Question

A hungry bear weighing 800 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. 50.0 Goodies The beam is uniform, weighs 240 N and is 6.00 m long; the basket weighs 76.0 N. (a) When the bear is at x = 1.40 m; find the tension in the wire and the components of the force exerted by the wall on the left end of the beam. (b) If the wire can withstand a maximum tension of 900 N, what is the maximum distance that the bear can walk before the wire breaks?

Answer 1

Sal Force on the beams are om Ty=TS inso Foreby Wall Tx Am Weight of bavet (wood Weight of bear two Weight of bean (W) 3m Since the beam is hot rotating » Nel tarane =o. Finding/ calculating tarane around told o. left side of beam 2. taking tarame downward as positive. : Wri (6) + Wall.a) +W(3) • Ty(6)=0 G606 + 800114) +24063) - Ty(6)=0 ² Ty 22.96 6 Ty 382.667 1 382.667 Sinso. Tension in the string (T) = 500 N

Also we can see that Jod is hat moving Unearly Wer force on horizontal direction =O EES x =-Tx DI fx 01 ST Ceso soocesso Ex 321 N x-component of force erolled by wall. Nor tarce in vertical direction. (haking down and es positive) Watwtwo - fy-Ty=0 C Fy oo+240 +76-382.687 Fy=733N. } y component of tarce exorted by Wall Again looking at the forces on bean Tm Tr so the بما 3a.

led when beare Works uplo a distance a tension in the string i Im (marimanteniend upto that point sr-o > Was twist Wolo) - Ty(6) > se -Imsinso) 6-3W - 6WB wa = (900x6) Sinso-3(240) - 6(76) к 3.70 Final answer is a) T = SOON, Fx=32IN, Fy=733N, 2-3.0m