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15. What is the tension T exerted by the hamstring muscles in the back of the thigh and the compressive force Fc in the knee joint due to the application of a horizontal force of 100 N to the ankle as shown in Figure P8.15? 8.15 Require that Στ = 0 about an axis through the knee joint and perpendicular to the page. This gives 100 N Στ = Fc ( 0 ) − T ( 4.00 cm ) + ( 100 N )( 42.0 cm ) = 0 , or the tension in the muscle is T = 1.05 × 103 N = 1.05 kN Then, ΣFx = − Fc − 100 N + 1.05 × 103 N=0 gives the compression force as Fc = 950 N 42.0 cm T 4.00 cm Fc 36. A cylindrical 5.00-kg reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel (Fig. P8.36). The bucket starts from rest and falls for 4.00 s. (a) What is the linear acceleration of the falling bucket? (b) How far does it drop? (c) What is the angular acceleration of the reel? FIGURE P8.36 8.36 The moment of inertia of the reel is R 1 1 2 I = MR 2 = ( 5.00 kg ) ( 0.600 m ) = 0.900 kg ⋅ m 2 2 2 T Applying Newton’s second law to the falling bucket gives T 29.4 N − T = ( 3.00 kg ) a t 3.00 kg Then, Newton’s second law for the reel gives ⎛ at ⎝R τ = T R = Iα = I ⎜ or T= ⎞ ⎟, ⎠ I a t ( 0.900 kg ⋅ m = R2 ( 0.600 m )2 at (1) 29.4 N 2 )a = t ( 2.50 kg ) a t . (a) Solving equations (1) and (2) simultaneously gives 2 a t = 5.35 m s downward 1 1 2 (b) Δy = v i t + a t t 2 = 0 + ( 5.35 m s 2 ) ( 4.00 s ) = 42.8 m 2 2 (2) (c) α = a t 5.35 m s 2 = = 8.91 rad s 2 R 0.600 m 60. A 12.0-kg object is attached to a cord that is wrapped around a wheel of radius r = 10.0 cm (Fig. P8.60). The acceleration of the object down the frictionless incline is measured to be 2.00 m/s2. Assuming the axle of the wheel to be frictionless, determine (a) the tension in the rope, (b) the moment of inertia of the wheel, and (c) the angular speed of the wheel 2.00 s after it begins rotating, starting from rest. FIGURE P8.60 8.60 (a) Consider the free-body diagram of the block given at the right. If the +x-axis is directed down the incline, ΣFx = max gives mg sin 37.0° − T = m at , or T = m ( g sin 37.0° − at ) n T at 12.0 kg +x T = ( 12.0 kg ) ⎡⎣( 9.80 m s 2 )sin 37.0° − 2.00 m s 2 ⎤⎦ = 46.8 N . (b) Now, consider the free-body diagram of the pulley. Choose an axis perpendicular to the page and passing through the center of the pulley, ⎛ at Στ = Iα gives T ⋅ r = I ⎜ ⎝ r ⎞ ⎟ , or ⎠ T ⋅ r 2 ( 46.8 N )( 0.100 m ) I= = = at 2.00 m s 2 2 0.234 kg ⋅ m 2 ⎛a (c) ω = ωi + α t = 0 + ⎜ t ⎝r 2 ⎞ ⎛ 2.00 m s ⎞ = t ⎟( 2.00 s ) = 40.0 rad s ⎟ ⎜ ⎠ ⎝ 0.100 m ⎠ mg 37.0° 65. In Figure P8.65 the sliding block has a mass of 0.850 kg, the counterweight has a mass of 0.420 kg, and the pulley is a uniform solid cylinder with a mass of 0.350 kg and an outer radius of 0.0300 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820 m/s toward the pulley when it passes through a photogate. (a) Use energy methods to predict its speed after it has moved to a second photogate, 0.700 m away. (b) Find the angular speed of the pulley at the same moment. FIGURE P8.65 8.65 Let mp be the mass of the pulley, m1 be the mass of the sliding block, and m2 be the mass of the counterweight. 1 (a) The moment of inertia of the pulley is I = m p R p2 and its angular velocity at 2 v , where v is the linear speed of the other masses. The any time is ω = Rp friction force retarding the sliding block is fk = μ k n = μk ( m1 g ) . Choose PEg = 0 at the level of the counterweight when the sliding mass reaches the second photogate. Then, from the work-kinetic energy theorem, ( Wnc = KEtrans + KErot + PEg ) − ( KE f trans + KErot + PEg ) i − fk ⋅ s = ⎛ v2 ⎞ 1 1 1 ( m1 + m2 )v 2f + ⎛⎜ mp Rp2 ⎞⎟⎜⎜ f2 ⎟⎟ + 0 2 2⎝ 2 ⎠⎝ R p ⎠ − or ⎛ v2 ⎞ 1 1 1 ( m1 + m2 )v i2 − ⎜⎛ mp Rp2 ⎟⎞⎜⎜ i2 ⎟⎟ − m2 gs , 2 2⎝ 2 ⎠⎝ R p ⎠ 1⎛ 1 ⎞ 2 1⎛ 1 ⎞ 2 ⎜ m1 + m2 + m p ⎟v f = ⎜ m1 + m2 + m p ⎟v i + m2 gs − μ k ( m1 g ) ⋅ s . 2⎝ 2 ⎠ 2⎝ 2 ⎠ This reduces to v f = v i2 + 2 ( m2 − μk m1 ) g s , 1 m1 + m2 + mp 2 and yields 2 2 m ⎞ 2 ( 0.208 kg )( 9.80 m s )( 0.700 m ) ⎛ v f = ⎜ 0.820 = 1.63 m s . ⎟ + s ⎠ 1.45 kg ⎝ (b) ω f = vf Rp = 1.63 m s = 54.2 rad s 0.0300 m