboy = 8.25 m (from support A)
support B = 7.0 m (from support A)
C) Again, the easiest method to solve this is to understand that on the verge of tipping, the forces and torque at support A is zero, i.e. FA = 0, TA = 0
So we can use pt. B as the pivot point and determine x:
x = distance of pt B from the right edge.
Since the center-of-mass of the beam is 4.5m (middle of beam), its distance from pt B = (4.5 - x)
viewing counter-clockwise torque as negative:
T = 0 = -Wbeam * xbeam + Fboy * xboy
T = 0 = -300(4.5-x) + 600 * x
900x = 1350
xboy = 1.50 m (from the right edge) OR
xboy = 7.50 m (from support A)
"That boy shouldn't be walking around in a beam. He could fall! :-) "
.
.
B) The easiest method to solve this is to understand that on the verge of tipping, the forces and torque at support A is zero, i.e. FA = 0, TA = 0
So we can use pt. B as the pivot point and determine x:
viewing counter-clockwise torque as negative:
T = 0 = -Wbeam * xbeam + Fboy * xboy
T = 0 = -300(7-4.5) + 600 * xboy
600 xboy = 750
xboy = 1.25 m (to the right of support B) OR
xboy = 8.25 m (from support A)
support B = 7.0 m (from support A)
C) Again, the easiest method to solve this is to understand that on the verge of tipping, the forces and torque at support A is zero, i.e. FA = 0, TA = 0
So we can use pt. B as the pivot point and determine x:
x = distance of pt B from the right edge.
Since the center-of-mass of the beam is 4.5m (middle of beam), its distance from pt B = (4.5 - x)
viewing counter-clockwise torque as negative:
T = 0 = -Wbeam * xbeam + Fboy * xboy
T = 0 = -300(4.5-x) + 600 * x
900x = 1350
xboy = 1.50 m (from the right edge) OR
xboy = 7.50 m (from support A)
"That boy shouldn't be walking around in a beam. He could fall! :-) "
.
.