A) \[f=mg\] [where f is the friction force]
B) \[F=N\] [where N is the normal force]
C) F will not produce torque
D) N will not produce torque
Correct Answer: D
Solution :
The various forces acting on the block are shown in figure. As the block remains stationary under the effect of these forces, so \[mg=f\] (force of friction) \[F=N\left( Normal\text{ }reaction \right)\] Torque due to F or mg is zero as the line of action of F or mg passes through centre C. Since the body is in equilibrium, so the sum of torque due to force of friction \[({{\vec{\tau }}_{f}})\] and torque due to normal reaction \[({{\vec{\tau }}_{N}})\] must be zero, i.e., \[{{\vec{\tau }}_{f}}+{{\vec{\tau }}_{N}}=0\]. As line of action of f and N may not pass through centre C of block, so Ty may not be zero, and \[\vec{N}\] may also not be zero. Therefore, f does not produce any torque is wrong and N does not produce a torque is also wrong. Hence, the incorrect statement is only.You need to login to perform this action.
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