Answer:
hello the required diagram is missing attached to the answer is the required diagram
7.9954 kip.ft
Explanation:
AB = 1550-Ib ( weight acting on AB )
BCD = 190 - Ib ( weight of cage )
169-Ib = weight of man inside cage
Attached is the free hand diagram of the question
calculate distance [tex]x![/tex]
= cos 75⁰ = [tex]\frac{x^!}{10ft}[/tex]
[tex]x! = 10 * cos 75^{o}[/tex] = 2.59 ft
calculate distance x
= cos 75⁰ = [tex]\frac{x}{30ft}[/tex]
x = 30 * cos 75⁰ = 7.765 ft
The resultant moment produced by all the weights about point A
∑ Ma = 0
Ma = 1550 * [tex]x![/tex] + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )
Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )
= 4014.5 + 1950.35 + 2030.535
= 7995.385 ft. Ib ≈ 7.9954 kip.ft
