A hungry bear weighing 740 N walks out on a beam in an attempt to retrieve a bas

Question

A hungry bear weighing 740 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 theta = 60.0 degree. The basket weighs 80.0 N. (a) Draw a free-body diagram for the beam. Choose File no file selected (b) When the bear is at x = 0.85 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. T = N F_x = N F_y = N (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

Given,

Weight of the bear, W = 740 N

Weight of the beam, w = 200 N

Length of the beam, L = 7 m

Weight of the basket, w' = 80 N

Distance of the bear, x = 0.85 m

Let us assume T be the tension in the string then,

The sum of the torque at the left end in counter clockwise direction is

l = 0, -(740 N) (0.85 m) - (200 N) (3.5 m ) - (80 N)(7 m) + (T sin60o) (7 m) = 0

T = 311.60 N

As the sum of forces acting on x direction is zero,

Fx = 0, (H - T cos 60o) = 0

Where H is the horizontal force on the hinge

H = 155.80 N

Similarly

Fy = 0

V - 740 N - 200 N - 80 N + 269.85 N = 0

V = 750.15 N

Let the maximum distance be dmax as the wire can withstand a maximum tension of 750 N

l = - (740 N)(dmax) - (200 N) (3.5 m) - (80 N) (7 m) + (750 N)sin60o(7 m) = 0

dmax = 4.44 m (>>1 m)

So the bear can go to the other end and get the basket

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