Pré-visualização12 páginas

is \u2c6 \u2c6 \u2c6i (0.34)(43.5 N) i (15 N) ik k Nf Fµ= = = & . 20. We use coordinates and weight-components as indicated in Fig. 5-18 (see Sample Problem 5-7 from the previous chapter). (a) In this situation, we take &f s to point uphill and to be equal to its maximum value, in which case fs, max = s NFµ applies, where µs = 0.25. Applying Newton\u2019s second law to the block of mass m = W/g = 8.2 kg, in the x and y directions, produces min 1 , maxsin 0 cos 0 s N F mg f ma F mg \u3b8 \u3b8 \u2212 + = = \u2212 = which (with \u3b8 = 20°) leads to ( )min 1 sin cos 8.6 N.sF mg \u3b8 µ \u3b8\u2212 + = (b) Now we take &f s to point downhill and to be equal to its maximum value, in which case fs, max = µsFN applies, where µs = 0.25. Applying Newton\u2019s second law to the block of mass m = W/g = 8.2 kg, in the x and y directions, produces min 2 , maxsin 0 cos 0 s N F mg f ma F mg \u3b8 \u3b8 = \u2212 = = \u2212 = which (with \u3b8 = 20°) leads to F mg smin sin cos2 46= + =\u3b8 µ \u3b8b g N. A value slightly larger than the \u201cexact\u201d result of this calculation is required to make it accelerate uphill, but since we quote our results here to two significant figures, 46 N is a \u201cgood enough\u201d answer. (c) Finally, we are dealing with kinetic friction (pointing downhill), so that sin 0 cos 0 k N F mg f ma F mg \u3b8 \u3b8 \u2212 \u2212 = = \u2212 = along with fk = µkFN (where µk = 0.15) brings us to F mg k= + =sin cos\u3b8 µ \u3b8b g 39 N . 21. The free-body diagrams for block B and for the knot just above block A are shown next. & T1 is the tension force of the rope pulling on block B or pulling on the knot (as the case may be), &T2 is the tension force exerted by the second rope (at angle \u3b8 = 30°) on the knot, &f is the force of static friction exerted by the horizontal surface on block B, NF & is normal force exerted by the surface on block B, WA is the weight of block A (WA is the magnitude of m gA & ), and WB is the weight of block B (WB = 711 N is the magnitude of m gB & ). For each object we take +x horizontally rightward and +y upward. Applying Newton\u2019s second law in the x and y directions for block B and then doing the same for the knot results in four equations: 1 ,max 2 1 2 0 0 cos 0 sin 0 s N B A T f F W T T T W \u3b8 \u3b8 \u2212 = \u2212 = \u2212 = \u2212 = where we assume the static friction to be at its maximum value (permitting us to use Eq. 6-1). Solving these equations with µs = 0.25, we obtain 2103 N 1.0 10 NAW = \u2248 × . 22. Treating the two boxes as a single system of total mass mC + mW =1.0 + 3.0 = 4.0 kg, subject to a total (leftward) friction of magnitude 2.0 + 4.0 = 6.0 N, we apply Newton\u2019s second law (with +x rightward): (4.0) total totalF f m a a \u2212 = \u2212 =12 0 6 0. . which yields the acceleration a = 1.5 m/s2. We have treated F as if it were known to the nearest tenth of a Newton so that our acceleration is \u201cgood\u201d to two significant figures. Turning our attention to the larger box (the Wheaties box of mass mW = 3.0 kg) we apply Newton\u2019s second law to find the contact force F' exerted by the Cheerios box on it. W W 4.0 (3.0)(1.5) F f m a F \u2032 \u2212 = \u2032 \u2212 = This yields the contact force F' = 8.5 N. 23. Let the tensions on the strings connecting m2 and m3 be T23, and that connecting m2 and m1 be T12, respectively. Applying Newton\u2019s second law (and Eq. 6-2, with FN = m2g in this case) to the system we have 3 23 3 23 2 12 2 12 1 1 k m g T m a T m g T m a T m g m a µ \u2212 = \u2212 \u2212 = \u2212 = Adding up the three equations and using 1 2 3, 2m M m m M= = = , we obtain 2Mg \u2013 2µk Mg \u2013 Mg = 5Ma . With a = 0.500 m/s2 this yields µk = 0.372. Thus, the coefficient of kinetic friction is roughly µk = 0.37. 24. (a) Applying Newton\u2019s second law to the system (of total mass M = 60.0 kg) and using Eq. 6-2 (with FN = Mg in this case) we obtain F \u2013 µkMg = Ma a= 0.473 m/s2. Next, we examine the forces just on m3 and find F32 = m3(a + µkg) = 147 N. If the algebra steps are done more systematically, one ends up with the interesting relationship: 32 3( / )F m M F= (which is independent of the friction!). (b) As remarked at the end of our solution to part (a), the result does not depend on the frictional parameters. The answer here is the same as in part (a). 25. The free-body diagrams for the two blocks are shown next. T is the magnitude of the tension force of the string, NAF & is the normal force on block A (the leading block), NBF & is the normal force on block B, &f A is kinetic friction force on block A, &f B is kinetic friction force on block B. Also, mA is the mass of block A (where mA = WA/g and WA = 3.6 N), and mB is the mass of block B (where mB = WB/g and WB = 7.2 N). The angle of the incline is \u3b8 = 30°. For each block we take +x downhill (which is toward the lower-left in these diagrams) and +y in the direction of the normal force. Applying Newton\u2019s second law to the x and y directions of first block A and next block B, we arrive at four equations: sin cos 0 sin cos 0 A A A NA A B B B NB B W f T m a F W W f T m a F W \u3b8 \u3b8 \u3b8 \u3b8 \u2212 \u2212 = \u2212 = \u2212 + = \u2212 = which, when combined with Eq. 6-2 ( A kA NAf Fµ= where µk A = 0.10 and B kB NBf Fµ= fB where µk B = 0.20), fully describe the dynamics of the system so long as the blocks have the same acceleration and T > 0. (a) These equations lead to an acceleration equal to 2sin cos 3.5 m/s .k A A k B B A B W W a g W W µ µ\u3b8 \u3b8 § ·§ ·+ = \u2212 =¨ ¸¨ ¸¨ ¸+© ¹© ¹ (b) We solve the above equations for the tension and obtain ( )cos 0.21 N.A B k B k A A B W WT W W µ µ \u3b8§ ·= \u2212 =¨ ¸ +© ¹ Simply returning the value for a found in part (a) into one of the above equations is certainly fine, and probably easier than solving for T algebraically as we have done, but the algebraic form does illustrate the µk B \u2013 µk A factor which aids in the understanding of the next part. 26. 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, FN is the magnitude of the normal force of the plane on block A, m gA & is the force of gravity on body A (where mA = 10 kg), and m gB & is the force of gravity on block B. \u3b8 = 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 fk = µkFN (where µk = 0.20). Newton\u2019s 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 \u3b8 \u3b8 \u2212 + = = \u2212 = \u2212 = = for the two bodies (where a = 0 is a consequence of the velocity being constant). We solve these for the mass of block B. m mB A k= \u2212 =sin cos .\u3b8 µ \u3b8b g 3 3 kg. 27. First, we check to see if the bodies start to move. We assume they remain at rest and compute the force of (static) friction which holds them there, and compare its magnitude with the maximum value µsFN. 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 body A, FN is the magnitude of the normal force of the plane on body A, m gA & is the force of gravity on body A (with magnitude WA = 102 N), and m gB & is the force of gravity on body B (with magnitude WB = 32 N). \u3b8 = 40° is the angle of incline. We are told the direction of &f but we assume it is downhill. If we obtain a negative result for f, then we know the force is actually up the plane. (a) For A we take the +x to be uphill and +y to be in the direction of the normal force. The x and y components