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The problem as stated in the book
"Calculate the tension [tex]F_{t}[/tex] in the wire that supports the 30kg beam shown in fig. 957 (attached), and the force [tex]F_{w}[/tex] exerted by the wall on the beam (give magnitude and direction)."
Getting the Tension in the string was easy.
[tex]\sum\tau = F_{ty} \cdot x_{1}  mg(\frac{x_{1}}{2}) = 0[/tex]
[tex]F_{ty} = 147N[/tex]
[tex]F_{t} = \frac{F_{ty}}{sin 50} = 1.9 \times 10^2N [/tex]
To get the [tex]F_{w}[/tex], I used the sum of forces.
[tex]\sum{F_{x}} = F_{tx}  F_{wx} = 0 \Rightarrow F_{tx} = F_{wx} = F_{t} \cdot cos 50 = 123.35 N[/tex]
[tex]\sum{F_{y}} = F_{ty} + F_{wy}  mg \Rightarrow F_{wy} = mg  F_{ty} = 102.11 N[/tex]
So now I have the two components for [tex]F_{w}[/tex] , I use pythagorous and solve for the resultant vector to get 160.13 N. The book, however, says the answer is [tex]1.9 \times 10^2 N[/tex]. Can anyone tell me what I'm doing wrong?
"Calculate the tension [tex]F_{t}[/tex] in the wire that supports the 30kg beam shown in fig. 957 (attached), and the force [tex]F_{w}[/tex] exerted by the wall on the beam (give magnitude and direction)."
Getting the Tension in the string was easy.
[tex]\sum\tau = F_{ty} \cdot x_{1}  mg(\frac{x_{1}}{2}) = 0[/tex]
[tex]F_{ty} = 147N[/tex]
[tex]F_{t} = \frac{F_{ty}}{sin 50} = 1.9 \times 10^2N [/tex]
To get the [tex]F_{w}[/tex], I used the sum of forces.
[tex]\sum{F_{x}} = F_{tx}  F_{wx} = 0 \Rightarrow F_{tx} = F_{wx} = F_{t} \cdot cos 50 = 123.35 N[/tex]
[tex]\sum{F_{y}} = F_{ty} + F_{wy}  mg \Rightarrow F_{wy} = mg  F_{ty} = 102.11 N[/tex]
So now I have the two components for [tex]F_{w}[/tex] , I use pythagorous and solve for the resultant vector to get 160.13 N. The book, however, says the answer is [tex]1.9 \times 10^2 N[/tex]. Can anyone tell me what I'm doing wrong?
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