A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a bas

Question

A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weights 200 Nm and is 6.00 m long, and it is suported by a wire at an angle of 60.0 degrees. The basket weighs 80.0 N

a) When the bear is at x=1.00 m find the tension in the wire supporting the beam and the compoents of the force exerted by the wall on the left end of the beam?

B) what 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

y^: Fy=0 = Fwy + T sin A - (M+m + mu)g

Fwy = (M+m + mu)g - T sinA

X: Fwx =T cos A

r: 0 = mgx+ MgL/2 + mu g L - Tsin A L

a) for x= 1

T= (mx+ML/2+ mu L)g/L sin A

= (700 + 200*3 + 80 * 6)/ 6 sin 60

= 343 N

Fwx = 343 cos 60 =171.5 N

Fwy = 683 N

b)T = 900

x = 1/mg(T sinA L - Mg L/2- mu g L)

substituitng the values gives

X= 5.14 m

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