Where we assume the static friction to be at its

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! " = " = " = " = where we assume the static friction to be at its maximum value (permitting us to use Eq. 6-1). Solving these equations with μ s = 0.25, we obtain 2 103 N 1.0 10 N A W = ! " .

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19. In Fig. 6-36, two blocks are connected over a pulley. The mass of block A is 10 kg, and the coefficient of kinetic friction between A and the incline is 0.20. Angle of the incline is 30°. Block A slides down the incline at constant speed. What is the mass of block B? The free-body diagrams are shown below. T is the magnitude of the tension force of the string, f is the magnitude of the force of friction on block A , F N is the magnitude of the normal force of the plane on block A , m g A r is the force of gravity on body A (where m A = 10 kg), and m g B r is the force of gravity on block B . θ = 30° is the angle of incline. For A we take the + x to be uphill and + y to be in the direction of the normal force; the positive direction is chosen downward for block B . Since A is moving down the incline, the force of friction is uphill with magnitude f k = μ k F N (where μ k = 0.20). Newton’s second law leads to sin 0 cos 0 0 k A A N A B B T f m g m a F m g m g T m a ! ! " + = = " = " = =
for the two bodies (where a = 0 is a consequence of the velocity being constant). We solve these for the mass of block B . ( ) sin cos 3.3 kg. B A k m m ! μ ! = " = 20. During an Olympic bobsled run, the Jamaican team makes a turn of radius 7.6 m at a speed of 96.6 km/h. What is their acceleration in terms of g? With v = 96.6 km/h = 26.8 m/s, Eq. 6-17 readily yields 2 2 2 (26.8 m/s) 94.7 m/s 7.6 m v a R = = = which we express as a multiple of g : 2 2 94.7 m/s 9.7 . 9.80 m/s a a g g g g ! " ! " = = = # \$ # \$ % & % & 21. A puck of mass slides in a circle of radius on a frictionless table while attached to a hanging cylinder of mass by a cord through a hole in the table (Fig. 6-44). What speed keeps the cylinder at rest? For the puck to remain at rest the magnitude of the tension force T of the cord must equal the gravitational force Mg on the cylinder. The tension force supplies the centripetal force that keeps the puck in its circular orbit, so T = mv 2 / r . Thus Mg = mv 2 / r . We solve for the speed: 2 (2.50 kg)(9.80 m/s )(0.200 m) 1.81 m/s. 1.50 kg Mgr v m = = =
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