Problem 11.70 Part A In a city park a nonuniform wooden beam 8.00 m long is susp

Question

Problem 11.70 Part A In a city park a nonuniform wooden beam 8.00 m long is suspended horizontally by a light steel cable at each end. The cable at the left-hand end makes an angle of 30.0° with the vertical and has tension 660 N. The cable at the right-hand end of the beam makes an angle of 50.0° with the vertical. As an employee of the Parks and Recreation Department, you are asked to find the weight of the beam. Express your answer to two significant figures and include the appropriate units Value Units Submit My Answers Give Up Part B Find the location of its center of gravity. Express your answer to three significant figures and include the appropriate units. dValue Units from the left end of the beam Submit My Answers Give Up

Explanation / Answer

Horizontal component of T of the left end will be,

=660sin30

=330,

now this will be equal to the horizontal component of the tension at the right end .

So,

Tsin50=330

T=430.784N

Now, Vertical components of these two tensions will be equal to the weight so,

(430.784cos50)+(660cos30) = W

W=848.4796N

Let the center of mass be at x m away from the left end,

So now balancing the torque,

848.4796x=(430.784cos50)*8

x=2.6108m

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