Asked by Jamie
A uniform beam of length L = 7.80 m and weight 4.05 102 N is carried by two workers, Sam and Joe, as shown in the figure below. Determine the force that each person exerts on the beam. (Sam is 1m from one end of beam and joe is 2m from the opposite end.)
Answers
Answered by
Elena
Points:
left end - A, Sam - B, center of mass - C , Joe - D , right end - E
Torques about the point B:
mg•BC –F(J) •BD =0
F(J) =mg•BC/BD= 405•2.9/4.8 = 244.7(N)
Torques about the point D
F(S) •BD - mg•CD = 0
F(S)= mg•CD/BD = 405•1.9/4.8=160.3 N
left end - A, Sam - B, center of mass - C , Joe - D , right end - E
Torques about the point B:
mg•BC –F(J) •BD =0
F(J) =mg•BC/BD= 405•2.9/4.8 = 244.7(N)
Torques about the point D
F(S) •BD - mg•CD = 0
F(S)= mg•CD/BD = 405•1.9/4.8=160.3 N
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