Answer to Question #215236 in Mechanics | Relativity for Jessy

Question

A uniform aluminum beam 9.00 m long, weighing 300 N, rests symmetrically on two supports 5.00 m apart (Fig. E11.12). 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 cm = 1.00 m horizontally. (b) From your diagram, how far beyond point B can the 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?

Answer 1

Apply static equilibrium conditions to the system. That is net torque acts on the system is zero.

(a) The boy, weight 600N, walks toward the right with a distance x from point A. The free-body diagram is shown below.

By taking positive clockwise torque about A as zero

$\sum τ_A = (600\;N)(x) +(300 \;N)(2.50\;m) -F_B(5.0\;m) \\ 0=600x+750N \cdot m -5F_B \\ 5F_B=600x + 750N \cdot m \\ F_B=120x+150N \cdot m$

And by taking positive clockwise torques about B

$\sum τ_B = F_A(5.0 \;m) -(300\;N)(2.50\;m) -(600\;N)(5.0\;m -x) \\ 0=5F_A -750\;N \cdot m -3000 \;N \cdot m + 600x \\ 5F_A = 3750 \;N \cdot m -600x \\ F_A = 750\;N \cdot m -120x$

(b) As the boy walks beyond the point B, then let force $F_A=0$ .

$F_A=750N \cdot m -120x \\ 0 = 750N \cdot m -120x$

Therefore the distance from point A

$x= \frac{750 \;N \cdot m}{120 \;N} = 6.25 \;m$

Thus it is 1.25 m beyond point B.

(c) The free-body diagram is shown below.

When the system is in equilibrium, the net torque acting on the system and the force acting on the system is zero.

Taking positive clockwise torques about point B

$\sum τ_B = (F_A)(7.0\;m)+(600\;N)(y) -(300\;N)(4.50 \;m -y)$

The point beyond point B can be determined by considering $F_A=0$ , then

$0=(0)(7.0\;m)+(600\;N)(y)-(300\;N)(4.50\;m -y) \\ 0= (600\;N)(y) -(300\;N)(4.50\;m -y) \\ 0= (600\;N)(y) -(300\;N)(4.50\;m) +(300\;N)(y) \\ 0 = (900\;N)(y) -(300\;N)(4.50\;m)$

Therefore the point is

$y=\frac{(300\;N)(4.50\;m)}{(900\;N)} = 1.50 \;m$

Thus point B is placed at a distance y = 1.50 m from the right end of the beam.