Solutions 6 - 5.2. IDENTIFY: Apply F = ma to each weight. r...

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5.2. IDENTIFY: Apply m = F a r r to each weight. SET UP: Two forces act on each mass : w down and () Tw = up. EXECUTE: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w . EVALUATE: The tension is the same in all three cases. 5.14.IDENTIFY: Apply m = F a r r to each block. 0 a = . SET UP: Take y + perpendicular to the incline and x + parallel to the incline. EXECUTE: The free-body diagrams for each block, A and B , are given in Figure 5.14. (a) For B , x x Fm a = gives 1 sin 0 α −= and 1 sin = . (b) For block A , x x a = gives 12 sin 0 TT w and 2 2s i n = . (c) y y a = for each block gives cos AB nnw == . (d) For 0 , 0 TT =→ and nn w . For 90 ° , 1 = , 2 2 = and 0 nn . EVALUATE: The two tensions are different but the two normal forces are the same. Figure 5.14a, b
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5.19.IDENTIFY: Apply m = F a r r to the load of bricks and to the counterweight. The tension is the same at each end of the rope. The rope pulls up with the same force () T on the bricks and on the counterweight. The counterweight accelerates downward and the bricks accelerate upward; these accelerations have the same magnitude.
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Solutions 6 - 5.2. IDENTIFY: Apply F = ma to each weight. r...

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