A hungry bear weighing 710 N walks out on a beam in an attempt to retrieve a bas

Question

A hungry bear weighing 710 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 7.00 m long, and it is supported by a wire at an angle of alpha = 60.0 degree. The basket weighs 80.0 N. Draw a free-body diagram for the beam. When the bear is at x = 1.10 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. If the wire can withstand a maximum tension of 825 N, what is the maximum distance the bear can walk before the wire breaks?

Explanation / Answer

2. Force balance on the beam
Tcos(theta) = R
Tsin(theta) + N = W + Mg + mg
Tsin(60) + N = 80 + 200 + 710 = 990

Using moment balance about the hinge
Tsin(60)*l - W*l - Mg*l/2 - mg*x = 0
Tsin(60)*7 - 80*7 - 200*3.5 - 710*1.1 = 0
T = 336.677 N
Fx = R = Tcos(60) = 168.3388 N
Fy = N = 990 - Tsin(60) = 698.429 N

Let max distance be x
Tsin(60)*l - W*l - Mg*l/2 - mg*x = 0
825*sin(60)*7 - 80*7 - 200*3.5 - 710*x = 0
x = 5.269 m

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