A) 100 N
B) 250 N
C) 450 N
D) 1000 N
Correct Answer: C
Solution :
[c]For vertical equilibrium, \[{{T}_{1}}\sin {{45}^{o}}=W\] |
\[\therefore \] \[{{T}_{1}}=\frac{W}{\sin {{45}^{o}}}\] |
For horizontal equilibrium, |
\[{{T}_{2}}={{T}_{1}}\cos {{45}^{o}}=\frac{W}{\sin {{45}^{o}}}\cos {{45}^{o}}=W\] |
and for the critical condition, \[{{T}_{2}}=F\] |
\[\therefore \] \[W={{T}_{2}}=F=450\,N\] |
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