A) 750 N
B) 500 N
C) 345 N
D) 250 N
Correct Answer: D
Solution :
Net force along the plane = \[P-mg\sin \theta \]= \[750-500\] = \[250\ N\] Limiting friction = \[{{F}_{l}}={{\mu }_{s}}R={{\mu }_{s}}mg\cos \theta \] \[=0.4\times 102\times 9.8\times cos30\text{ }=\text{ }346N\] As net external force is less than limiting friction therefore friction on the body will be 250 N.You need to login to perform this action.
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