question_answer

A boy of 30 kg weight sitting on a horse whips it The horse speeds up at an average acceleration of\[\text{2 m}/{{\text{s}}^{\text{2}}}\]. (a) If the boy does not slide back, what is the force of friction exerted by the horse on the boy? (b) If the boy slides back during the acceleration, what can be, said about tee coefficient of static friction between the horse and the boy? Take \[\text{g}=10\text{m}/{{\text{s}}^{\text{2}}}\].

Answer:

                (a) As the boy does not slide back, its acceleration = ace. of the horse. As friction is the only horizontal force, it must act along the acceleration. \[\therefore \]  \[{{f}_{s}}=M\,\,a=30\left( 2.0 \right)=60N\]. (b) If the boy slides back, the horse could not exert a friction of 60 N on the boy. Maximum force of static friction \[{{f}_{s}}={{\mu }_{s}}R={{\mu }_{s}}\left( M\,g \right)\] \[\therefore \]  \[{{\mu }_{s}}\left( 30 \right)\left( 10 \right)\le 60N\] \[\therefore \]  Max. value of coeff. of static friction \[={{\mu }_{s}}=\frac{60}{300}=0.20\]