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Physics 4 - 4.13 Mode-i Use the particle model for the...

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Unformatted text preview: 4.13. Mode-i: Use the particle model for the object. Select {a} We are told that for on unknown force {cell it Fa) acting on an unlotown moss [cell it mu) the acceleration efttle mess is ID misi. According to Newton’s second law. Fn== mutll] rots-’1. The force then becomes m. Newton’s 'tccoed law gins This means a is 5 misz. .[h] The force is FE and the mess is now firm”. Newton‘s second law gives Pg. =-;-mno = mn(lfl 111/3!) This means a = 20 111st. :_[c) Asimilar procedure gives a. = [O misz. III) A similar procedure gives a = 2.5 ntfs’. 4.34. Visualize: J‘ Solve: {d} There is an object on an inclined surface. The net force is down the plane so the acceleration is down the plane. The net force includes both the frictional force and the component of the weight. The direction of the force of kinetic friction implies that the object is moving upward. The description could be: “A car is skidding up an embankment." 4.43. Visualize: ‘5 Weight fi- Hmm ":1 The drag fume due to air is opposite. the mutiun- 4.45. Visualize: _ :r' "M -_ Weight Fr There arc nc contact force: cm the rack. Weight is the only force acting on the rock. 4.49. Visualize: (a) First contact Loses contact ‘11 HI .5 f Magnified view of :15 - hall in contact with E, ' ground 5 E 1 1 5 U +. {h} g g g {e} 3 Weight fir Normal force if {d} The hall accelerates downward until the instant when it makes contact with the ground. Once it makes contact, it begins to compress and to slow down. The compression takes a short but nonzero distance. as shown in the motion diagram. The point of maximum compression is the turning point. where the ball has an instantaneous speed of v = I] mfs and reverses direction. The ball then expands and speeds up until it loses contact with the ground. The motion diagram shows that the acceleration vector 5 points upward the entire time that the ball is in contact with the ground. An upward acceleration implies that there is a net upward force an on the hall. The only two forces on the bail are its weight downward and the normal force of the ground upward. To have a net force upward requires a :- w. So the bail bounces because the normal force of the ground amends the weight. causing a net upward force during the entire time that the ball is in contact with the ground. This net upward force siows the ball. toms it, and accelerates it upward until it loses contact with the ground. Once contact with the ground is lost. the nonnal force vanishes and the ball is simply,I in free fall. —|- 5.2?. Made]: You can model the beam as a particle and assume F m = UN to calculate the tensions in the sus- pension ropes. Visualize: Physical rapresentatlen Known m =1lllllsg Tm” =fi{m N Find T] and T2 Salve: The beam attached to the ropes will remain in stch equilibrium out}.' if both inequalitim T; c Tm and Ir"2 a TM hold, where Tm is the maximum sustained tension. The equilibrium equations in vector and component form are: E; =fi+ l+fi=fiN (Fa). = a. +1"... + =0N (FML': 1‘]! + T_ +11; = e N Using the free‘body diagram yields: _ —Ii[sinl5‘I +Jt"j._sint5‘2 =0N ficosfl. -t-'i“:.,:cosl5i2 —w=0N ”the mathematical model is reduced to a simple algebraic system of two equations with two unknowns, f; and T2. Substituting 0, = 20°. 93 = 30°. and n' = mg 2 9800 N, the simultaneous equations become —'I] sin 20“ + T2 sin 30” = 0 N T, cos20°+ T: cos30° = 9300 N You can solve this system of equations by simple substitution. The result is 1‘", = 639? N and I; = 4376 N . The rope en the left side {mpe l} breaks since the tension in this rope is larger than 6000 N. Once the left mpe is broken, the right rope will also break because now the whole weight of the beam will be applied to the rope on the right side. Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than in the right rope. 5.33. Model: We can assume the foot is a single particle in equilibrium under the combined effects of gravity, the tensions in the upper and lower sections of the traction rope. and the opposing traction foroe of the leg itself. We can also treat the hanging mass as a particle in equilibrium Since the pulleys are frictionless, the tension is the same everywhere in the rope. Because all pulleys are in equilibrium, their net force is acre. So the}.r do not contribute to T. Visualize: Physical representation 3" T Known Hanging m = E kg mass Find 11; T {a} {b} Solve: {a} From the free-body diagram for the mass. the tension in the rope is r = it: = mg = (s kg)(s.s me?) = 53.3 N (b) Using Newton‘s first law her the vertical direction on the pulley attached to the foot, (Fm)? = 21-; = TsinB— Tsin15°— w.” = o N - .. 4 k 9.3 misi- T511115 + ‘l'l'm ___ sin 15“ + ”smug ___ 9.259 +( SK ) T T 53.8 N => 3 = sin“I 0.926 = 613“ (c) Using Newton's first law for the horizontal direction. (Fm)! = SE = Tcosfl + Tcoslfi“ — Fm = o N = FM 2 Tcosfl + Tcoslfi“ = T(cosfi7.3° + coslfi") = {53.3 N](U.3T73+ 0.9659} = (53.3 N]( 1.344} = 79.0 N :e sintl = = 0.259 + 0.667 = 0.926 5.4T. Model: We will model the sled and friend as a particle. and use the model of kinetic friction because the sled is in motion- V'maliu: Pictorial and physical representations The net force on the sled is zero {note the constant speed of the sled). That means the component of the pulling latte along the +.r-direction is equal to the magnitude of the kinetic force of friction in the flit-direction. Also note that {Fm}, = D N. since the sled is not moving along the y—asis- g'Solve: Newton’s second law is (F...L=EF.=H.+ u;+tfi1.+ (PM); UN +0 Isl—11+ FM cusEl=UN [FH)F=EE1=n}-+“ly+m)r +{FM}I=H _pflg+flN+Fwfl Sifl9=UN The .r-component equation using the kinetic friction model j; = Jnth reduces to An = cha cost}' The y~component equation gives 11 = mg — Fm] sin 9 We see that the normal force is smaller than the 1weight because cha has a component in a direction opposite to the direction of the weight. In other words, PM is partly lifting the sled. From the .r-component equation. pt can now he obtained as F... was? (75 N)(cos3£l°) _____ = —_—__1 _ a = 0-12 mg~Fpufl 51113 {fit} kg}(9.8 I'ni’s )-—(?5 N)(51I13U ) #1:: Assess: A quick glance at the various pt values in Table 5.1 suggests that a value of 0.12 for pk is reasonable. ...
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