A hungry bear weighing 680 N walks out on a beam in an attempt to retrieve a bas

Question

A hungry bear weighing 680 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 5.40 m long; the basket weighs 80.0 N.

(a) Draw a free-body diagram for the beam. (Do this on paper. Your instructor may ask you to turn in this sketch.)

(b) When the bear is at x = 0.80 m, find the tension in the wire and the components of the force exerted by the wall on the left end of the beam. (tension) ( Fx) ( Fy) (c) If the wire can withstand a maximum tension of 900 N, what is the maximum distance the bear can walk before the wire breaks?

Explanation / Answer

Fx = Tcos(60)
Fy + Tsin(60) = 680+ 200 + 80
Tsin(60)*5.5 = 750*.80 + 200*2.70 + 80*5.4(sum of the moments about the left end of beam)

solving these three equations for T,Fx, and Fy (in that order) you get
T=353.87 Fx=147.17, Fy=710.75
to find the max distance, solve the moment equation again substituting 680 for T and x for .80
x=2.502m

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