A) 25 N
B) 30 N
C) 35 N
D) 40 N
Correct Answer: A
Solution :
The frictional force\[{{F}_{S}}\]on block A, when it tends to move is\[{{F}_{S}}={{\mu }_{S}}R={{\mu }_{S}}{{W}_{A}}=0.25\times 100=25\,N\] For equilibrium of block A, \[{{T}_{1}}={{F}_{S}}=25\,N\] ...(i) and for equilibrium of block B \[{{T}_{1}}={{T}_{2}}\cos {{45}^{o}}\]and \[{{W}_{B}}={{T}_{2}}\sin {{45}^{o}}\] ...(ii) From Eq (ii) \[{{W}_{B}}={{T}_{2}}\sin {{45}^{o}}=\left( \frac{{{T}_{1}}}{\cos {{45}^{o}}} \right)\sin {{45}^{o}}\] \[={{T}_{1}}\tan {{45}^{o}}={{T}_{1}}\] \[\therefore \] \[{{W}_{B}}={{T}_{1}}=25N\]
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