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Unformatted text preview: 79. (a) The free-body diagram for the person (shown as an L-shaped block) is shown below. The force that she exerts on the rock slabs is not directly shown (since the diagram should only show forces exerted on her), but it is related by Newton’s third law) to the normal forces FN 1 and FN 2 exerted horizontally by the slabs onto her shoes and back, respectively. We will show in part (b) that FN1 = FN2 so that we there is no ambiguity in saying that the magnitude of her push is FN2. The total upward force due to (maximum) static friction is f = f 1 + f 2 where f1 = μ s1 FN 1 and f 2 = μ s 2 FN 2 . The problem gives the values μs1 = 1.2 and μs2 = 0.8. (b) We apply Newton’s second law to the x and y axes (with +x rightward and +y upward and there is no acceleration in either direction). FN 1 − FN 2 = 0 f1 + f 2 − mg = 0 The first equation tells us that the normal forces are equal FN1 = FN2 = FN. Consequently, from Eq. 6-1,
f1 = μs 1 FN f 2 = μ s 2 FN we conclude that f1 = Therefore, f1 + f2 – mg = 0 leads to μs 1 μs 2 f2 . μs 1 + 1 f 2 = mg μs 2
which (with m = 49 kg) yields f2 = 192 N. From this we find FN = f 2 / μ s 2 = 240 N. This is equal to the magnitude of the push exerted by the rock climber. (c) From the above calculation, we find f1 = μs1 FN = 288 N which amounts to a fraction
f1 288 = = 0.60 W 49 9.8 or 60% of her weight. b gb g ...
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