Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

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(1) 120 N. Idea: take the two blocks together. The wall is the only thing outside them that can push or rub vertically, so its friction must hold up both. The applied force F is horizontal, and the weights are vertical. For the pair A+B in vertical equilibrium: f_wall=W_A+W_B=20+100=120 N, upward. The friction between A and B is internal to this system and cancels out; the coefficients 0.1 and 0.15 only tell you how large F must be for this to be possible, not what the friction actually is. This is the key point about static friction: it is not μ N in general. μ N is only a ceiling, and the actual value is whatever equilibrium requires — here, 120 N.

Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

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