Physics 16 Problem Set 5 Solutions

Physics 16 Problem Set 5 Solutions 5.62. IDENTIFY: Apply r r ∑ F = ma to each object . Constant speed means a = 0 . SET UP: The free-body diagrams are sketched in Figure 5.62. T1 is the tension in the lower chain, T2 is the tension in the upper chain and T = F is the tension in the rope. EXECUTE: The tension in the lower chain balances the weight and so is equal to w. The lower pulley must have no net force on it, so twice the tension in the rope must be equal to w and the tension in the rope, which equals F, is w 2 . Then, the downward force on the upper pulley due to the rope is also w, and so the upper chain exerts a force w on the upper pulley, and the tension in the upper chain is also w. EVALUATE: The pulley combination allows the worker to lift a weight w by applying a force of only w/ 2 . Figure 5.62 5.65. IDENTIFY: SET UP: Apply r r ∑ F = ma to each block. Constant speed means a = 0 . When the blocks are moving, the friction force is f k and when they are at rest, the friction force is f s . EXECUTE: (a) The tension in the cord must be m2 g in order that the hanging block move at constant speed. This tension must overcome friction and the component of the gravitational force along the incline, so m2 g = ( m1 g sin α + µk m1 g cos α ) and m2 = m1 (sin α + µk cos α ) . (b) In this case, the friction force acts in the same direction as the tension on the block of mass m1 , so m2 g = ( m1 g sin α − µk m1 g cos α ) , or m2 = m1 (sin α − µk cos α ) . (c) Similar to the analysis of parts (a) and (b), the largest m2 could be is m1 (sin α + µs cos α ) and the smallest m2 could be is m1 (sin α − µs cos α ) . EVALUATE: In parts (a) and (b) the friction force changes direction when the direction of the motion of m1 changes. In part (c), for the largest m2 the static friction force on m1 is directed down the incline 5.66. and for the smallest m2 the static friction force on m1 is directed up the incline. IDENTIFY: The system is in equilibrium. Apply Newton’s 1st law to block A, to the hanging weight and to the knot where the cords meet. Target variables are the two forces. (a) SET UP: The free-body diagram for the hanging block is given in Figure 5.66a. EXECUTE: ∑ Fy = ma y T3 − w = 0 T3 = 12.0 N Figure 5.66a SET UP: The free-body diagram for the knot is given in Figure 5.66b. EXECUTE: ∑ Fy = ma y T2 sin 45.0° − T3 = 0 T3 12.0 N = sin 45.0° sin 45.0° T2 = 17.0 N T2 = Figure 5.66b ∑F x = max T2 cos 45.0° − T1 = 0 T1 = T2 cos 45.0° = 12.0 N SET UP: The free-body diagram for block A is given in Figure 5.66c. EXECUTE: ∑ Fx = max T1 − f s = 0 f s = T1 = 12.0 N Figure 5.66c EVALUATE: Also can apply ∑F y = ma y to this block: n − wA = 0 n = wA = 60.0 N Then µs n = (0.25)(60.0 N) = 15.0 N; this is the maximum possible value for the static friction force. We see that f s < µ s n; for this value of w the static friction force can hold the blocks in place. (b) SET UP: We have all the same free-body diagrams and force equations as in part (a) but now the static friction force has its largest possible value, f s = µs n = 15.0 N. Then T1 = f s = 15.0 N. EXECUTE: From the equations for the forces on the knot 15.0 N = 21.2 N T2 cos 45.0° − T1 = 0 implies T2 = T1 / cos 45.0° = cos 45.0° T2 sin 45.0° − T3 = 0 implies T3 = T2 sin 45.0° = (21.2 N)sin 45.0° = 15.0 N And finally T3 − w = 0 implies w = T3 = 15.0 N. EVALUATE: Compared to part (a), the friction is larger in part (b) by a factor of (15.0/12.0) and w is larger by this same ratio. 5.87. IDENTIFY: Let the tensions in the ropes be T1 and T2 . Figure 5.87a Consider the forces on each block. In each case take a positive coordinate direction in the direction of the acceleration of that block. SET UP: The free-body diagram for m1 is given in Figure 5.87b. EXECUTE: ∑ Fx = max T1 = m1a1 Figure 5.87b SET UP: The free-body diagram for m2 is given in Figure 5.87c. EXECUTE: ∑ Fy = ma y m2 g − T2 = m2 a2 Figure 5.87c This gives us two equations, but there are 4 unknowns ( T1 , T2 , a1 , and a2 ) so two more equations are required. SET UP: The free-body diagram for the moveable pulley (mass m) is given in Figure 5.87d. EXECUTE: ∑ Fy = ma y mg + T2 − 2T1 = ma Figure 5.87d But our pulleys have negligible mass, so mg = ma = 0 and T2 = 2T1. Combine these three equations to eliminate T1 and T2 : m2 g − T2 = m2 a2 gives m2 g − 2T1 = m2 a2 . And then with T1 = m1a 1 we have m2 g − 2m1a1 = m2 a2 . SET UP: There are still two unknowns, a1 and a2 . But the accelerations a1 and a2 are related. In any time interval, if m1 moves to the right a distance d, then in the same time m2 moves downward a distance d / 2. One of the constant acceleration kinematic equations says x − x0 = v0 xt + 12 axt 2 , so if m2 moves half the distance it must have half the acceleration of m1 : a2 = a1 / 2, or a1 = 2a2 . EXECUTE: This is the additional equation we need. Use it in the previous equation and get m2 g − 2m1 (2a2 ) = m2 a2 . a2 (4m1 + m2 ) = m2 g 2m2 g m2 g . and a1 = 2a2 = 4m1 + m2 4m1 + m2 EVALUATE: If m2 → 0 or m1 → ∞, a1 = a2 = 0. If m2 >> m1 , a2 = g and a1 = 2 g . a2 = 5.114. IDENTIFY: The tension F in the string must be the same as the weight of the hanging block, and must also provide the resultant force necessary to keep the block on the table in uniform circular motion. SET UP: The acceleration of the block is arad = v 2 / r , directed toward the hole. v2 , so v = gr M m. r EVALUATE: The larger M is the greater must be the speed v, if r remains the same. 5.115. EXECUTE: Mg = F = m IDENTIFY: Apply r r ∑ F = ma to the circular motion of the bead. Also use Eq.(5.16) to relate arad to the period of rotation T. SET UP: The bead and hoop are sketched in Figure 5.115a. The bead moves in a circle of radius R = r sin β . The normal force exerted on the bead by the hoop is radially inward. Figure 5.115a The free-body diagram for the bead is sketched in Figure 5.115b. EXECUTE: ∑ Fy = ma y n cos β − mg = 0 n = mg / cos β ∑F x = max n sin β = marad Figure 5.115b Combine these two equations to eliminate n: ⎛ mg ⎞ ⎜ ⎟ sin β = marad ⎝ cos β ⎠ sin β arad = cos β g arad = v 2 / R and v = 2π R / T , so arad = 4π 2 R / T 2 , where T is the time for one revolution. R = r sin β , so arad = 4π 2 r sin β T2 sin β 4π 2 r sin β = cos β T 2g This equation is satisfied by sin β = 0, so β = 0, or by Use this in the above equation: T 2g 1 4π 2 r = 2 , which gives cos β = 2 4π r cos β T g (a) 4.00 rev/s implies T = (1/ 4.00) s = 0.250 s (0.250 s) 2 (9.80 m/s 2 ) and β = 81.1°. 4π 2 (0.100 m) (b) This would mean β = 90°. But cos90° = 0, so this requires T → 0. So β approaches 90° as the hoop rotates very fast, but β = 90° is not possible. (c) 1.00 rev/s implies T = 1.00 s Then cos β = T 2g (1.00 s) 2 (9.80 m/s 2 ) equation then says cos β = = 2.48, which is not possible. The 4π 2 r 4π 2 (0.100 m) r r only way to have the ∑ F = ma equations satisfied is for sin β = 0. This means β = 0; the bead sits The cos β = at the bottom of the hoop. EVALUATE: β → 90° as T → 0 (hoop moves faster). The largest value T can have is given by T 2 g/(4π 2 r ) = 1 so T = 2π r / g = 0.635 s. This corresponds to a rotation rate of (1/ 0.635) rev/s = 1.58 rev/s. For a rotation rate less than 1.58 rev/s, β = 0 is the only solution and the bead sits at the bottom of the hoop. Part (c) is an example of this. 5.118. IDENTIFY: Apply r r r ∑ F = ma to the car. It has acceleration a rad , directed toward the center of the circular path. SET UP: The analysis is the same as in Example 5.24. ⎛ ⎛ v2 ⎞ (12.0 m/s) 2 ⎞ EXECUTE: (a) FA = m ⎜ g + ⎟ = (1.60 kg) ⎜ 9.80 m/s 2 + ⎟ = 61.8 N. R⎠ 5.00 m ⎠ ⎝ ⎝ ⎛ ⎛ v2 ⎞ (12.0 m/s) 2 ⎞ (b) FB = m ⎜ g − ⎟ = (1.60 kg) ⎜ 9.80 m/s 2 − ⎟ = −30.4 N. , where the minus sign indicates R⎠ 5.00 m ⎠ ⎝ ⎝ that the track pushes down on the car. The magnitude of this force is 30.4 N. EVALUATE: FA > FB . FA − 2mg .