In the figure shown,a horizontal force F_1 is | vedclass

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(C) For the block in vertical equilibrium,the forces are the normal reaction $N$ (upwards),the vertical force $F_2$ (upwards),and the weight $mg$ (dow

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The maximum value of the static frictional force decreases.

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(C) For the block in vertical equilibrium,the forces are the normal reaction $N$ (upwards),the vertical force $F_2$ (upwards),and the weight $mg$ (downwards).$N + F_2 = mg$$N = mg - F_2$As the magnitude of the vertical force $F_2$ increases,the normal reaction $N$ decreases.The maximum value of static friction (limiting friction) is given by $f_{max} = \mu_s N = \mu_s(mg - F_2)$.As $F_2$ increases,the normal reaction $N$ decreases,which in turn causes the maximum value of static friction $f_{max}$ to decrease.When $f_{max}$ becomes equal to the applied horizontal force $F_1$,the block begins to slide. Therefore,the correct statement is that the maximum value of the static frictional force decreases.

In the figure shown,a horizontal force $F_1$ is applied on a block,but the block does not slide. Then,as the magnitude of the vertical force $F_2$ is increased from zero,the block begins to slide. Which of the following statements is correct?

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