5.1.

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IDENTIFY: a = 0 for each object. Apply ΣFy = ma y to each weight and to the pulley.

SET UP: Take + y upward. The pulley has negligible mass. Let Tr be the tension in the rope and let Tc

be the tension in the chain.

EXECUTE: (a) The free-body diagram for each weight is the same and is given in Figure 5.1a.

ΣFy = ma y gives Tr = w = 25.0 N.

(b) The free-body diagram for the pulley is given in Figure 5.1b. Tc = 2Tr = 50.0 N.

EVALUATE: The tension is the same at all points along the rope.

Figure 5.1a, b

5.2.

5.3.

IDENTIFY: Apply Σ F = ma to each weight.

SET UP: Two forces act on each mass: w down and T ( = w) 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.

IDENTIFY: Both objects are at rest and a = 0. Apply Newton’s first law to the appropriate object. The maximum tension Tmax is at the top of the chain and the minimum tension is at the bottom of the chain.

SET UP: Let + y be upward. For the maximum tension take the object to be the chain plus the ball. For the

minimum tension take the object to be the ball. For the tension T three-fourths of the way up from the bottom of the chain, take the chain below this point plus the ball to be the object. The free-body diagrams in each of these three cases are sketched in Figures 5.3a, 5.3b and 5.3c. mb + c = 75.0 kg + 26.0 kg = 101.0 kg. mb = 75.0 kg. m is the mass of three-fourths of the chain: m = 3 (26.0 kg) = 19.5 kg.

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EXECUTE: (a) From Figure 5.3a, Σ Fy = 0 gives Tmax − mb + c g = 0 and

Tmax = (101.0 kg)(9.80 m/s 2 ) = 990 N. From Figure 5.3b, Σ Fy = 0 gives Tmin − mb g = 0 and

Tmin = (75.0 kg)(9.80 m/s2 ) = 735 N.

(b) From Figure 5.3c, Σ Fy = 0 gives T − (m + mb ) g = 0 and T = (19.5 kg + 75.0 kg)(9.80 m/s 2 ) = 926 N.

EVALUATE: The tension in the chain increases linearly from the bottom to the top of the chain.

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