A uniform, aluminum beam 9.00 mm long, weighting 300 nn, rests symmetrically on two supports 5.00 mm apart. a boy weighing 600 nn starts at point aa and walks toward the right.

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26-08-2022
Physics

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A uniform, aluminum beam 9.00 mm long, weighting 300 nn, rests symmetrically on two supports 5.00 mm apart. a boy weighing 600 nn starts at point aa and walks toward the right.

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tripatsingh039 tripatsingh039 · Answer 1 · 2022-09-02T05:21:18-04:00

Set up the system under static equilibrium conditions. That is, there is no net torque acting on the system.

(a) The youngster, who weighs 600 N, moves x steps away from point A in a direction toward the right. Below is a diagram of a free body.By assuming zero positive clockwise torque about A

A =(600N)(x)+(300N)(2.50m)-F B(5.0m) 0=600x+750N cdot m 5F B 5F B=600x + 750N cm F B=120x+150N cm

0=600x+750N⋅m−5F B 5F B = 600x+ 750N m F B = 120x + 150N m

Additionally, by applying positive clockwise torques to B

sum B = F A(5.0 m) -(300 m)(2.50 m) -(600 m)(5.0 m -x) 5F A = 3750; N cm -600x; F A = 750; N cm -120x; 0=5F A -750; N cm -3000; N cm + 600x; B = F A (5.0m), (300N), (2.5m), (600N), (5.0m x), 0 = 5F A 750N m 3000N m + 600x 5F A = 3750N m 600x F A = 750N m 120x

Complete question is:

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

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