FarazAli
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The problem as stated in the book
"Calculate the tension F_{t} in the wire that supports the 30-kg beam shown in fig. 9-57 (attached), and the force F_{w} exerted by the wall on the beam (give magnitude and direction)."
Getting the Tension in the string was easy.
\sum\tau = F_{ty} \cdot x_{1} - mg(\frac{x_{1}}{2}) = 0
F_{ty} = 147N
F_{t} = \frac{F_{ty}}{sin 50} = 1.9 \times 10^2N
To get the F_{w}, I used the sum of forces.
\sum{F_{x}} = F_{tx} - F_{wx} = 0 \Rightarrow F_{tx} = F_{wx} = F_{t} \cdot cos 50 = 123.35 N
\sum{F_{y}} = F_{ty} + F_{wy} - mg \Rightarrow F_{wy} = mg - F_{ty} = 102.11 N
So now I have the two components for F_{w} , I use pythagorous and solve for the resultant vector to get 160.13 N. The book, however, says the answer is 1.9 \times 10^2 N. Can anyone tell me what I'm doing wrong?
"Calculate the tension F_{t} in the wire that supports the 30-kg beam shown in fig. 9-57 (attached), and the force F_{w} exerted by the wall on the beam (give magnitude and direction)."
Getting the Tension in the string was easy.
\sum\tau = F_{ty} \cdot x_{1} - mg(\frac{x_{1}}{2}) = 0
F_{ty} = 147N
F_{t} = \frac{F_{ty}}{sin 50} = 1.9 \times 10^2N
To get the F_{w}, I used the sum of forces.
\sum{F_{x}} = F_{tx} - F_{wx} = 0 \Rightarrow F_{tx} = F_{wx} = F_{t} \cdot cos 50 = 123.35 N
\sum{F_{y}} = F_{ty} + F_{wy} - mg \Rightarrow F_{wy} = mg - F_{ty} = 102.11 N
So now I have the two components for F_{w} , I use pythagorous and solve for the resultant vector to get 160.13 N. The book, however, says the answer is 1.9 \times 10^2 N. Can anyone tell me what I'm doing wrong?