A hungry bear weighing 750 N walks out on a beam in an attempt to retrieve a bas

Question

A hungry bear weighing 750 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 6.00 m long; the basket weighs 80.0 N.

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(b) When the bear is at x = 1.10 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) Answer is in N

(c) If the wire can withstand a maximum tension of 850 N, what is the maximum distance the bear can walk before the wire breaks?

Thanks in advance

Explanation / Answer

b) Write a torque balance around the point where the beam meets the wall:

(750N)*(1.10m) + (200N)*(3.0m) + (80N)*(6.0m) = T*sin(60)*(6.0m)

Solve for T:

T = 366.6N

Fy = 366.6*sin(60) = 317.5N

Fx = 366.6*cos(60) = 183.3N

c) Using the same torque balance equation, replace the 1.0 m with x, set T = 850N, and solve for x:

(750N)*x + (200N)*(3.0m) + (80N)*(6.0m) = (850N)*sin(60)*(6.0m)

x = 4.45m

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