homework 6 solutions

# homework 6 solutions - Chapter 6 1 The greatest...

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Unformatted text preview: Chapter 6 1, The greatest deceleration (of magnitude a) is provided by the maximum friction force (Eq. 6-1, with F N = mg in this case). Using Newton’s second law, we ﬁnd aiﬁ,max/m =ysg. Equation 2—16 then gives the shortest distancefto stop: |Ax| = 2/2a = 36 m. In this calculation, it is important to ﬁrst convert v to 13 m/s. ~ 5. In addition to the forces already shown in Fig. 6-17, a free-body diagram would include an upward normal force JEN exerted by the ﬂoor on the block, a downward mg representing the gravitational pull exerted by Earth, and an assumed-leftward j7 for the kinetic or static friction. We choose +x rightward and +y upward. We apply Newton’s second law to these axes: _ F - f = ma P + FN —- mg = 0 where F = 6.0 N and m = 2.5 kg is the mass of the block. (a) In this case, P = 8.0 N leads to FN= (2.5 kg)(9.8 m/sz) — 8.0 N = 16.5 N. Using Eq. 6—1, this implies fsmx =,usFN =6.6 N , which is larger than the 6.0 N rightward force. Thus, the block (which was initially at rest) does not move. Putting a = 0 into the ﬁrst of our equations above yields a static friction force of f = P = 6.0 N. (b) In this case, P = 10 N, the normal force is FN = (2.5 kg)(9.8 m/sz) — 10 N = 14.5 N. Using Eq. 6-1, this implies f5,max = ,uSFN = 5.8 N , which is less than the 6.0 N rightward force — so the block does move. Hence, we are dealing not with static but with kinetic friction, which Eq. 6-2 reveals to be fk = ,ukFN = 3 .6 N . (c) In this last case, P = 12 N leads to FN = 12.5 N and thus to f3,max =,usFN = 5.0N , which (as expected) is less than the 6.0 N rightward force. Thus, the block moves. The kinetic friction force, then, is fk = ,ukFN = 3 . 1 N . 11. (a) The free-body diagram for the crate is shown below. mg" T is the tension force of the rope on the crate, FN is the normal force of the floor on the crate, mg is the force of gravity, and is the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. We assume the crate is motionless. The equations for the x and the y components of the force according to Newton’s second law are: Tcos 6-f= 0 Tsin6+FN —-mg = 0 where 6: 15° is the angle between the rope and the horizontal. The ﬁrst equation gives f = T cos 6 and the second gives F N = mg — T sin 6. If the crate is to remain at rest, f must be less than ,us F N, or T cos 6 < ,us (mg — T sin6). When the tension force is sufﬁcient to just start the crate moving, we must have Tcos 6: ,us (mg— Tsin 6). We solve for the tension: (0.50) (68 kg) (9.8 m/sz) T: %= /_—————= 304N~ 3.0x102N. cos 6+ y, sin 6 cos 15° + 0.50 sin 15° (b) The second law equations for the moving crate are Tcos 6—f‘=.,"ma FN+ Tsin 6—mg=0. Now f =kaN, and the second equation gives F N = mg ‘— Tsin6, which yields f = ,uk (mg — T sin 6). This expression is substituted for f in the ﬁrst equation to obtain Tcos 6— ,uk (mg— Tsin 6) = ma, so the acceleration is T(cos 6+ ,uk sin 6) ___ ___,_,__..__._——— m "#kg- a Numerically, it is given by a = W — (0.35)(9.8 m/sz) = 1.3 m/sz. 6 g 29. (a) Free—body diagrams for the blocks A and C, considered as a single object, and for the block B are shown below. T ——> FgB T is the magnitude of the tension force of the rope, F N is the magnitude of the normal force of the table on block- A, f is the magnitude of the force of friction, WAC is the combined weight of blocks A and C (the magnitude of force Fg AC shown in the ﬁgure), and W3 is the weight of block B (the magnitude of force F g B shown). Assume the blocks are not moving. For the blocks on the table we take the x axis to be to the right and the y axis to be upward. From Newton’s second law, we have x component: T— f = 0 y component: F N— WAC = 0. For block B take the downward direction to be positive. Then Newton’s second law for that block is W3 — T = 0. The third equation gives T = W; and the ﬁrst gives f = T = W3. The second equation gives F N = WAC. If sliding is not to occur, f must be less than ,u, F N, or WB < ,us WAC. The smallest that WAC can be with the blocks still at rest is WAC = WB/ys = (22 N)/(O.20) = 110 N. Since the weight of block A is 44 N, the least weight for C is (110 — 44) N = 66 N. (b) The second law equations become T—f =(WA/g)a FN— WA = 0 W3 — T = (WE/gm. In addition, f = ,ukFN. The second equation gives F N = WA, so f = ,ukWA. The third gives T = WB — (WE/gm. Substituting these two expressions into the ﬁrst equation, we obtain WB ‘ (WE/g)a # ,UkWA = (WA/g)a. Therefore, ' WA+WB 44N+22N = 2.3 m/sz. shown below. At the top (the highest point in the circular motion) the seat pushes up on the student with a force of magnitude F Mop, while the Earth pulls down with a force of magnitude mg. Newton’s second law for the radial direction gives I’I’IV2 R At the bottom of the ride, F Nabomm is the magnitude of the upward force exerted by the seat. The net force toward the center of the circle is (choosing upward as the positive direction): mg—FN,top : 2 mv FN,bottom —mg 2 R ' The Ferris wheel is “steadily rotating” so the value Fe = mv2 / R is the same everywhere. The apparent weight of the student is given by F N . FNrtoP /’ \ t0 \ ’ \p l/ \ I / / \ \ \\ / \ I \ / \ l / \ l ' l / \\ \ F’ I -> \ I I mg \ \ Mbotrom I l l \ l l \ / \ I \ / \ I \ \ \ I / / \ / N _, I \ // bottom \ \ / ‘ / mg’ (a) At the top, we are told that F Mop = 556 N and mg = 667 N. This means that the seat is pushing up with a force that is smaller than the student’s weight, and we say the student experiences a decrease in his “apparent weight” at the highest point. Thus, he feels “light.” (b) From (a), we ﬁnd the centripetal force to be 2 P; = m]: =mg—FNW =667 N—556N=111N. Thus, the normal force at the bottom is 2 FNMom 4”]: +mg=Fc+mg=lllN+667N=778N. 2 (c) If the speed is doubled, Fc’ = mfg) =4(111N)=444 N. Therefore, at the highest point we have F’ =mg—Fc’=667N—444N=223N. N ,top ((1) Similarly, the normal force at the lowest point is now found to be Fkbonom .=Fc'+mg=444N+667 N=1111Nz1.ll><103 N. Note: The apparent weight of the student is the greatest at the bottom and smallest at the top of the ride. The speed vzﬂlgR would result in FN,top =0 , giving the student a endden sensation of “weightlessness” at the top of the ride. 70. (a) We note that R (the horizontal distance from the bob to the axis of rotation) is the circumference of the circular path divided by 27F, therefore, R = 0.94/27Z'= 0.15 m. The angle that the cord makes with the horizontal is now easily found: 0: cos—‘(R/L) = cos—1(0.15 m/0.90 m) = 80°. The vertical component of the force of tension in the string is Tsint9 and must equal the downward pull of gravity (mg). Thus, T: ’_"g =0.40N. sm6’ Note that we are using T for tension (not for the period). (b) The horizontal component of that tension must supply the centripetal force (Eq. 6-18), so we have T 0056’ = mvz/R. This gives speed v = 0.49 m/s. This divided into the circumference gives the time for one revolution: 0.94/0.49 = 1.9 s. ...
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• Spring '08
• Richardr

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