A hungry bear weighing 705 N walks out on a beam in an attempt to retrieve a bas

Question

A hungry bear weighing 705 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam (see figure below). The beam is uniform, weighs 200 N, and is 5.50 m long, and it is supported by a wire at an angle of = 60.0°. The basket weighs 80.0 N.

(b) When the bear is at x = 0.80 m, find the tension in the wire supporting the beam and the components of the force exerted by the wall on the left end of the beam.

(c) If the wire can withstand a maximum tension of 750 N, what is the maximum distance the bear can walk before the wire breaks?
m

Explanation / Answer

B)

Fx = Tcos(60)

Fy + Tsin(60) = 705 + 200 + 80

T*sin(60)*5.5 = 705*.8 + 200*2.75 + 80*5.5 (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=326.25 N

Fx=163.12

Fy=702.45

C)

To find the max distance, solve the moment equation again substituting 750 for T and:

T*sin(60)*5.5 = 705*x + 200*2.75 + 80*5.5

750*5.5*sin(60) = 705*x + 200*2.75 + 80*5.5

3572.35 = 705*x + 990

x = 3.66 m

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